diff --git a/.devcontainer.json b/.devcontainer.json
index ed2623030..ae8eddb74 100644
--- a/.devcontainer.json
+++ b/.devcontainer.json
@@ -27,7 +27,7 @@
},
// Respect the project's designated dependencies, and create codechat config file
- "postCreateCommand": "pip install -r requirements.txt && git config core.hooksPath .githooks && touch codechat_config.yaml",
+ "postCreateCommand": "bash scripts/setup.sh",
// Port forwarding
// ---------------
@@ -71,18 +71,18 @@
"other": "off"
},
"editor.snippetSuggestions": "top",
- "xml.validation.enabled": false,
+ "xml.validation.enabled": true,
+ "pretext-tools.schema.versionName": "Custom",
+ "pretext-tools.schema.customPath": "/workspaces/library/schema/tbil.rng",
+ "redhat.telemetry.enabled": false,
"CodeChat.CodeChatServer.Command": "CodeChat_Server",
- "files.eol": "\n",
- "files.associations": {
- "*.sage": "python"
- }
+ "files.eol": "\n"
},
"extensions": [
- "ms-vscode.live-server",
"oscarlevin.pretext-tools",
"CodeChat.codechat",
- "streetsidesoftware.code-spell-checker"
+ "streetsidesoftware.code-spell-checker",
+ "lov3.sagemath-enhanced"
]
}
}
diff --git a/.github/workflows/pretext-cli.yml b/.github/workflows/pretext-cli.yml
index ff5fc3445..118591121 100644
--- a/.github/workflows/pretext-cli.yml
+++ b/.github/workflows/pretext-cli.yml
@@ -17,6 +17,7 @@ jobs:
build:
runs-on: ubuntu-latest
container: oscarlevin/pretext:full
+ if: github.ref == format('refs/heads/{0}', github.event.repository.default_branch)
steps:
- name: Checkout source
@@ -25,23 +26,15 @@ jobs:
- name: install deps
run: pip install -r requirements.txt
- - name: quick build instructor versions
- id: quickbuild
- if: github.ref != format('refs/heads/{0}', github.event.repository.default_branch) && github.event_name != 'workflow_dispatch'
- run: |
- pretext build precalculus-web-instructor -q
- pretext build calculus-web-instructor -q
- pretext build linear-algebra-web-instructor -q
-
- name: build all deploy targets
- if: steps.quickbuild.outcome == 'skipped'
+ # if: steps.quickbuild.outcome == 'skipped'
run: pretext build --deploys
- name: stage deployment
run: pretext deploy --stage-only
- name: build and stage all banks
- if: steps.quickbuild.outcome == 'skipped'
+ # if: steps.quickbuild.outcome == 'skipped'
run: |
python scripts/build_bank.py precalculus --full
python scripts/build_bank.py calculus --full
@@ -53,10 +46,76 @@ jobs:
name: deploy
path: output/stage
- deploy-cloudflare:
+ deploy-ghpages:
runs-on: ubuntu-latest
needs: build
- if: vars.CLOUDFLARE_PROJECT_NAME != '' && github.ref != format('refs/heads/{0}', github.event.repository.default_branch)
+ if: github.ref == format('refs/heads/{0}', github.event.repository.default_branch)
+ permissions:
+ contents: read
+ pages: write
+ id-token: write
+ environment:
+ name: github-pages
+ url: ${{ steps.deployment.outputs.page_url }}
+ steps:
+ - name: Download website artifact
+ uses: actions/download-artifact@v4
+ with:
+ name: deploy
+ path: deploy
+ - name: Setup GitHub Pages
+ id: check
+ uses: actions/configure-pages@v4
+ - name: Upload artifact
+ uses: actions/upload-pages-artifact@v3
+ with:
+ path: deploy
+ - name: Deploy to Github Pages
+ id: deployment
+ uses: actions/deploy-pages@v4
+
+ preview:
+ runs-on: ubuntu-latest
+ container: oscarlevin/pretext:full
+ if: github.ref != format('refs/heads/{0}', github.event.repository.default_branch)
+ outputs:
+ markdown: ${{ steps.changes.outputs.markdown }}
+
+ steps:
+ - name: Checkout source
+ uses: actions/checkout@v4
+ with:
+ fetch-depth: 0
+
+ - name: install deps
+ run: pip install -r requirements.txt
+
+ - name: Set necessary Git permission
+ run: git config --global --add safe.directory $(pwd)
+
+ - name: build previews of all changed files
+ id: changes
+ run: python -m scripts.quickbuild_changes
+
+ - name: stage deployment
+ run: pretext deploy --stage-only
+
+ # - name: build and stage all banks
+ # run: |
+ # python scripts/build_bank.py precalculus --full
+ # python scripts/build_bank.py calculus --full
+ # python scripts/build_bank.py linear-algebra --full
+
+ - name: Bundle output/stage as artifact
+ uses: actions/upload-artifact@v4
+ with:
+ name: deploy
+ path: output/stage
+
+ deploy-cloudflare:
+ runs-on: ubuntu-latest
+ needs: preview
+ if: github.ref != format('refs/heads/{0}', github.event.repository.default_branch)
permissions:
contents: read
deployments: write
@@ -70,8 +129,8 @@ jobs:
path: deploy
- name: Create 404.html
run: echo "404 page not found" >> deploy/404.html
- - name: Delete CheckIt builds not supported by CloudFlare
- run: find . -name "bank.json" -type f -delete
+ # - name: Delete CheckIt builds not supported by CloudFlare
+ # run: find . -name "bank.json" -type f -delete
- name: Publish to Cloudflare
id: cloudflare_publish
uses: cloudflare/pages-action@v1
@@ -87,37 +146,11 @@ jobs:
if: github.event_name == 'pull_request'
with:
script: |
+ cf_url = "${{ steps.cloudflare_publish.outputs.url }}"
github.rest.issues.createComment({
issue_number: context.issue.number,
owner: context.repo.owner,
repo: context.repo.repo,
- body: '## 🚀 Preview available 🚀\n\n${{ steps.cloudflare_publish.outputs.url }}\n\n- [Precalculus](${{ steps.cloudflare_publish.outputs.url }}/preview/precalculus/instructor/frontmatter)\n\n- [Calculus](${{ steps.cloudflare_publish.outputs.url }}/preview/calculus/instructor/frontmatter)\n\n- [Linear Algebra](${{ steps.cloudflare_publish.outputs.url }}/preview/linear-algebra/instructor/frontmatter)'
+ body: `${{ needs.preview.outputs.markdown }}`
})
- deploy-ghpages:
- runs-on: ubuntu-latest
- needs: build
- if: vars.PTX_ENABLE_DEPLOY_GHPAGES == 'yes' && github.ref == format('refs/heads/{0}', github.event.repository.default_branch)
- permissions:
- contents: read
- pages: write
- id-token: write
- environment:
- name: github-pages
- url: ${{ steps.deployment.outputs.page_url }}
- steps:
- - name: Download website artifact
- uses: actions/download-artifact@v4
- with:
- name: deploy
- path: deploy
- - name: Setup GitHub Pages
- id: check
- uses: actions/configure-pages@v4
- - name: Upload artifact
- uses: actions/upload-pages-artifact@v3
- with:
- path: deploy
- - name: Deploy to Github Pages
- id: deployment
- uses: actions/deploy-pages@v4
diff --git a/requirements.txt b/requirements.txt
index 0a31b9dcb..4093cd034 100644
--- a/requirements.txt
+++ b/requirements.txt
@@ -1,3 +1,3 @@
-pretext == 2.6.2
+pretext == 2.10.1
pelican == 4.9.1
checkit-dashboard == 0.2.6
diff --git a/schema/tbil.rng b/schema/tbil.rng
new file mode 100644
index 000000000..67fd96836
--- /dev/null
+++ b/schema/tbil.rng
@@ -0,0 +1,133 @@
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/scripts/preview_outcome.py b/scripts/preview_outcome.py
index 72c64a994..4a764f467 100644
--- a/scripts/preview_outcome.py
+++ b/scripts/preview_outcome.py
@@ -8,7 +8,7 @@
from checkit.bank import Bank
from checkit import static
-def main(book:str, outcome:str, amount:int = 20):
+def build_preview(book:str, outcome:str, amount:int = 20):
# check that this outcome exists
exercise_path = Path("source", book, "exercises")
sage_library_path = Path("source", book, "sage")
@@ -89,9 +89,15 @@ def main(book:str, outcome:str, amount:int = 20):
# copy assets
shutil.copytree(b.build_path(), os.path.join(docs_path, "assets"), dirs_exist_ok=True)
+ return sandbox_bank_path
+
+
+def main(book:str, outcome:str, amount:int = 20):
+ sandbox_bank_path = build_preview(book, outcome, amount)
subprocess.run(['python', '-m', 'http.server', '-d', str(sandbox_bank_path/'docs')])
+
if __name__ == "__main__":
parser = argparse.ArgumentParser(description='Preview a CheckIt outcome.')
parser.add_argument('book', choices=['precalculus', 'calculus', 'linear-algebra'])
diff --git a/scripts/quickbuild_changes.py b/scripts/quickbuild_changes.py
new file mode 100644
index 000000000..076163a55
--- /dev/null
+++ b/scripts/quickbuild_changes.py
@@ -0,0 +1,77 @@
+from pretext.project import Project
+from git import Repo
+from pathlib import Path
+from lxml import etree
+import os
+import shutil
+from . import preview_outcome
+
+def main():
+ repo = Repo()
+ commit_head = repo.commit("HEAD")
+ commit_origin_main = repo.commit("origin/main")
+ commit_merge_base = repo.merge_base(commit_head, commit_origin_main)[0]
+ diff_index = commit_merge_base.diff(commit_head)
+ changed_files = [Path(item.a_path) for item in diff_index if not item.deleted_file and not item.renamed_file]
+ preview_links = []
+
+ p = Project.parse()
+ BOOKS = ['precalculus', 'calculus', 'linear-algebra']
+
+ # for each .ptx file, try to build subtree of document
+ for f in [f for f in changed_files if f.suffix == ".ptx"]:
+ book = None
+ for b in BOOKS:
+ b_path = Path("source") / b
+ if b_path in f.parents:
+ book = b
+ if book is not None:
+ t = p.get_target(f"{book}-web-instructor")
+ root = etree.parse(f).getroot()
+ xml_id = root.get(r"{http://www.w3.org/XML/1998/namespace}id")
+ print(f"Building book `{book}` with ID `{xml_id}`")
+ title_ele = root.find("title")
+ if title_ele is None:
+ if xml_id is None:
+ path = f"/preview/{book}/instructor"
+ else:
+ path = f"/preview/{book}/instructor/{xml_id}.html"
+ else:
+ path = f"/preview/{book}/instructor/{xml_id}.html"
+ t.build(xmlid=xml_id, no_knowls=True, generate=(xml_id is not None))
+ preview_links.append({
+ "file": f,
+ "path": path
+ })
+ # for each CheckIt file, build its preview
+ for b in BOOKS:
+ EXERCISE_FILES = [f for f in changed_files if Path("source", b, "exercises", "outcomes") in f.parents]
+ # collect changed outcomes
+ changed_outcomes = []
+ for f in EXERCISE_FILES:
+ if f.parent.name not in changed_outcomes:
+ changed_outcomes.append(f.parent.name)
+ # build changed outcomes
+ for o in changed_outcomes:
+ sandbox_bank_path = preview_outcome.build_preview(b, o)
+ output_path = Path("output", f"{b}-web-instructor", "exercises", o)
+ output_path.mkdir(parents=True)
+ shutil.copytree(sandbox_bank_path / "docs", output_path, dirs_exist_ok=True)
+ preview_links.append({
+ "file": f,
+ "path": f"/preview/{b}/instructor/exercises/{o}/"
+ })
+
+
+
+ # create Javascript template for markdown output
+ markdown = f"## 🚀 Preview available 🚀\n\n<${{cf_url}}>\n\n"
+ for link in preview_links:
+ # ${{cf_url}} is provided by action Javascript
+ markdown += f"- \\`{link["file"]}\\`: <${{cf_url}}{link["path"]}>\n"
+ with open(os.environ["GITHUB_OUTPUT"], 'a') as fh:
+ fh.write(f"markdown<
+- Lewis, D., Clontz, S., and Estis, J. (2021). Team-Based Inquiry Learning. _PRIMUS_ 31 (2), 223-238.
- Team-Based Learning (TBL) is a cooperative learning strategy blending elements of flipped learning, inquiry-based learning, and problem-based learning. Although used quite frequently in other disciplines, use of this strategy in mathematics has been limited. In this article, we describe how TBL can be implemented in math courses with adherence to essential elements of TBL and introduce modifications specific to mathematics instruction. In particular, we introduce a particular style of TBL, which we term Team-Based Inquiry Learning, that satisfies the defining pillars of inquiry-based learning.
-- Lewis, D. and Estis, J. 2020. Improving Mathematics Content Mastery and Enhancing Flexible Problem Solving through Team-Based Inquiry Learning. Teaching & Learning Inquiry 8 (2), 165-183.
+- Lewis, D. and Estis, J. (2020). Improving Mathematics Content Mastery and Enhancing Flexible Problem Solving through Team-Based Inquiry Learning. _Teaching & Learning Inquiry_ 8 (2), 165-183.
- This article examines how student learning is affected by the use of team-based inquiry learning, a novel pedagogy in mathematics that uses team-based learning to implement inquiry-based learning. We conducted quasi-experimental and observational studies in intermediate level mathematics courses, finding that team-based inquiry learning led to increased content mastery and that students took a more flexible approach to solving problems. We also found that in the courses using this pedagogy, women (but not men) had a reduction in communication apprehension over the course of a semester. We conclude that team-based inquiry learning effectively enhances student learning and problem solving, preparing students for future academic success and fostering career readiness.
-- Lewis, D., Clontz, S., Parrish, C., Estis, J., and Chaudhury, S.R. To Appear. Supporting Instructors in Implementing Team-Based Inquiry Learning. Proceedings of the 26th Annual Conference on Research in Undergraduate Mathematics Education.
+- Lewis, D., Clontz, S., Parrish, C., Estis, J., and Chaudhury, S.R. 2024. Supporting Instructors in Implementing Team-Based Inquiry Learning. _Proceedings of the 26th Annual Conference on Research in Undergraduate Mathematics Education_, 1152-1157.
- Team-Based Inquiry Learning (TBIL) is a novel active learning pedagogy designed to facilitate the use of inquiry-based learning in lower division courses. This preliminary report examines supports provided by the TBIL project to instructors, as well as the fidelity of implementation of TBIL by participants of the project. Initial findings suggest that classroom-ready materials and ongoing support, both synchronous and asynchronous, were most helpful to faculty in their TBIL implementations.
@@ -40,14 +43,21 @@ If you're planning to use TBIL in your classroom, you can get a link to join our
- Kate Owens† has created [these excellent slides](https://docs.google.com/presentation/d/1824oMy2HetUVz4AlEZFufjftg72gh0Zqqu4RzOP08PE/edit?usp=sharing) overviewing TBIL and resources to implement it in your classroom.
+- Drew Lewis presented a poster "[Transforming Lower Division Mathematics Through Team-Based Inquiry Learning](https://drive.google.com/file/d/1rruYnUEJKJZROXcXGkjl7aXdtc2v5FSs/view?usp=sharing)" at the IUSE Summit (June 2022)
+
+
- Drew Lewis gave a colloquium presentation on "[Team-Based Inquiry Learning](https://drive.google.com/file/d/19zgcQU37PqaxUka8AgBL9sh2tBz2VO56/view?usp=sharing)" at the University of Tennessee Department of Mathematics (October 2022)
- Kate Owens† & Abby Noble† presented "[How we juggled all the things: SBG+TBIL+CheckIt+LMS](https://docs.google.com/presentation/d/1tjZNkLplPyYpanPpkzfN0a-1n8FSTdNgMAh7RYkFFbs/edit?usp=sharing)" at the Fall 2023 Southeast Sectional Meeting of the American Mathematics Society
- Jordan Kostiuk† presented on "[TBIL & Student Motivation](https://drive.google.com/file/d/1PtJmaPv24viRJPOUWGqvRu-Dx3jQvSB3/view?usp=sharing)" at the 2023 AAC&U STEM Conference
+- Julie Estis presented a poster "[Transforming Lower Division Mathematics Through Team-Based Inquiry Learning: A Model for Training Faculty, Developing Resources, and Measuring Impact](https://drive.google.com/file/d/1fMj8_3Ea946Rvshmh6Z0oveKw7rk_N3c/view?usp=sharing)" at the 2023 Team-Based Learning Collaborative Annual Conference
+
- Manoj Lamichhane† presented "Implementation of Team Based Inquiry Learning (TBIL) in Calculus Classes at Florida Polytechnic University" at the Third International Conference on Applications of Mathematics to Nonlinear Sciences (May 2023)
+- Raj Chaudhury presented a poster "[Actors, Scripts and Orchestration: a framework for understanding Team-Based Inquiry Learning](https://drive.google.com/file/d/17f2bvjhRxPK6xFbu_BemDwdKe72LKxl4/view?usp=drive_link)" at the 2023 Physics Education Research Conference.
+
- Francesca Gandini† & Joe Hibdon† presented "Making Student Thinking Visible with Team-Based Inquiry Learning" at the 2024 Joint Mathematics Meeting
- Drew Lewis presented "[Team-Based Inquiry Learning in Lower Division Mathematics Courses](https://docs.google.com/presentation/d/11Ssh8PXw233QkBK3M17togEX0y6PE2_yaC9j4Z0OPYM/edit?usp=sharing)" at the 2024 Joint Mathematics Meeting
diff --git a/source/calculus/exercises/outcomes/AD/AD5/generator.sage b/source/calculus/exercises/outcomes/AD/AD5/generator.sage
index 18b1eaf88..605f2839d 100644
--- a/source/calculus/exercises/outcomes/AD/AD5/generator.sage
+++ b/source/calculus/exercises/outcomes/AD/AD5/generator.sage
@@ -1,21 +1,428 @@
class Generator(BaseGenerator):
def data(self):
- x = var("x")
- a = randrange(-8,4)
- b = a + randrange(2,6)
- c = 3*choice([-1,1])
- derivative = expand(c*(x-a)*(x-b))
- function = integrate(derivative,x)+randrange(1,10)*choice([-1,1])
- if c > 0:
- scenario = "increasing"
- else:
- scenario = "decreasing"
- return {
- "derivative": derivative,
- "function": function,
- "a": a,
- "fa": function(x=a),
- "b": b,
- "fb": function(x=b),
- scenario: True,
- }
\ No newline at end of file
+
+
+ cases = ["case1", "case2", "case3", "case4"]
+
+ scenario = choice(cases)#"case4"#
+
+
+ if scenario == "case1": #x^3/(x+stuff)^3
+ k = randrange(1,4)*choice([-1,1])
+ a = 2*k^3
+ c1 = randrange(1,4)
+ c2 = randrange(1,4)
+ sign = choice([-1,1])
+ f(x) = c1*sign*x^3/expand(c2*(x^3-a))
+ fp(x) = f(x).derivative(x)
+ fpp(x) = f(x).derivative(x,2)
+ hasymp = c1*sign/c2
+ vasymp = (a)^(1/3)
+ xint = 0
+ yint = 0
+
+
+ cv1 = min(0, a^(1/3))
+ cv2 = max(0, a^(1/3))
+ cp1 = 0
+
+ IP = [-1*k, 0, a^(1/3)]
+ IP.sort()
+ cc1 = IP[0]
+ cc2 = IP[1]
+ cc3 = IP[2]
+ ip1 = min(-1*k, 0)
+ ip2 = max(-1*k, 0)
+
+
+
+ if sign*a^(1/3) > 0:
+ change1 = "decreasing"
+ change2 = "decreasing"
+ change3 = "decreasing"
+ sign11 = "< 0"
+ sign12 = "< 0"
+ sign13 = "< 0"
+ concave1 = "concave down"
+ concave2 = "concave up"
+ concave3 = "concave down"
+ concave4 = "concave up"
+ sign21 = "< 0"
+ sign22 = "> 0"
+ sign23 = "< 0"
+ sign24 = "> 0"
+
+ if sign*a^(1/3) < 0:
+ change1 = "increasing"
+ change2 = "increasing"
+ change3 = "increasing"
+ sign11 = "> 0"
+ sign12 = "> 0"
+ sign13 = "> 0"
+ concave1 = "concave up"
+ concave2 = "concave down"
+ concave3 = "concave up"
+ concave4 = "concave down"
+ sign21 = "> 0"
+ sign22 = "< 0"
+ sign23 = "> 0"
+ sign24 = "< 0"
+
+ return{
+ scenario:True,
+ "case":"case1",
+ "f":f(x),
+ "fp":factor(fp(x)),
+ "fpp":factor(fpp(x)),
+ "xint":(xint, f(xint)),
+ "yint":(yint, f(yint)),
+ "hasymp":hasymp,
+ "vasymp":vasymp,
+ "cv1":cv1,
+ "cv2":cv2,
+ "cp1":(cp1, f(cp1)),
+ "cc1":cc1,
+ "cc2":cc2,
+ "cc3":cc3,
+ "ip1":(ip1, f(ip1)),
+ "ip2":(ip2, f(ip2)),
+ "change1":change1,
+ "change2":change2,
+ "change3":change3,
+ "sign11":sign11,
+ "sign12":sign12,
+ "sign13":sign13,
+ "concave1":concave1,
+ "concave2":concave2,
+ "concave3":concave3,
+ "concave4":concave4,
+ "sign21":sign21,
+ "sign22":sign22,
+ "sign23":sign23,
+ "sign24":sign24,
+ "minx":-8,
+ "maxx":8,
+ "miny":-8,
+ "maxy":8,
+ "k":k,
+
+ }
+
+
+
+
+
+
+
+
+
+ if scenario == "case2": #x/(x^2+stuff)
+ k = randrange(1,4)
+ a = sqrt(3)*k
+ c1 = randrange(1,4)
+ c2 = randrange(1,4)
+
+ f(x) = c1*x/expand(c2*(x^2 + a^2))
+ fp(x) = f(x).derivative(x)
+ fpp(x) = f(x).derivative(x,2)
+
+ hasymp = 0
+
+
+
+ xint = 0
+ yint = 0
+
+
+ cv1 = -a
+ cv2 = a
+ cp1 = -a
+ cp2 = a
+
+ cc1 = -3*k
+ cc2 = 0
+ cc3 = 3*k
+ ip1 = -3*k
+ ip2 = 0
+ ip3 = 3*k
+
+
+
+ change1 = "decreasing"
+ change2 = "increasing"
+ change3 = "decreasing"
+ sign11 = "< 0"
+ sign12 = "< 0"
+ sign13 = "< 0"
+ concave1 = "concave down"
+ concave2 = "concave up"
+ concave3 = "concave down"
+ concave4 = "concave up"
+ sign21 = "< 0"
+ sign22 = "> 0"
+ sign23 = "< 0"
+ sign24 = "> 0"
+
+
+
+ return{
+ scenario:True,
+ "case":"case2",
+ "f":f(x),
+ "fp":factor(fp(x)),
+ "fpp":factor(fpp(x)),
+ "xint":(xint, f(xint)),
+ "yint":(yint, f(yint)),
+ "hasymp":hasymp,
+ "cv1":cv1,
+ "cv2":cv2,
+ "cp1":(cp1, f(cp1)),
+ "cp2":(cp2, f(cp2)),
+ "cc1":cc1,
+ "cc2":cc2,
+ "cc3":cc3,
+ "ip1":(ip1, f(ip1)),
+ "ip2":(ip2, f(ip2)),
+ "ip3":(ip3, f(ip3)),
+ "change1":change1,
+ "change2":change2,
+ "change3":change3,
+ "sign11":sign11,
+ "sign12":sign12,
+ "sign13":sign13,
+ "concave1":concave1,
+ "concave2":concave2,
+ "concave3":concave3,
+ "concave4":concave4,
+ "sign21":sign21,
+ "sign22":sign22,
+ "sign23":sign23,
+ "sign24":sign24,
+ "minx":-40,
+ "maxx":40,
+ "miny":f(cp1)*1.2,
+ "maxy":f(cp2)*1.2,
+ "k":k,
+ "lb":f(cp1),
+ "ub":f(cp2),
+
+ }
+
+
+ if scenario == "case3": #x/(x^2+stuff)
+ k = randrange(1,4)
+ a = sqrt(3)*k
+ c1 = randrange(1,4)
+ c2 = randrange(1,4)
+
+ f(x) = c1*x/expand(c2*(x^2 - a^2))
+ fp(x) = f(x).derivative(x)
+ fpp(x) = f(x).derivative(x,2)
+
+ hasymp = 0
+
+
+
+ xint = 0
+ yint = 0
+
+
+ cv1 = -a
+ cv2 = a
+
+ cc1 = -a
+ cc2 = 0
+ cc3 = a
+ ip1 = 0
+
+
+
+ change1 = "decreasing"
+ change2 = "decreasing"
+ change3 = "decreasing"
+ sign11 = "< 0"
+ sign12 = "< 0"
+ sign13 = "< 0"
+ concave1 = "concave down"
+ concave2 = "concave up"
+ concave3 = "concave down"
+ concave4 = "concave up"
+ sign21 = "< 0"
+ sign22 = "> 0"
+ sign23 = "< 0"
+ sign24 = "> 0"
+
+ return{
+ scenario:True,
+ "case":"case3",
+ "f":f(x),
+ "fp":factor(fp(x)),
+ "fpp":factor(fpp(x)),
+ "xint":(xint, f(xint)),
+ "yint":(yint, f(yint)),
+ "hasymp":hasymp,
+ "vasymp1":-1*a,
+ "vasymp2":a,
+ "cv1":cv1,
+ "cv2":cv2,
+ "cc1":cc1,
+ "cc2":cc2,
+ "cc3":cc3,
+ "ip1":(0, 0),
+ "change1":change1,
+ "change2":change2,
+ "change3":change3,
+ "sign11":sign11,
+ "sign12":sign12,
+ "sign13":sign13,
+ "concave1":concave1,
+ "concave2":concave2,
+ "concave3":concave3,
+ "concave4":concave4,
+ "sign21":sign21,
+ "sign22":sign22,
+ "sign23":sign23,
+ "sign24":sign24,
+ "minx":-30,
+ "maxx":30,
+ "miny":-1,
+ "maxy":1,
+ "k":k,
+ "a":a,
+
+ }
+
+
+
+
+
+ if scenario == "case4": #x^(n+k)/n + x^k/n
+ n = 2*randrange(1,3)+1
+ k = randrange(1,n)
+ sign = choice([-1,1])
+ b = sign*(n+k)
+
+
+ f(x) = x^((n+k)/n) + b*x^(k/n)
+ fp(x) = f(x).derivative(x)
+ fpp(x) = f(x).derivative(x,2)
+
+
+ xint = 0
+ yint = 0
+
+
+ parity = "odd"
+
+ if k%2 == 1:
+ cv1 = min(-1*k*sign,0)
+ cp1 = -1*k*sign
+ cv2 = max(-1*k*sign,0)
+ cp2 = 0
+ cc1 = min((n-k)*sign, 0)
+ cc2 = max((n-k)*sign, 0)
+ ip1 = (n-k)*sign
+ change1 = "decreasing"
+ change2 = "decreasing"
+ change3 = "increasing"
+ sign11 = "< 0"
+ sign12 = "< 0"
+ sign13 = "> 0"
+ lb = f(cp1)
+ if sign < 0:
+ change1 = "decreasing"
+ change2 = "decreasing"
+ change3 = "increasing"
+ sign11 = "< 0"
+ sign12 = "< 0"
+ sign13 = "> 0"
+ concave1 = "concave up"
+ concave2 = "concave down"
+ concave3 = "concave up"
+ sign21 = "> 0"
+ sign22 = "< 0"
+ sign23 = "> 0"
+
+
+
+ if k%2 == 0:
+ cv1 = min(-1*k*sign, 0)
+ cp1 = min(-1*k*sign, 0)
+ cv2 = max(-1*k*sign, 0)
+ cp2 = max(-1*k*sign, 0)
+ parity = "even"
+ cc1 = min((n-k)*sign, 0)
+ cc2 = max((n-k)*sign, 0)
+ ip1 = (n-k)*sign
+ change1 = "increasing"
+ change2 = "decreasing"
+ change3 = "increasing"
+ sign11 = "> 0"
+ sign12 = "< 0"
+ sign13 = "> 0"
+ concave1 = "concave down"
+ concave2 = "concave down"
+ concave3 = "concave up"
+ sign21 = "< 0"
+ sign22 = "< 0"
+ sign23 = "> 0"
+ lb = r"-\infty"
+ if sign < 0:
+ concave1 = "concave down"
+ concave2 = "concave up"
+ concave3 = "concave up"
+ sign21 = "< 0"
+ sign22 = "> 0"
+ sign23 = "> 0"
+
+
+ return{
+ scenario:True,
+ "case":"case4",
+ "parity":"parity",
+ parity:True,
+ "f":f(x),
+ "fp":factor(fp(x)),
+ "fpp":factor(fpp(x)),
+ "xint":(xint, f(xint)),
+ "yint":(yint, f(yint)),
+ "xint2":(-1*b, f(-1*b)),
+ "cp1":(cp1, f(cp1)),
+ "cp2":(cp2, f(cp2)),
+ "cv1":cv1,
+ "cv2":cv2,
+ "cc1":cc1,
+ "cc2":cc2,
+ "ip1":(ip1, f(ip1)),
+ "change1":change1,
+ "change2":change2,
+ "change3":change3,
+ "sign11":sign11,
+ "sign12":sign12,
+ "sign13":sign13,
+ "concave1":concave1,
+ "concave2":concave2,
+ "concave3":concave3,
+ "sign21":sign21,
+ "sign22":sign22,
+ "sign23":sign23,
+ "minx":-10,
+ "maxx":10,
+ "miny":min(f(ip1),0)-10,
+ "maxy":max(f(ip1),0)+10,
+ "k":k,
+ "lb":lb,
+ "sign":sign,
+ "n":n,
+ "b":b,
+
+ }
+
+
+
+
+
+
+
+
+
diff --git a/source/calculus/exercises/outcomes/AD/AD5/template.xml b/source/calculus/exercises/outcomes/AD/AD5/template.xml
index 8ad7dc7ad..a111c3887 100644
--- a/source/calculus/exercises/outcomes/AD/AD5/template.xml
+++ b/source/calculus/exercises/outcomes/AD/AD5/template.xml
@@ -2,55 +2,69 @@
-Explain how to find the following for the function f(x)={{function}}.
+ For \displaystyle f(x) = {{f}} , identify the regions for which f(x) is increasing and decreasing (if any). Additionally, identify and classify all local extrema.
-
-
-
-The open intervals where f(x) is increasing or decreasing.
-
-
-
-
-
-The function increases on (-\infty,{{a}}),
-decreases on ({{a}},{{b}}), and increases on
-({{b}},\infty).
-
-
-
-
-
-
-The function decreases on (-\infty,{{a}}),
-increases on ({{a}},{{b}}), and decreases on
-({{b}},\infty).
-
-
-
-
-
-
-
-The local extrema of f(x).
-
-
-
-
-
-The point (x,y)=({{a}},{{fa}}) is a local maximum,
-and the point (x,y)=({{b}},{{fb}}) is a local minimum.
-
-
-
-
-
-
-The point (x,y)=({{a}},{{fa}}) is a local minimum,
-and the point (x,y)=({{b}},{{fb}}) is a local maximum.
-
-
-
-
-
\ No newline at end of file
+
+
+
+
+ We have that \displaystyle f'(x) = {{fp}}.
+ When x < {{cv1}}, f'(x) {{sign11}} and y is {{change1}}.
+ When {{cv1}} < x < {{cv2}}, f'(x) {{sign12}} and y is {{change2}}.
+ When {{cv2}} < x, f'(x) {{sign13}} and y is {{change3}}. There is a critical point {{cp1}} which is neither a max nor min.
+
+
+
+
+
+
+ We have that \displaystyle f'(x) = {{fp}}.
+ When x < {{cv1}}, f'(x) {{sign11}} and y is {{change1}}.
+ When {{cv1}} < x < {{cv2}}, f'(x) {{sign12}} and y is {{change2}}.
+ When {{cv2}} < x, f'(x) {{sign13}} and y is {{change3}}. There is a critical point {{cp1}} which is a local minimum and a critical point {{cp2}} is a local maximum.
+
+
+
+
+
+
+ We have that \displaystyle f'(x) = {{fp}}.
+ When x < {{cv1}}, f'(x) {{sign11}} and y is {{change1}}.
+ When {{cv1}} < x < {{cv2}}, f'(x) {{sign12}} and y is {{change2}}.
+ When {{cv2}} < x, f'(x) {{sign13}} and y is {{change3}}. There are no critical points.
+
+
+
+
+
+
+
+
+ We have that \displaystyle f'(x) = {{fp}}.
+ When x < {{cv1}}, f'(x) {{sign11}} and y is {{change1}}.
+ When {{cv1}} < x < {{cv2}}, f'(x) {{sign12}} and y is {{change2}}.
+ When {{cv2}} < x, f'(x) {{sign13}} and y is {{change3}}. There is a critical point {{cp1}} which is a local minimum.
+
+
+
+
+ We have that \displaystyle f'(x) = {{fp}}.
+ When x < {{cv1}}, f'(x) {{sign11}} and y is {{change1}}.
+ When {{cv1}} < x < {{cv2}}, f'(x) {{sign12}} and y is {{change2}}.
+ When {{cv2}} < x, f'(x) {{sign13}} and y is {{change3}}. There are critical points {{cp1}} which is a local maximum and {{cp2}} which is a local minimum.
+
-For each of the following functions, describe the open intervals
-where it is concave up or concave down, and any inflection points.
+ For \displaystyle f(x) = {{f}} , identify the regions where f(x) is concave up and concave down (if any) as well as all inflection points.
-
-
-
-
f(x)={{f}}
-
-
-
f''(x)={{fpp}}
-
-
-
-
-
-{{interval}} is concave {{concavity}}
-
-
-
-
-
-
-{{point}}
-{{#inflection}}is{{/inflection}}
-{{^inflection}}is not{{/inflection}}
-an inflection point
-
-
-
-
-
-
-
-
-
\ No newline at end of file
+
+
+
+
+ We have that \displaystyle f''(x) = {{fpp}}.
+ When x < {{cc1}}, f''(x) {{sign21}} and y is {{concave1}}.
+ When {{cc1}} < x < {{cc2}}, f''(x) {{sign22}} and y is {{concave2}}.
+ When {{cc2}} < x < {{cc3}}, f''(x) {{sign23}} and y is {{concave3}}.
+ When {{cc3}} < x, f''(x) {{sign24}} and y is {{change4}}. There are inflection points {{ip1}} and {{ip2}}.
+
+
+
+
+
+ We have that \displaystyle f''(x) = {{fpp}}.
+ When x < {{cc1}}, f''(x) {{sign21}} and y is {{concave1}}.
+ When {{cc1}} < x < {{cc2}}, f''(x) {{sign22}} and y is {{concave2}}.
+ When {{cc2}} < x < {{cc3}}, f''(x) {{sign23}} and y is {{concave3}}.
+ When {{cc3}} < x, f''(x) {{sign24}} and y is {{change4}}. There are inflection points {{ip1}}, {{ip2}} and {{ip3}}.
+
+
+
+
+
+
+ We have that \displaystyle f''(x) = {{fpp}}.
+ When x < {{cc1}}, f''(x) {{sign21}} and y is {{concave1}}.
+ When {{cc1}} < x < {{cc2}}, f''(x) {{sign22}} and y is {{concave2}}.
+ When {{cc2}} < x < {{cc3}}, f''(x) {{sign23}} and y is {{concave3}}.
+ When {{cc3}} < x, f''(x) {{sign24}} and y is {{change4}}. There is an inflection point {{ip1}}.
+
+
+
+
+
+
+
+ We have that \displaystyle f''(x) = {{fpp}}.
+ When x < {{cc1}}, f''(x) {{sign21}} and y is {{concave1}}.
+ When {{cc1}} < x < {{cc2}}, f''(x) {{sign22}} and y is {{concave2}}.
+ When {{cc2}} < x, f''(x) {{sign23}} and y is {{concave3}}.
+ There is an inflection point {{ip1}}.
+
- Consider the region bounded by y={{ f }}, y=0, x={{a}}, x={{b}}.
+ Consider the region \mathbf{R} bound by the curves y={{line}}, x={{quad}}, as well as the lines x={{xline}}, y={{yline}}
-
+
+
+
+
+
+ Sketch the region \mathbf{R} and the two lines.
+
+
+
+
+
+
+
+
+
+
+
+
+ Consider the solid of revolution generated by rotating \mathbf{R} across y={{yline}}. Express this volume as a (sum of) integral(s) with respect to x.
+
Find an integral or sum of integrals which computes the volume of the solid formed by rotating this region about the x-axis.
-
Using the washer/disk method, we obtain \int_{ {{a}} }^{ {{b}} } \pi \left({{f}}\right)^2dx and using the shell method, we obtain
- \int_0^{ {{low}} } 2\pi y\left( {{b}}-{{a}}\right)dy +\int_{ {{low}} }^{ {{high}} } 2\pi y\left( {{b}}-\left({{g}}\right)\right)dy.
+
+
+ Consider the solid of revolution generated by rotating \mathbf{R} across y={{yline}}. Express this volume as a (sum of) integral(s) with respect to y.
+
+
+
+
+ The volume is expressible as \int_{ {{a}} }^{ {{b}} } 2\pi\left( {{poly}} \right)({{radiusy}})dy
+
+
-
-
Find an integral or sum of integrals which computes the volume of the solid formed by rotating this region about the y-axis.
-
Using the washer/disk method, we obtain
- \int_0^{ {{low}} } \pi\left({{b}}^2-{{a}}^2\right)dy + \int_{ {{low}} }^{ {{high}} } \pi\left({{b}}^2-\left({{g}}\right)^2\right)dy and using the shell method, we obtain
- \int_{ {{a}} }^{ {{b}} }2\pi x \left({{f}}\right)dx.
-
+
+
+
+
+ Consider the solid of revolution generated by rotating \mathbf{R} across x={{xline}}. Express this volume as a (sum of) integral(s) with respect to x.
+
+
+
+
+ The volume is expressible as = \int_{ {{x1}} }^{ {{x2}} } 2\pi\left( {{f}} \right)({{radiusx}})dx +\int_{ {{x2}} }^{ {{x3}} } 2\pi\left( {{g}} \right)({{radiusx}})dx.
+
+
+
+
+
+
+ Consider the solid of revolution generated by rotating \mathbf{R} across x={{xline}}. Express this volume as a (sum of) integral(s) with respect to y.
+
+
+
+
+ The volume is expressible as \int_{ {{a}} }^{ {{b}} } \pi\left(\left( {{Big}} \right)^2 - \left({{Small}}\right)^2\right)dy
+
+
+
+
+
+
diff --git a/source/calculus/exercises/outcomes/AI/AI5/generator.sage b/source/calculus/exercises/outcomes/AI/AI5/generator.sage
index abc0f32a9..c152a2700 100644
--- a/source/calculus/exercises/outcomes/AI/AI5/generator.sage
+++ b/source/calculus/exercises/outcomes/AI/AI5/generator.sage
@@ -13,26 +13,26 @@ class Generator(BaseGenerator):
d(h)=randint(10, 30)/20*h+randint(1,5)
- scenario=randint(0,3)
+ scenario=randint(0,1)
- if scenario==0:
- solid="rectangular prism"
- base="sqaure"
- measuretype="side length"
- b=randint(5,10)
- height=randint(1,5)*5
- A(h)=b^2
+ #if scenario==0:
+ # solid="rectangular prism"
+ # base="sqaure"
+ # measuretype="side length"
+ # b=randint(5,10)
+ # height=randint(1,5)*5
+ # A(h)=b^2
- if scenario==1:
- solid="cylinder"
- base="circular"
- measuretype="radius"
- b=randint(2,6)
- height=randint(1,5)*5
- A(h)=pi*b^2
+ #if scenario==1:
+ # solid="cylinder"
+ # base="circular"
+ # measuretype="radius"
+ # b=randint(2,6)
+ # height=randint(1,5)*5
+ # A(h)=pi*b^2
- if scenario==2:
+ if scenario==0:
solid="pyramid"
base="square"
measuretype="side length"
@@ -40,7 +40,7 @@ class Generator(BaseGenerator):
height=randint(1,5)*5
A(h)=(b*(1-h/height))^2
- if scenario==3:
+ if scenario==1:
solid="cone"
base="circular"
measuretype="radius"
diff --git a/source/calculus/exercises/outcomes/LT/LT1/template.xml b/source/calculus/exercises/outcomes/LT/LT1/template.xml
index fe2f7b120..f78b343d8 100644
--- a/source/calculus/exercises/outcomes/LT/LT1/template.xml
+++ b/source/calculus/exercises/outcomes/LT/LT1/template.xml
@@ -84,10 +84,10 @@ Sketch the graph of a function {{fn}}(x) that meets all of the following
-
There is a jump or infinite discontinuity at x={{x}}, but the function approaches the point ({{x}},{{y1}}) as x approaches {{x}} from the left.
+
There is a jump or infinite discontinuity at x={{x}}, but the function approaches the point ({{x}},{{y1}}) as x approaches {{x}} from the left, but a different value from the right.
-
There is a jump or infinite discontinuity at x={{x}}, but the function approaches the point ({{x}},{{y1}}) as x approaches {{x}} from the right.
+
There is a jump or infinite discontinuity at x={{x}}, but the function approaches the point ({{x}},{{y1}}) as x approaches {{x}} from the right, but a different value from the left.
There is a jump discontinuity at x={{x}} with the graph approaching the point ({{x}},{{y1}}) as x approaches {{x}} from the left and approaching the point ({{x}},{{y2}}) as x approaches {{x}} from the right.
diff --git a/source/calculus/exercises/outcomes/TI/TI3/generator.sage b/source/calculus/exercises/outcomes/TI/TI3/generator.sage
index 0c92a0cd8..e2afa39d5 100644
--- a/source/calculus/exercises/outcomes/TI/TI3/generator.sage
+++ b/source/calculus/exercises/outcomes/TI/TI3/generator.sage
@@ -1,32 +1,50 @@
+#Functions to display powers of trig functions
+def print_cosp(self,*args): return f"\\cos ^{{{args[1]}}}({latex(args[0])})"
+def deriv_cosp(self,*args,**kwds):
+ if args[1]==1:
+ return -1*sinp(args[0],1)*args[0].derivative(args[kwds['diff_param']])
+ else:
+ return args[1]*-1*cosp(args[0],args[1]-1)*sinp(args[0],1)*args[0].derivative(args[kwds['diff_param']])
+
+cosp = function("cosp",nargs=2,print_latex_func=print_cosp,derivative_func=deriv_cosp)
+def print_sinp(self,*args): return f"\\sin ^{{{args[1]}}}({latex(args[0])})"
+def deriv_sinp(self,*args,**kwds):
+ if args[1]==1:
+ return cosp(args[0],1)*args[0].derivative(args[kwds['diff_param']])
+ else:
+ return args[1]*sinp(args[0],args[1]-1)*cosp(args[0],1)*args[0].derivative(args[kwds['diff_param']])
+
+sinp = function("sinp",nargs=2,print_latex_func=print_sinp, derivative_func=deriv_sinp)
+
class Generator(BaseGenerator):
+
def data(self):
- # integral with odd power
+ # Task 1, integral with odd power
x=var("x")
- even=2*randint(1,6)
-
- odd=choice([5,7])
+ n=randint(1,6)
+ m=randint(2,3)
- z = var("z")
- if odd==5:
- hint = "hint_2"
- else:
- hint = "hint_3"
+ k=var('k')
+ z=var('z')
+ hint=f"(1-z)^{m}={latex(sum(binomial(m,k)*(-z)^k,k,0,m))}"
- trigs=[sin, cos]
- shuffle(trigs)
- f=randint(1,5)*(trigs[0](x))^even*(trigs[1](x))^odd
+ trigs=[sinp, cosp]
+ shuffle(trigs)
- F=f.integral(x)
+ a = randint(1,6)
+ f = a*trigs[0](x,2*n)*trigs[1](x,2*m+1)
+ F = sum(a*binomial(m,k)*(-1)^k*1/(2*n+2*k+1)*trigs[0](x,2*n+2*k+1)*trigs[0](x,1).derivative(x)/trigs[1](x,1),k,0,m)
+
- # integral with even powers
+ # Task 2, integral with even powers
a = randrange(2,6)
m = randrange(2,5)
n = randrange(2,5)
k = 2^randrange(2,5)
- g = k*cos(a*x)^(2*m)*sin(a*x)^(2*n)
+ g = k*cosp(a*x,2*m)*sinp(a*x,2*n)
also_g = k/2^(m+n)*(1+cos(2*a*x))^m*(1-cos(2*a*x))^n
@@ -35,5 +53,5 @@ class Generator(BaseGenerator):
"F": F,
"g": g,
"also_g": also_g,
- hint: True,
+ "hint": hint,
}
diff --git a/source/calculus/exercises/outcomes/TI/TI3/template.xml b/source/calculus/exercises/outcomes/TI/TI3/template.xml
index 0377ef9f6..783f780d2 100644
--- a/source/calculus/exercises/outcomes/TI/TI3/template.xml
+++ b/source/calculus/exercises/outcomes/TI/TI3/template.xml
@@ -3,14 +3,8 @@
-Use trigonometric identities and
-{{#hint_2}}
-(1-z)^2=1-2z+z^2
-{{/hint_2}}
-{{#hint_3}}
-(1-z)^3=1-3z+3z^2-z^3
-{{/hint_3}}
-to explain and demonstrate how to find
+Using trigonometric identities and the fact that {{hint}},
+explain and demonstrate how to find
\displaystyle \int {{f}} dx.
@@ -24,8 +18,8 @@ to explain and demonstrate how to find
Explain and demonstrate how to use double-angle identities to
-rewrite \displaystyle \int {{g}} dx. (Do not solve the
-rewritten integral.)
+rewrite \displaystyle \int {{g}} dx into a form
+with smaller exponents. (Do not solve the rewritten integral.)
@@ -34,4 +28,4 @@ rewritten integral.)
-
\ No newline at end of file
+
diff --git a/source/calculus/exercises/outcomes/TI/TI4/template.xml b/source/calculus/exercises/outcomes/TI/TI4/template.xml
index 6528e6f92..04d31656d 100644
--- a/source/calculus/exercises/outcomes/TI/TI4/template.xml
+++ b/source/calculus/exercises/outcomes/TI/TI4/template.xml
@@ -9,7 +9,8 @@ Consider these given integrals of trigonometric functions.
-Explain and demonstrate how to write each of the following improper integrals
-as a limit, and why this limit converges or diverges.
+For each of the following integrals, identify if the integral is proper or improper. If improper, then compute the integral by finding the antiderivative, using the fundamental theorem of calculus, and computing the limit.
A graph of a function that has not been rendered properly.
-
+
+
+
Imagine moving along the graph toward the missing portion from the left, so that you are climbing up and to the right toward the obscured area of the graph. What y-value are you approaching?
@@ -41,9 +43,17 @@
2
2.5
+
+
+
+ B. 1
+
+
+
+
Think of the same process, but this time from the right. You're falling down and to the left this time as you come close to the missing portion. What y-value are you approaching?
@@ -53,8 +63,15 @@
2
2.5
+
+
+
+ B. 1
+
+
+
@@ -75,11 +92,19 @@
+
What is the value of f(2)?
+
+
+
+ Does not exist.
+
+
+
What is the y-value that is approached as we move toward x = 2 from the left?
@@ -90,8 +115,15 @@
2
2.5
+
+
+
+ B. 1
+
+
+
What is the y-value that is approached as we move toward x = 2 from the right?
@@ -102,14 +134,18 @@
2
2.5
+
+
+
+ B. 1
+
+
-
When studying functions in algebra, we often focused on the value of a function given a specific x-value. For instance, finding f(2) for some function f(x). In calculus, and here in and , we have instead been exploring what is happening as we approach a certain value on a graph. This concept in mathematics is known as finding a limit.
-
@@ -164,7 +200,8 @@
-->
-
+
+
What is the limit as x approaches 0 in ?
@@ -174,7 +211,13 @@
The limit is 0
The limit is not defined
-
+
+
+
+ D.
The limit is not defined
+
+
+
@@ -204,6 +247,7 @@
+
Which of the following best matches the definition of right and left limits? (Note that DNE is short for "does not exist.")
For which x-values does the overall limit exist? Select all. If the limit exists, find it. If it does not, explain why.
@@ -279,10 +366,16 @@
2
4
+
+
+
+ At A. -3 and D. 4, the limit exist.
+
+
-
+
Sketch the graph of a function f(x) that meets all of the following criteria. Be sure to scale your axes and label any important features of your graph.
@@ -299,9 +392,9 @@
\displaystyle \lim_{x\to -1^-} f(x)=-1 but \displaystyle \lim_{x\to -1^+} f(x)\neq-1.
-
+
-
+
Suppose that:
@@ -320,12 +413,14 @@
The part of the theorem that starts with “Suppose…” forms the assumptions of the theorem, while the part of the theorem that starts with “Then…” is the conclusion of the theorem. What are the assumptions of the Squeeze Theorem? What is the conclusion?
+
The assumptions of the Squeeze Theorem can be restated informally as “the function g is squeezed between the functions f and h around a.” Explain in your own words how the two assumptions result into a “squeezing effect.”
+
@@ -350,16 +445,27 @@
A pictorial example of the Squeeze Theorem.
+
Match the functions f(x), g(x), h(x) in the picture to the functions \cos(x), 1, \dfrac{\sin(x)}{x}.
+
+
+ f(x)= \cos x, g(x)= \dfrac{\sin(x)}{x}, and h(x)= 1
+
+
Using trigonometry, one can show algebraically that \cos(x) \leq \dfrac{\sin(x)}{x} \leq 1 for x values close to zero. Moreover, \displaystyle\lim_{x \to 0} \cos(x) = \cos(0)=1 (we say that cosine is a continuous function). Use these facts and the Squeeze Theorem, to find the limit \displaystyle\lim_{x\to 0} \dfrac{\sin(x)}{x} .
Based on the values of , what is the best approximation for \displaystyle\lim_{x\to 7} f(x)?
@@ -78,7 +81,15 @@
the limit is approximately 0.1667
the limit is approximately 6.9999
+
+
+
+
+
+ D. The limit is approximately 0.1667
+
+
@@ -156,6 +167,11 @@ the function as x tends to 2?
The limit can be approximated to be 1 because the values appear to approach 1 and the graph appears to approach 1, but we should zoom in on the graph to be sure.
The limit cannot be approximated because the function might not exist at x = 2.
+
+
+ C. The limit can be approximated to be 1 because the values appear to approach 1 and the graph appears to approach 1, but we should zoom in on the graph to be sure.
+
+
@@ -223,6 +239,11 @@ the function as x tends to 2?
\displaystyle \lim_{x\rightarrow 1^-}f(x) = 0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = -0.5, and \displaystyle\lim_{x\rightarrow 1}f(x) does not exist
+ B. \displaystyle \lim_{x\to 1^-}f(x) = -0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = 0.5, and \displaystyle\lim_{x\rightarrow 1}f(x) does not exist
+
+
@@ -231,6 +252,7 @@ the function as x tends to 2?
Consider the following function f(x)= 3x^3 + 2x^2- 5x+20.
+
Of the following options, at which values given would you evaluate f(x) to best determine \displaystyle \lim_{x \to 2}f(x) numerically?
@@ -239,11 +261,23 @@ the function as x tends to 2?
1.8, 1.9, 2.0, 2.1, 2.2
1.0, 1.5, 2.0, 2.5, 3.0
-
+
+
+
+ B. 1.98, 1.99, 2.0, 2.01, 2.02
+
+
+
Use the values that you chose in part (a) to calculate an approximation for \displaystyle\lim_{x \to 2}f(x).
+
+
+
+ 41.22, 41.61, 42, 42.39, 42.78
+
+
Which value best describes the limit that you obtained in part (b)?
@@ -254,7 +288,11 @@ the function as x tends to 2?
The approximate value is 41.75
The approximate value is 42
-
+
+
+ D. The approximate value is 42.
+
+
@@ -341,13 +379,26 @@ the function as x tends to 2?
The limit does not exist because you are dividing by zero when x = 0 for f(x).
-
+
+
+
+ C. The limit does not exist because the function is oscillating between -1 and 1.
+
+
+
Would your conclusion that resulted from change if the function was f(x) = \cos(1/x) or f(x) = \tan(1/x)?
-
+
+
+
+ No, both f(x) = \tan(1/x) and f(x) = \cos(1/x) oscillate as x tends to 0.
+
+
+
+
Use technology to complete the following table of values.
@@ -364,7 +415,9 @@ Then explain how to use it to make an educated guess as to the value of
the limit \displaystyle \lim_{ x\to -3 } \dfrac{ x^{2} - x - 12 }{ x^{2} + 16 \, x + 39 }
-
+
@@ -48,11 +48,9 @@
-
How would you use the language of limits to clarify the definition of continuity?
-
@@ -300,14 +298,12 @@
-
When \displaystyle\lim_{x \to a} f(x) exists but is not equal to f(a), we say that f has a removable discontinuityremovable discontinuity at x = a. This is because if f(a) were redefined to be equal to \displaystyle\lim_{x \to a} f(x), the redefined function would be continuous at x = a, thus removing the discontinuity.
When the left and right limit exist separately, but are not equal, the discontinuity is not removable and is called a jump discontinuityjump discontinuity.
diff --git a/source/calculus/source/01-LT/06.ptx b/source/calculus/source/01-LT/06.ptx
index 3484bf89c..2a3ebb4bf 100644
--- a/source/calculus/source/01-LT/06.ptx
+++ b/source/calculus/source/01-LT/06.ptx
@@ -161,9 +161,7 @@ Which of the following best describes the limit as x approaches zero in t
-
If x=a is a vertical asymptote for the function f(x) , the function f(x) is not defined at x=a. As f(a) does not exist, the function is NOT continuous at x=a. Moreover, the function's output tends to plus or minus infinity and so the limit is not equal to a number.
-
@@ -184,7 +182,7 @@ Which of the following best describes the limit as x approaches zero in t
-
Informally, we say that the limit of "\dfrac{1}{0}" is infinite. Notice that this could be either positive or negative infinity, depending on how whether the outputs are becoming more and more positive or more and more negative as we approach zero.
+
Informally, we say that the limit of "\dfrac{1}{0}" is infinite. Notice that this could be either positive or negative infinity, depending on how whether the outputs are becoming more and more positive or more and more negative as we approach zero.
@@ -206,9 +204,7 @@ Consider the rational function f(x) = \dfrac{2}{x-3} . Which of the follo
-
When considering a ratio of functions f(x)/g(x) , the inputs a where g(a)=0 are not in the domain of the ratio. If g(a)=0 but f(a) is not equal to 0, then x=a is a vertical asymptote.
-
@@ -229,7 +225,8 @@ Consider the function f(x)=\dfrac{x^2-1}{x-1}. The line x=1 is NOT
-
Recall the definition of a hole from . In we have a hole at x=1.
+
Recall the definition of a hole from . In we have a hole at x=1.
+
@@ -411,7 +408,7 @@ Explain and demonstrate how to find the value of each limit.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/02-DF/01.ptx b/source/calculus/source/02-DF/01.ptx
index 5e3ae2643..51fa6c36d 100644
--- a/source/calculus/source/02-DF/01.ptx
+++ b/source/calculus/source/02-DF/01.ptx
@@ -443,7 +443,7 @@ The rate of change of f(x) when x=-1 is positive
Exercises
The first derivative encodes information about the rate of change of the original function. In particular,
+
The first derivative encodes information about the rate of change of the original function. In particular,
If f'\gt 0 , then f is increasing;
If f'\lt 0 , then f is decreasing;
@@ -168,7 +168,7 @@ The second derivative is the derivative of the derivative. It encodes informatio
If f'' \gt 0 , then f' is increasing;
If f'' \lt 0 , then f' is decreasing;
If f'' = 0 , then f' has a horizontal tangent line (and it might have a max or min or it might just be changing pace).
-
+
Consider the function f(x)=-x^2+2x-4. Earlier you saw that f'(x)=-2x+2 and f''(x)=-2. What does this tell you about the graph of f(x) for x \gt 1?
@@ -281,7 +281,7 @@ The second derivative is the derivative of the derivative. It encodes informatio
Videos
-
+
Video for DF2
@@ -289,7 +289,7 @@ The second derivative is the derivative of the derivative. It encodes informatio
Exercises
- Exercises available at
+ Exercises available at
-
\ No newline at end of file
+
diff --git a/source/calculus/source/02-DF/03.ptx b/source/calculus/source/02-DF/03.ptx
index aca384a8f..ce0836bf9 100644
--- a/source/calculus/source/02-DF/03.ptx
+++ b/source/calculus/source/02-DF/03.ptx
@@ -293,7 +293,7 @@ Find the values of the parameters a,b,c for the quadratic polynomial
Exercises
- Exercises available at
+ Exercises available at
\ No newline at end of file
diff --git a/source/calculus/source/02-DF/04.ptx b/source/calculus/source/02-DF/04.ptx
index 9d8c6df81..075988438 100644
--- a/source/calculus/source/02-DF/04.ptx
+++ b/source/calculus/source/02-DF/04.ptx
@@ -400,7 +400,7 @@ In answering the following questions, be sure to explicitly denote which derivat
Exercises
-
\ No newline at end of file
+
diff --git a/source/calculus/source/02-DF/06.ptx b/source/calculus/source/02-DF/06.ptx
index f95357563..d39940312 100644
--- a/source/calculus/source/02-DF/06.ptx
+++ b/source/calculus/source/02-DF/06.ptx
@@ -247,7 +247,7 @@ you are using in your work.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/02-DF/07.ptx b/source/calculus/source/02-DF/07.ptx
index ab0b28f4f..7ea52a355 100644
--- a/source/calculus/source/02-DF/07.ptx
+++ b/source/calculus/source/02-DF/07.ptx
@@ -178,7 +178,7 @@ Explain how to use implicit differentiation to find
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/02-DF/08.ptx b/source/calculus/source/02-DF/08.ptx
index 9ae89e091..3274c9298 100644
--- a/source/calculus/source/02-DF/08.ptx
+++ b/source/calculus/source/02-DF/08.ptx
@@ -317,7 +317,7 @@ You are given the graphs of f(v), f'(v) below.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/03-AD/01.ptx b/source/calculus/source/03-AD/01.ptx
index ea5e2d0d2..af967803f 100644
--- a/source/calculus/source/03-AD/01.ptx
+++ b/source/calculus/source/03-AD/01.ptx
@@ -260,7 +260,7 @@ and the marginal profit is -12 \, x^{2} - 20 \, x + 56.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/03-AD/02.ptx b/source/calculus/source/03-AD/02.ptx
index c23110222..876b9a7e2 100644
--- a/source/calculus/source/03-AD/02.ptx
+++ b/source/calculus/source/03-AD/02.ptx
@@ -51,43 +51,16 @@ Notice that this is obtained by writing the tangent line to f(x) at (a
Sketch the tangent line L(x) on the same plane as the graph of \ln(x). What do you notice?
-
+
+
+ x = var('x')
+ f = ln(x)
+ p = plot(f,(x,0.1,5), gridlines=True, axes_labels=('$x$','$y$'), thickness=2, aspect_ratio = 1.25)
+ p
+
+
+
The graph of \ln(x)
+
@@ -280,7 +253,7 @@ The estimate is greater than the actual value.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/03-AD/03.ptx b/source/calculus/source/03-AD/03.ptx
index 5df37d29b..d6e7c9eec 100644
--- a/source/calculus/source/03-AD/03.ptx
+++ b/source/calculus/source/03-AD/03.ptx
@@ -53,11 +53,14 @@ recall that when y is a function of x, which in turn is a function
+
A spherical balloon is being inflated at a constant rate of 20 cubic inches per second.
How fast is the radius of the balloon changing at the instant the balloon's diameter is 12 inches?
Is the radius changing more rapidly when d = 12 or when d = 16?
Why?
+
+
@@ -120,7 +123,6 @@ recall that when y is a function of x, which in turn is a function
-
@@ -346,7 +348,7 @@ recall that when y is a function of x, which in turn is a function
Exercises
diff --git a/source/calculus/source/03-AD/05.ptx b/source/calculus/source/03-AD/05.ptx
index ad48b23c5..cf4566c50 100644
--- a/source/calculus/source/03-AD/05.ptx
+++ b/source/calculus/source/03-AD/05.ptx
@@ -243,7 +243,7 @@ and the point (x,y)=(1,6) is a local maximum.-->
Exercises
-
+
+
+
+ x = var('x')
+ f = 2*exp(4*x-3)
+ ticks=[[],[]]
+ p = plot(f,(x,.2,1), xmin = -10, xmax = 1, ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.2)
+ p
+
+
+
+
+ x = var('x')
+ f = x+.75
+ ticks=[[],[]]
+ p = plot(f,(x,.05,1), xmin = -10, xmax = 1, ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.25)
+ p
+
+
+
+
+ x = var('x')
+ f = (x-1.5)^3+2.5
+ ticks=[[],[]]
+ p = plot(f,(x,.25,1.25), ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.25)
+ p
+
+
+
+
@@ -65,7 +94,6 @@
-
On the leftmost curve in ,
as we move from left to right,
the slopes of the tangent lines will increase.
@@ -73,7 +101,6 @@
and this explains why we say this function is
increasing at an increasing rate.
-
We must be extra careful with our language when dealing with negative numbers.
@@ -93,7 +120,35 @@
From left to right, three functions that are all decreasing.
-
+
+
+
+ x = var('x')
+ f = -(x-1.5)^3+.5
+ ticks=[[],[]]
+ p = plot(f,(x,.25,1.25), ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.25)
+ p
+
+
+
+
+ x = var('x')
+ f = -2*x + 3
+ ticks = [[],[]]
+ p = plot(f, (x,.25, 1.25), ymin = -1, ymax = 3, gridlines=True, thickness=2, ticks=ticks, aspect_ratio = 0.25)
+ p
+
+
+
+
+ x = var('x')
+ f = -(x)^3+2.5
+ ticks=[[],[]]
+ p = plot(f,(x,.25,1.25), ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.25)
+ p
+
+
+
@@ -131,7 +186,7 @@
-
Recall the terminology of concavity: when a curve bends upward, we say its shape is concave up. When a curve bends downwards, we say its shape is concave down.
+
Recall the terminology of concavity: when a curve bends upward, we say its shape is concave up. When a curve bends downwards, we say its shape is concave down.
-
-
+
+
+ x = var('x')
+ f = -(x)^3+2.5
+ ticks=[[],[]]
+ p = plot(f,(x,.25,1.25), ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.25)
+ p
+
+
+
@@ -46,7 +55,15 @@
-
+
+
+ x = var('x')
+ f = (x-1.5)^3+2.5
+ ticks=[[],[]]
+ p = plot(f,(x,.25,1.25), ymin = -1, ymax = 3, gridlines=True,thickness=2, ticks=ticks, aspect_ratio=.25)
+ p
+
+
@@ -239,7 +256,7 @@ Note that some of the graphs below are definitely hard... might want to tell thi
Exercises
- Exercises available at
+ Exercises available at
\ No newline at end of file
diff --git a/source/calculus/source/03-AD/08.ptx b/source/calculus/source/03-AD/08.ptx
index a8c12f3fe..1f6ebe0c5 100644
--- a/source/calculus/source/03-AD/08.ptx
+++ b/source/calculus/source/03-AD/08.ptx
@@ -29,18 +29,252 @@
Which of the following diagrams best illustrates how the box is created?
\ No newline at end of file
diff --git a/source/calculus/source/04-IN/02.ptx b/source/calculus/source/04-IN/02.ptx
index 8a065e911..c2e8f7806 100644
--- a/source/calculus/source/04-IN/02.ptx
+++ b/source/calculus/source/04-IN/02.ptx
@@ -31,6 +31,26 @@
axes for plotting y = v(t);
at right, for plotting
y = s(t).
+
+
+
+ x = var('x')
+ f = 0
+ ticks = [True,[a for a in (1..8)]]
+ p = plot(f,(x,0,2.25), thickness=0, ymin = 0.25, ymax = 8.25, gridlines='minor', axes_labels=('hrs','mph'), axes_labels_size=1, aspect_ratio=.25, ticks=ticks)
+ p
+
+
+
+
+ x = var('x')
+ f = 0
+ ticks = [True,[a for a in (1..8)]]
+ p = plot(f,(x,0,2.25), thickness=0, ymin = 0.25, ymax = 8.25, gridlines='minor', axes_labels=('hrs','miles'), axes_labels_size=1, aspect_ratio=.25, ticks=ticks)
+ p
+
+
+
@@ -130,7 +150,27 @@
The graph of y = v(t).
-
+
+
+ x = var('x')
+ f = (x-1)^3 + 2.5
+ p = plot(f, (x,-.25,2.25), thickness=2, ymin=0, ymax = 3.5, gridlines=True, axes_labels=('hrs','mph'), axes_labels_size=1, aspect_ratio=.6)
+ a = text("$y = v(t)$",(1, 2.65), fontsize=12, color='black')
+ p + a
+
+
+
@@ -214,51 +254,64 @@
Explain how to approximate the area under the curve
- f(x) = -\frac{1}{5} (x-4)(x-10)(x-12)
+ f(x) = \frac{1}{5} (x-4)(x-10)(x-12)
on the
interval [4,10] using a right Riemann
sum with 3 subintervals.
- The graph of the function f(x) = -1/5 (x-4)(x-10)(x-12)
+ The graph of the function f(x) = 1/5 (x-4)(x-10)(x-12)
crosses the x-axis upward at (4,0) and downward at (10,0)
with a maximum at about (6.3, 9.7).
-
- \begin{tikzpicture}
- \begin{axis}[
- axis lines=middle,
- grid=both,
- xmin=3,
- xmax=11,
- ymin=-1,
- ymax=12,
- xlabel=$x$,
- ylabel=$y$,
- xtick={3,4,...,10},
- ytick={0,1,...,12},
- legend pos=north east,
- ]
- \addplot[domain=3:11, smooth, thick, blue] {1/5*(x-4)*(x-10)*(x-12)};
- \addlegendentry{{\tiny $f(x)=\frac{1}{5}(x-4)(x-10)(x-12)$}}
- \end{axis}
- \end{tikzpicture}
-
+
+ x = var('x')
+ f = 0.2*(x-4)*(x-10)*(x-12)
+ p = plot(f,(x,3,10.5),gridlines=True,ymin = -1, ymax = 11, thickness=2, axes_labels=('$x$','$y$'))
+ p
+
@@ -425,7 +464,7 @@
Exercises
- Exercises available at
+ Exercises available at
\ No newline at end of file
diff --git a/source/calculus/source/04-IN/03.ptx b/source/calculus/source/04-IN/03.ptx
index 14561a3cf..8146b1a32 100644
--- a/source/calculus/source/04-IN/03.ptx
+++ b/source/calculus/source/04-IN/03.ptx
@@ -235,7 +235,7 @@ Find each indefinite integral.
Exercises
In there are more than one solution that satisfies the differential equation. In fact their is a family of functions that satisfies the differential equation, that is
f(x) = \frac{x^3}{3} + x^2 + c_1,
where c_1 is an arbitrary constant yet to be defined. To find c_1 we have to have some initial value for the differential equation, y(x_0) = y_0, where the point (x_0,y_0) is the starting point for the differential equation. In general this section we will focus on solving initial value problems (differential equation with an initial condition) of the form,
\frac{dy}{dx} = f(x), \;\;\; y(x_0) = y_0.
-
@@ -88,28 +86,27 @@
-
-
So far we have only been going from velocity to position of an object. Recall that to find the acceleration of an object, you can take the derivative of the velocity of an object. Let use say we have the acceleration of a falling object in m/s^2 given by a(t) = -9.8.
-
-
What is the velocity of the falling object, if the initial velocity is given by v(0) = 0 m/s.
+
+
So far we have only been going from velocity to position of an object. Recall that to find the acceleration of an object, you can take the derivative of the velocity of an object. Let use say we have the acceleration of a falling object in m/s^2 given by a(t) = -9.8.
+
+
+
What is the velocity of the falling object, if the initial velocity is given by v(0) = 0 m/s.
v(t) = -9.8t m
v(t) = -9.8t m/s
v(t) = 9.8t m/s
v(t) = 9.8t+1 m
-
-
-
-
What is the position of the object, if the initial position is given by s(0) = 10 m.
+
+
+
+
What is the position of the object, if the initial position is given by s(0) = 10 m.
Approximate the area under the curve f(x)=(x-1)^2+2 on the interval [1,5] using a left Riemann sum with four uniform subdivisions. Draw your rectangles on the graph.
-
+
+
+ x = var('x')
+ f = (x-1)^2 + 2
+ p = plot(f,(x,0,5), gridlines=True, thickness=2, axes_labels=('$x$','$f(x)$'), aspect_ratio = .25)
+ p
+
+
@@ -96,21 +111,44 @@
\displaystyle \int_0^2 \left(x^2+3\right) \, dx
-
+
+
+ x = var('x')
+ f = x^2 + 3
+ ticks = [True, [1..12]]
+ p = plot(f,(x,-3,3), ymin = 0, gridlines=True, thickness=2, axes_labels=('$x$','$y$'),ticks=ticks)
+ p
+
+
A geometrical interpretation of \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_a^bf(x)dx
() defines \int_a^bf(x)dx as the net area
@@ -28,7 +27,6 @@ between the graph of y=f(x) and the x-axis. By net area, we mean t
Improving approximations of \int_0^5(x-2)(x-4)dx
-
@@ -159,7 +157,7 @@ from x=-1 to x=9?
Exercises
- Exercises available at
+ Exercises available at
\ No newline at end of file
diff --git a/source/calculus/source/04-IN/08.ptx b/source/calculus/source/04-IN/08.ptx
index 54a1d5135..56ca66010 100644
--- a/source/calculus/source/04-IN/08.ptx
+++ b/source/calculus/source/04-IN/08.ptx
@@ -60,9 +60,30 @@
We now look for a general way of writing definite integrals for the area between two given curves, f(x) and g(x). Consider this area, illustrated in .
Consider the integral \displaystyle \int x^9\ln(x) \,dx. Suppose we proceed using integration by parts. We choose u=\ln(x) and dv=x^9\,dx.
+
What is du?
@@ -169,7 +170,6 @@
Neither integral seems harder than the other one.
-
@@ -373,7 +373,7 @@
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/05-TI/03.ptx b/source/calculus/source/05-TI/03.ptx
index 8abf2fbf5..6dc467f2a 100644
--- a/source/calculus/source/05-TI/03.ptx
+++ b/source/calculus/source/05-TI/03.ptx
@@ -223,12 +223,10 @@ If \cos's power is odd, rewrite the integral as
-
We might also use some other trigonometric identities to manipulate our integrands, listed in
.
-
@@ -268,7 +266,7 @@ We might also use some other trigonometric identities to manipulate our integran
Exercises
diff --git a/source/calculus/source/05-TI/06.ptx b/source/calculus/source/05-TI/06.ptx
index 3b131d608..cd0978b3e 100644
--- a/source/calculus/source/05-TI/06.ptx
+++ b/source/calculus/source/05-TI/06.ptx
@@ -348,7 +348,7 @@ Using some algebra, it's possible to find values for A through G t
Exercises
diff --git a/source/calculus/source/05-TI/08.ptx b/source/calculus/source/05-TI/08.ptx
index 6773817b6..9906fb647 100644
--- a/source/calculus/source/05-TI/08.ptx
+++ b/source/calculus/source/05-TI/08.ptx
@@ -565,7 +565,7 @@ we let \displaystyle \int_c^b f(x)\, dx denote \int_c^b f(x)\, dx=\li
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/06-AI/01.ptx b/source/calculus/source/06-AI/01.ptx
index ca119d768..798ccbafb 100644
--- a/source/calculus/source/06-AI/01.ptx
+++ b/source/calculus/source/06-AI/01.ptx
@@ -274,7 +274,7 @@ For example, the following calculuation confirms the previous activity:
Exercises
diff --git a/source/calculus/source/06-AI/03.ptx b/source/calculus/source/06-AI/03.ptx
index 010ba0239..47cd0b3c4 100644
--- a/source/calculus/source/06-AI/03.ptx
+++ b/source/calculus/source/06-AI/03.ptx
@@ -174,7 +174,6 @@ Which of these formulas is most appropriate to find this illustration's cross-se
-
Generally when solving problems without the aid of technology, it's useful to draw your region in
two dimensions, choose whether to use a horizontal or vertical line segment, and draw its rotation
@@ -195,7 +194,6 @@ When the shape is a cylindrical shell, this is called the shell methodx or y.
V=\int_a^b 2\pi r(x)h(x)\,dx,\hspace{2em}V=\int_a^b 2\pi r(y)h(y)\,dy.
-
@@ -369,7 +367,7 @@ Which of these integrals is equal to the volume of the solid of revolution?
Exercises
diff --git a/source/calculus/source/07-CO/01.ptx b/source/calculus/source/07-CO/01.ptx
index 744b0e11c..c0d692912 100644
--- a/source/calculus/source/07-CO/01.ptx
+++ b/source/calculus/source/07-CO/01.ptx
@@ -263,7 +263,7 @@ show that this is the vector equation for the line y=-3x+4.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/07-CO/02.ptx b/source/calculus/source/07-CO/02.ptx
index c3240123e..afa025868 100644
--- a/source/calculus/source/07-CO/02.ptx
+++ b/source/calculus/source/07-CO/02.ptx
@@ -204,7 +204,7 @@ corresponding to the parameter t=-3.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/07-CO/03.ptx b/source/calculus/source/07-CO/03.ptx
index 21ecb36c3..da1dbad6d 100644
--- a/source/calculus/source/07-CO/03.ptx
+++ b/source/calculus/source/07-CO/03.ptx
@@ -310,7 +310,7 @@ with t replaced with x.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/07-CO/04.ptx b/source/calculus/source/07-CO/04.ptx
index 8d2366356..65799cd9d 100644
--- a/source/calculus/source/07-CO/04.ptx
+++ b/source/calculus/source/07-CO/04.ptx
@@ -221,7 +221,7 @@ Find a Cartesian form of each of the given polar equations.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/07-CO/05.ptx b/source/calculus/source/07-CO/05.ptx
index 539c07c2b..7abf0db31 100644
--- a/source/calculus/source/07-CO/05.ptx
+++ b/source/calculus/source/07-CO/05.ptx
@@ -348,7 +348,7 @@ show that this is the vector equation for the line y=-3x+4.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/07-CO/06.ptx b/source/calculus/source/07-CO/06.ptx
index b98d33c1a..b0c936503 100644
--- a/source/calculus/source/07-CO/06.ptx
+++ b/source/calculus/source/07-CO/06.ptx
@@ -419,7 +419,7 @@ show that this is the vector equation for the line y=-3x+4.
Exercises
As seen in the previous activity, having too few terms may prevent us from finding a unique way to continue creating a sequence of numbers. In fact, we need sufficiently many terms to uniquely continue a sequence of numbers (and how many terms is sufficient depends on which sequence of numbers you are trying to generate). Sometimes, we do not want to write out all of the terms needed to allow for this. Therefore, we will want to find short-hand notation that allows us to do so.
- For each of the following sequences, determine which of the properties: monotonic, bounded and convergent, the sequence satisfies. If a sequence is convergent, determine to what it converges.
-
- \left\{\displaystyle 3n\right\}_{n=0}^\infty.
-
-
- \left\{\displaystyle \frac{n^3}{3^n}\right\}_{n=0}^\infty.
-
-
- \left\{\displaystyle \frac{n}{n+3}\right\}_{n=1}^\infty.
-
-
- \left\{\displaystyle \frac{(-1)^n}{n+3}\right\}_{n=1}^\infty.
-
-
-
+For each of the following sequences, determine which of the properties: monotonic, bounded and convergent, the sequence satisfies. If a sequence is convergent, determine to what it converges.
+
+
+
+
diff --git a/source/calculus/source/08-SQ/03.ptx b/source/calculus/source/08-SQ/03.ptx
index 194bc4c38..beaaef77e 100644
--- a/source/calculus/source/08-SQ/03.ptx
+++ b/source/calculus/source/08-SQ/03.ptx
@@ -452,7 +452,7 @@ When the series (the sequence of partial sums) diverges, we say the series is Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/08-SQ/04.ptx b/source/calculus/source/08-SQ/04.ptx
index 6a9b2ae86..4e0def4e4 100644
--- a/source/calculus/source/08-SQ/04.ptx
+++ b/source/calculus/source/08-SQ/04.ptx
@@ -207,7 +207,7 @@ seen without any modification of the original series.
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/08-SQ/05.ptx b/source/calculus/source/08-SQ/05.ptx
index c969b6c63..b88b0b29b 100644
--- a/source/calculus/source/08-SQ/05.ptx
+++ b/source/calculus/source/08-SQ/05.ptx
@@ -388,7 +388,7 @@ The series \displaystyle \sum_{n=1}^\infty \displaystyle\frac{1}{n^p} con
Exercises
Consider a power series \displaystyle\sum c_n(x-a)^n. Recall from that if
+
Consider a power series \displaystyle\sum a_n(x-c)^n. Recall from that if
- \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| & < 1
+ \displaystyle \lim_{n\to \infty} \left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| & < 1
- then \displaystyle\sum c_n(x-a)^n converges.
+ then \displaystyle\sum a_n(x-c)^n converges.
- For what values of x is \displaystyle |x|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1?
+ For what values of x is \displaystyle |x|\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1?
- For what values of x is \displaystyle |x-2|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1?
+ For what values of x is \displaystyle |x-2|\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1?
- For what values of x is \displaystyle \left|x+\frac{1}{2}\right|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1?
+ For what values of x is \displaystyle \left|x+\frac{1}{2}\right|\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1?
0\leq x < \infty.
All real numbers.
@@ -219,18 +216,18 @@
- Given the power series \displaystyle\sum c_n(x-a)^n, the center of convergencecenter of convergence is x=a. The radius of convergenceradius of convergence is r=\frac{1}{\displaystyle\lim_{n\to\infty} \left| \frac{c_{n+1}}{c_n} \right|}. If \displaystyle\lim_{n\to\infty} \left| \frac{c_{n+1}}{c_n} \right|=0, we say that r=\infty.
+ Given the power series \displaystyle\sum a_n(x-c)^n, the center of convergencecenter of convergence is x=a. The radius of convergenceradius of convergence is r=\frac{1}{\displaystyle\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|}. If \displaystyle\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=0, we say that r=\infty.
- The interval of convergenceinterval of convergence represents all possible values of x for which \displaystyle\sum c_n(x-a)^n converges, which is of the form:
+ The interval of convergenceinterval of convergence represents all possible values of x for which \displaystyle\sum a_n(x-c)^n converges, which is of the form:
-
(a-r, a+r)
-
[a-r, a+r)
-
(a-r, a+r]
-
[a-r, a+r]
+
(c-r, c+r)
+
[c-r, c+r)
+
(c-r, c+r]
+
[c-r, c+r]
- Depending on if \displaystyle\sum c_n(x-a)^n converges when x=a-r or x=a+r.
+ Depending on if \displaystyle\sum a_n(x-c)^n converges when x=c-r or x=c+r.
If r=\infty, the interval of convergence is all real numbers.
@@ -327,7 +324,7 @@
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/09-PS/03.ptx b/source/calculus/source/09-PS/03.ptx
index 2ad791530..84d88daa2 100644
--- a/source/calculus/source/09-PS/03.ptx
+++ b/source/calculus/source/09-PS/03.ptx
@@ -611,7 +611,7 @@ Let's investigate the power series
Exercises
- Exercises available at
+ Exercises available at
diff --git a/source/calculus/source/09-PS/04.ptx b/source/calculus/source/09-PS/04.ptx
index 5274818f8..5033533d2 100644
--- a/source/calculus/source/09-PS/04.ptx
+++ b/source/calculus/source/09-PS/04.ptx
@@ -492,7 +492,7 @@ This gives rise to the following notion.
Exercises
+ A vector is a quantity which has both magnitude/length and direction. We will denote a vector using boldface type, as \textbf{v}. Vectors are generally depicted visually as an arrow beginning at an initial point (
+ A below) and ending at a terminal point (B below).
+
+
+
+
+ \begin{tikzpicture}
+ \coordinate (a) at (0,0);
+ \coordinate (b) at (4,3);
+ \fill (a) circle (0.1);
+ \draw[->] (a)--(b) node[midway, yshift=7.5pt] {$\textbf{v}$};
+ \node at (0,-.5) {$A$};
+ \node at (4.25,2.5) {$B$};
+ \end{tikzpicture}
+
+
+
+
+
+
+
+
+
+
+ Vectors can be decomposed into components- horizontal and vertical. See the diagram below.
+
+ If the length of the horizontal leg is a, and the length of the vertical leg is b, which is the best description for the magnitude of \textbf{v}, denoted ||\textbf{v}||?
+
+
+
+ a + b
+
+
+
+
+ \sqrt{b^2 - a^2}
+
+
+
+
+ \sqrt{a^2 + b^2}
+
+
+
+
+ a^2 + b^2
+
+
+
+
+
+
+
+
+
+ Which expression best identifies the direction of the vector, denoted \theta in the diagram?
+
+
+
+
+ \sin^{-1}\left(\dfrac{a}{b}\right)
+
+
+
+
+ \sin^{-1}\left(\dfrac{b}{a}\right)
+
+
+
+
+ \cos^{-1}\left(\dfrac{a}{b}\right)
+
+
+
+
+ \cos^{-1}\left(\dfrac{b}{a}\right)
+
+
+
+
+ \tan^{-1}\left(\dfrac{a}{b}\right)
+
+
+
+
+ \tan^{-1}\left(\dfrac{b}{a}\right)
+
+
+
+
+
+
+
+
+
+ Use trigonometry and/or part (a) to find expressions for the length of a and the length of b in terms of the direction and magnitude of \textbf{v}.
+
+
+
+
+
+
+
+ A vector \textbf{v} with horizontal component x and vertical component y is written in component form as \langle x,y\rangle.
+
+
+
+
+
+
+ Consider the points A(1,2) and B(6,9).
+
+
+
+
+
+ Write the component form of the vector beginning at B and ending at A.
+
+
+
+
+ \langle7,5\rangle
+
+
+
+
+
+
+ \langle-7,-5\rangle
+
+
+
+
+
+
+ \langle5,7\rangle
+
+
+
+
+
+
+ \langle-5,-7\rangle
+
+
+
+
+
+
+
+
+
+
+ What is the direction of this vector?
+
+
+
+
+
+
+ Write the component form of the vector beginning at A and ending at B.
+
+
+
+
+ \langle7,5\rangle
+
+
+
+
+
+
+ \langle-7,-5\rangle
+
+
+
+
+
+
+ \langle5,7\rangle
+
+
+
+
+
+
+ \langle-5,-7\rangle
+
+
+
+
+
+
+
+
+
+
+ What is the direction of this vector?
+
+
+
+
+
+
+ What do you notice about the vectors \textbf{BA} from part (a) and \textbf{AB} from part (c)?
+
+
+
+
+
+
+
+
+ Now consider the points A(3,2) and B(1,1)
+
+
+
+
+
+ Find the vector \textbf{AB}.
+
+
+
+
+
+
+ Consider the additional point C(2,2). Find the vector \textbf{AC}.
+
+
+
+
+ \langle2,1\rangle
+
+
+
+
+
+
+ \langle1,2\rangle
+
+
+
+
+
+
+ \langle-1,-2\rangle
+
+
+
+
+
+
+ \langle-2,-1\rangle
+
+
+
+
+
+
+
+
+
+
+ Sketch \textbf{AB} and \textbf{AC} on the same coordinate grid. What relationship do you see between \textbf{AC} and \textbf{AB}?
+
+
+
+
+
+
+ Now consider the point D(-1,-1). Find the vector \textbf{AD}.
+
+
+
+
+
+
+ Sketch \textbf{AD} on the same grid from part (f). What relationship do see between \textbf{AB} and \textbf{AD}?
+
+
+
+
+
+
+
+
+ Vectors can be added and subtracted like regular numbers. Addition happens component-wise; for example, adding the vectors \langle1,3\rangle + \langle-4,2\rangle gives the resultant vector\langle1-4,3+2\rangle = \langle-3,5\rangle
+
+
+
+
+
+
+
+ Let \textbf{u} = \langle-2,1\rangle and \textbf{v} = \langle0,3\rangle. Find the following vectors.
+
+
+
+
+
+ \textbf{u} + \textbf{v}
+
+
+
+
+ \langle2,2\rangle
+
+
+
+
+
+
+ \langle-2,4\rangle
+
+
+
+
+
+
+ \langle1,1\rangle
+
+
+
+
+
+
+ \langle-1,3\rangle
+
+
+
+
+
+
+
+
+
+
+ \textbf{u} - \textbf{v}
+
+
+
+
+ \langle-2,-2\rangle
+
+
+
+
+
+
+ \langle2,2\rangle
+
+
+
+
+
+
+ \langle-5,1\rangle
+
+
+
+
+
+
+ \langle5,-1\rangle
+
+
+
+
+
+
+
+
+
+
+ 2\textbf{u} - \dfrac{1}{2}\textbf{v}
+
+
+
+
+ \langle2,5\rangle
+
+
+
+
+
+
+ \langle-4,\frac{5}{2}\rangle
+
+
+
+
+
+
+ \langle1,-\frac{5}{2}\rangle
+
+
+
+
+
+
+ \langle-4,-\frac{1}{2}\rangle
+
+
+
+
+
+
+
+
+
+
+
+
+ This activity will develop the parallelogram rule, a method for visualizing the addition or subtraction of two vectors.
+
+
+ Consider the vectors \textbf{a} = \langle2,0\rangle and \textbf{b} = \langle1,1\rangle for the tasks below.
+
+
+
+
+
+ Find \textbf{a} + \textbf{b}.
+
+
+
+
+
+
+ On a coordinate grid, draw the vectors \textbf{a} and \textbf{b} so that their tails are touching.
+
+
+
+
+
+
+ Create a parallelogram by translating \textbf{a} and \textbf{b}, then include \textbf{a} + \textbf{b} on the parallelogram. What do you notice about its position on the diagram?
+
+
+
+
+
+
+ Compute \textbf{a} - \textbf{b}. How, if at all, does it fit in the diagram from (c)?
+
+
+
+
+
+
+ Vector Properties
+
+ Let \textbf{a}, \textbf{b} and \textbf{c} be three vectors, and let k,r be scalar values. The following properties hold:
+
+ (Multiplicative Zero for Vectors)
+
+ 0\textbf{a} = \textbf{0}
+
+
+
+
+
+
+
+
+
+ A convenient thing about working with vectors is that the same ideas that work in two dimensions work in any number of dimensions. The number of dimensions is equal to the number of components in the vector. For example, the vector \langle1,4\rangle is two-dimensional because there are only two components, while the vector \langle0,1,2,3,4,5\rangle is six-dimensional because there are six components.
+
+
+
+
+
+
+ For the vectors \textbf{a} = \langle-3,5,1\rangle, \textbf{b} = \langle1,0,2}, find the following.
+
+ In order to avoid confusion, we talk about a three-dimensional coordinate system using the right-hand rule: when using your right hand, point your index finger out, thumb up, and middle finger perpendicular to both. Your index finger represents the y-axis, your middle finger represents the x-axis, and your thumb represents the z-axis.
+
+ Recall from that the magnitude or length of a two-dimensional vector \textbf{v} = \langle a,b\rangle can be found by computing \sqrt{a^2 + b^2}.
+
+
+
+
+
+ For a three-dimensional vector \textbf{u} = \langle a,b,c\rangle, how might you compute ||\textbf{u}||?
+
+
+
+
+
+
+ Find the length of the vector \textbf{a} = \langle0,3,4\rangle.
+
+
+
+
+
+
+ What is the length of the vector \dfrac{1}{||\textbf{a}||}\textbf{a}?
+
+
+
+
+
+
+
+
+ A unit vector is a vector with unit length, i.e. length exactly 1.
+
+
+
+
+
+
+
+ For each vector, find its associated unit vector.
+
+
+
+
+
+
+ \langle7,6\rangle
+
+
+
+
+
+
+
+
+ \langle-1,0,8\rangle
+
+
+
+
+
+
+
+
+ \langle0,1\rangle
+
+
+
+
+
+
+
+
+ \langle1,1,1\rangle
+
+
+
+
+
+
+
+
+
+ The unit vector \textbf{i} is the vector whose first component 1 and all others are 0, \textbf{j} is the one whose second component is 1 with all others 0, and \textbf{k} is the one whose third component is 1 with all others 0.
+
+
+
+
+
+
+ Write each vector as a sum of the standard unit vectors \textbf{i}, \textbf{j}, \textbf{k}.
+
+
+
+
+
+
+ \langle-4,3\rangle
+
+
+
+
+
+ -4\textbf{j} + 3\textbf{k}
+
+
+
+
+
+
+ -4\textbf{i} + 3\textbf{k}
+
+
+
+
+
+
+ -4\textbf{i} + 3\textbf{j}
+
+
+
+
+
+
+ -4\textbf{j} + 3\textbf{i}
+
+
+
+
+
+
+
+
+
+
+
+ \langle3,1,-5\rangle
+
+
+
+
+
+ 3\textbf{j} + \textbf{i} - 5\textbf{k}
+
+
+
+
+
+
+ 3\textbf{j} + \textbf{k} - 5\textbf{i}
+
+
+
+
+
+
+ 3\textbf{i} + \textbf{j} - 5\textbf{k}
+
+
+
+
+
+
+ 3\textbf{i} + \textbf{k} - 5\textbf{j}
+
+
+
+
+
+
+ 3\textbf{k} + \textbf{i} - 5\textbf{j}
+
+
+
+
+
+
+ 3\textbf{k} + \textbf{j} - 5\textbf{i}
+
+
+
+
+
+
+
+
+
+
+
+ \langle-1,1,0\rangle
+
+
+
+
+
+
+
+
+
+ The speed of an object is the magnitude of its associated velocity vector. A football is thrown by a player with an initial speed of 82 miles per hour and an angle of elevation of 25^{\circ}. Determine the velocity vector in miles per hour and express in component form and in terms of the standard unit vectors. Assume the ball is thrown in the xz-plane.
+
+
+
+
+
+ Which of the following best expresses the horizontal component of the vector?
+
+
+
+
+ 82\sin(25^{\circ})
+
+
+
+
+
+
+ 25\cos(82^{\circ})
+
+
+
+
+
+
+ 82\cos(25^{\circ})
+
+
+
+
+
+
+ 25\sin(82^{\circ})
+
+
+
+
+
+
+ 82\sin(65^{\circ})
+
+
+
+
+
+
+ 82\cos(65^{\circ})
+
+
+
+
+
+
+
+
+
+
+ Which option best describes the vertical component of the vector?
+
+
+
+
+ 82\sin(25^{\circ})
+
+
+
+
+
+
+ 25\cos(82^{\circ})
+
+
+
+
+
+
+ 82\cos(25^{\circ})
+
+
+
+
+
+
+ 25\sin(82^{\circ})
+
+
+
+
+
+
+ 82\sin(65^{\circ})
+
+
+
+
+
+
+ 82\cos(65^{\circ})
+
+
+
+
+
+
+
+
+
+
+ Use your answers to (a) and (b) to write the velocity vector in component form.
+
+
+
+
+
+
+ Use your answers to (a) and (b) to write the velocity vector using the standard unit vectors. Remember that the ball is thrown in the xz-plane.
+
+
+
+
+
+
+ Why would we consider the ball to be thrown in the xz-plane rather than the xy-plane or the yz-plane?
+
+
+
+
+
+
+
+
+ In order to fly from Oklahoma City to Dayton, OH, a plane must fly 64^{\circ} east of north. A plane leaves Oklahoma City at a ground speed of 525 miles per hour. A 30 mph wind blows from the north during the flight.
+
+
+
+
+
+ What is the ground speed of the plane when the wind begins to blow?
+
+
+
+
+
+
+ What is the new bearing of the plane?
+
+
+
+
+
+
+
+
+ An object is said to be in static equilibrium if the resultant force of the forces that act on it is zero. Let \textbf{F}_1 = \langle10,6,3\rangle, \textbf{F}_2 = \langle0,4,9\rangle, and \textbf{F}_3 = \langle 10,-3,-9\rangle be three forces acting on an object. Find the resultant force necessary to place the object in static equilibrium.
+
+
+
+
+
\ No newline at end of file
diff --git a/source/calculus/source/10-VE/main.ptx b/source/calculus/source/10-VE/main.ptx
new file mode 100644
index 000000000..96e002a30
--- /dev/null
+++ b/source/calculus/source/10-VE/main.ptx
@@ -0,0 +1,8 @@
+
+
+
+ Vectors (VE)
+
+
+
+
\ No newline at end of file
diff --git a/source/calculus/source/10-VE/outcomes/01.ptx b/source/calculus/source/10-VE/outcomes/01.ptx
new file mode 100644
index 000000000..c833541bc
--- /dev/null
+++ b/source/calculus/source/10-VE/outcomes/01.ptx
@@ -0,0 +1,4 @@
+
+
+Identify, create, combine, modify, and apply vectors in two- or three-dimensional space
+
\ No newline at end of file
diff --git a/source/calculus/source/10-VE/outcomes/02.ptx b/source/calculus/source/10-VE/outcomes/02.ptx
new file mode 100644
index 000000000..bab0e056b
--- /dev/null
+++ b/source/calculus/source/10-VE/outcomes/02.ptx
@@ -0,0 +1,4 @@
+
+
+Compute and apply the dot product and cross product of vectors.
+
\ No newline at end of file
diff --git a/source/calculus/source/10-VE/outcomes/03.ptx b/source/calculus/source/10-VE/outcomes/03.ptx
new file mode 100644
index 000000000..d2e42fefb
--- /dev/null
+++ b/source/calculus/source/10-VE/outcomes/03.ptx
@@ -0,0 +1,4 @@
+
+
+ Determine information about lines and planes in space, including: distance, angles between objects, and equations in various forms.
+
\ No newline at end of file
diff --git a/source/calculus/source/10-VE/outcomes/main.ptx b/source/calculus/source/10-VE/outcomes/main.ptx
new file mode 100644
index 000000000..e1898e375
--- /dev/null
+++ b/source/calculus/source/10-VE/outcomes/main.ptx
@@ -0,0 +1,23 @@
+
+>
+
+
+What are vectors and how can we use them?
+
+
+By the end of this chapter, you should be able to...
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/source/calculus/source/10-VE/readiness.ptx b/source/calculus/source/10-VE/readiness.ptx
new file mode 100644
index 000000000..c0474b775
--- /dev/null
+++ b/source/calculus/source/10-VE/readiness.ptx
@@ -0,0 +1,106 @@
+
+
+
+
+ Readiness Assurance
+
+ Before beginning this chapter, you should be able to...
+
+
+
+
Use function notation and evaluate functions
+
+
+
Review: Khan Academy
+
+
+
Practice:
+
+
Evaluate functions
+
Evaluate functions from their graphs
+
Function notation word problems
+
+
+
+
+
+
+
Find the domain of a function
+
+
+
+
Review: Khan Academy
+
+
+
Practice: Determine the domain of functions
+
+
+
+
+
Determine vertical asymptotes, horizontal asymptotes, and holes (removable discontinuities) of rational functions
+
+
+
Review:
+
+
Discontinuities of rational functions
+
Finding horizonal asymptotes
+
+
+
+
Practice:
+
+
Rational functions: zeros, asymptotes, and undefined points
+
Finding horizonal asymptotes
+
+
+
+
+
+
+
Perform basic operations with polynomials
+
+
+
Review:
+
+
Adding and subtracting polynomials
+
Multiplying polynomials
+
+
+
+
+
Practice:
+
+
Add polynomials
+
Subtract polynomials
+
Multiply binomials by polynomials
+
+
+
+
+
+
+
Factor quadratic expressions
+
+
+
Review: Khan Academy
+
+
+
Practice: Factoring quadratics intro
+
+
+
+
+
+
+
Represent intervals using number lines, inequalities, and interval notation
+
+
+
+
Review: Varsity Tutor
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/source/calculus/source/main.ptx b/source/calculus/source/main.ptx
index 96e3e13e9..3a335c8d4 100644
--- a/source/calculus/source/main.ptx
+++ b/source/calculus/source/main.ptx
@@ -27,6 +27,8 @@
+
+
diff --git a/source/calculus/source/meta/frontmatter.ptx b/source/calculus/source/meta/frontmatter.ptx
index 1194c7570..e31fb87d2 100644
--- a/source/calculus/source/meta/frontmatter.ptx
+++ b/source/calculus/source/meta/frontmatter.ptx
@@ -4,7 +4,7 @@
-
+ TBIL Communitysclontz+tbil@southalabama.edu
@@ -67,12 +67,15 @@
-
-
-
-
+
+
+
+
+
+
+ Attribution and Acknowledgements
This work includes materials used under license from the following works:
@@ -90,7 +93,7 @@ This work includes materials used under license from the following works:
-
+
diff --git a/source/common/authors/clontz.ptx b/source/common/authors/clontz.ptx
index a93335a5e..37304054c 100644
--- a/source/common/authors/clontz.ptx
+++ b/source/common/authors/clontz.ptx
@@ -1,5 +1,5 @@
-
+Steven ClontzDepartment of Mathematics and StatisticsUniversity of South Alabama
diff --git a/source/common/authors/lewis.ptx b/source/common/authors/lewis.ptx
index 337f87e23..86cd5eaff 100644
--- a/source/common/authors/lewis.ptx
+++ b/source/common/authors/lewis.ptx
@@ -1,5 +1,5 @@
-
+Drew Lewisdrew.lewis@gmail.com
diff --git a/source/common/copyright.ptx b/source/common/copyright.ptx
index a871d6ac8..58111799f 100644
--- a/source/common/copyright.ptx
+++ b/source/common/copyright.ptx
@@ -1,6 +1,6 @@
-
+20212024Steven Clontz and Drew Lewis
diff --git a/source/common/resource-library.ptx b/source/common/resource-library.ptx
index 8e5874021..7f293ad17 100644
--- a/source/common/resource-library.ptx
+++ b/source/common/resource-library.ptx
@@ -1,4 +1,4 @@
-
+TBIL Resource Library
This book is a part of the
TBIL Resource Library, which includes a growing
@@ -10,4 +10,4 @@
Any opinions, findings, and conclusions or recommendations expressed in this material
are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
- For each of the following matrices, explain why it is not in reduced row
- echelon form.
+ For each of the following matrices, mark the entries
+ where pivots should appear, and either label the matrix as
+ in reduced row echelon form, or explain what should be done
+ to put the matrix in reduced row echelon form and why.
-
+
{{label}} = {{matrix}}
- {{label}}={{matrix}} is not in reduced row echelon form because
-
- there is a non-zero entry above or below a pivot.
-
-
- the pivots are not descending to the right.
-
-
- a leading term has a value besides 1.
-
+
+ Is in reduced row echelon form
+
+
+ Not in reduced row echelon form because
+ there is a non-zero entry above or below a pivot, which may be fixed
+ by {{rowop}}.
+
+
+ Not in reduced row echelon form because
+ the pivots are not descending to the right, which may be fixed
+ by {{rowop}}.
+
+
+ Not in reduced row echelon form because
+ a leading term has a value besides 1, which may be fixed
+ by {{rowop}}.
+
-
+
Use technology to compute
\operatorname{RREF}{{A}}.
-
{{rref}}
+
{{Arref}}
Show step by step how to compute
\operatorname{RREF}{{A}} without technology.
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/01-LE/02.ptx b/source/linear-algebra/source/01-LE/02.ptx
index cf5ef39ff..226e67d27 100644
--- a/source/linear-algebra/source/01-LE/02.ptx
+++ b/source/linear-algebra/source/01-LE/02.ptx
@@ -644,64 +644,96 @@ in many applications we use it for in .
-
-
+
+
+
+
+ Instructor Note
+
+ In the following 4-part activity, students make decisions about which row operations are most helpful for them in producing the RREF.
+ The matrices appearing in parts (b) and (c) are obtained by applying specific choices of row operations to the matrix appearing previously; the matrix in part (d) is the one that we started with.
+
-Consider the matrix
-\left[\begin{array}{cccc}2 & 6 & -1 & 6 \\ 1 & 3 & -1 & 2 \\ -1 & -3 & 2 & 0 \end{array}\right].
-Which row operation is the best choice for the first move in converting to RREF?
+ Our experience in facilitating these activities is that student reasoning in part (a) is largely grounded in the production of zeroes, no matter how they appear; because of this, all of the choices in part (a) may appear to be equally valid to them.
+ In contrast, choices made in parts (b) and (c) are more constrained as our students recognize the utility of using the marked pivot intentionally to clear entries above and below.
+ By revisiting the original matrix in part (d) after marking the pivot position, the instructor can highlight how choices are more intentionally guided by the pivots.
+ In turn, this helps students develop a more strategic approach, which is codified in the observation following the activity.
+
+
+
+
+
+
+ Consider the matrix:
+ \left[\begin{array}{cccc}2 & 6 & -1 & 6 \\ 1 & 3 & -1 & 2 \\ -1 & -3 & 2 & 0 \end{array}\right].
+ Which row operation is the best choice for the first move in converting to RREF?
+
+ Consider the matrix:
+ \left[\begin{array}{cccc} \markedPivot{1} & 3 & -1 & 2 \\ 2 & 6 & -1 & 6 \\ -1 & -3 & 2 & 0 \end{array}\right].
+ Which row operation is the best choice for the next move in converting to RREF?
-Consider the matrix
-\left[\begin{array}{cccc} \markedPivot{1} & 3 & -1 & 2 \\ 2 & 6 & -1 & 6 \\ -1 & -3 & 2 & 0 \end{array}\right].
-Which row operation is the best choice for the next move in converting to RREF?
+ Consider the matrix:
+ \left[\begin{array}{cccc}\markedPivot{1} & 3 & -1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right].
+ Which row operation is the best choice for the next move in converting to RREF?
-Consider the matrix
-\left[\begin{array}{cccc}\markedPivot{1} & 3 & -1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right].
-Which row operation is the best choice for the next move in converting to RREF?
+ Consider the matrix:
+ \left[\begin{array}{cccc}2 & 6 & -1 & 6 \\ 1 & 3 & -1 & 2 \\ -1 & -3 & 2 & 0 \end{array}\right].
+ Mark the position where the first pivot should be.
+ Which row operation is the best choice for the first move in converting to RREF?
+
+
@@ -792,66 +824,59 @@ Can \RREF(A) be used to find \RREF(B)?
-
+
Free browser-based technologies for mathematical computation
are available online.
-
-
- Go to .
-
-
-
- In the dropdown on the right, you can select a number of different languages.
- Select "Octave" for the Matlab-compatible syntax used by this text.
-
-
-
Type rref([1,3,2;2,5,7]) and then press the Evaluate button to compute the \RREF of
- \left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right].
-
-
-
-
+
+
+
+Go to .
+
+
+
+
+In the dropdown on the right, you can select a number of different languages.
+Select "Octave" for the Matlab-compatible syntax used by this text.
+
+
+
+
+Type rref([1,4,6;2,5,7]) and then press the Evaluate button to compute the \RREF of
+\left[\begin{array}{ccc} 1 & 4 & 6 \\ 2 & 5 & 7 \end{array}\right].
+
+
+
+
Now try using whitespace to write out the matrix and compute \RREF instead:
-
diff --git a/source/linear-algebra/source/01-LE/03.ptx b/source/linear-algebra/source/01-LE/03.ptx
index 49270f4dc..e0bbcb1c8 100644
--- a/source/linear-algebra/source/01-LE/03.ptx
+++ b/source/linear-algebra/source/01-LE/03.ptx
@@ -454,7 +454,7 @@ different solutions. We'll learn how to find such solution sets in
-
Exercises available at .
+
Exercises available at .
diff --git a/source/linear-algebra/source/01-LE/04.ptx b/source/linear-algebra/source/01-LE/04.ptx
index 52cb12b4d..587aebe0d 100644
--- a/source/linear-algebra/source/01-LE/04.ptx
+++ b/source/linear-algebra/source/01-LE/04.ptx
@@ -432,7 +432,7 @@ a
-
Exercises available at .
+
Exercises available at .
diff --git a/source/linear-algebra/source/02-EV/01.ptx b/source/linear-algebra/source/02-EV/01.ptx
index 3d02d22d4..ddffb3d6d 100644
--- a/source/linear-algebra/source/02-EV/01.ptx
+++ b/source/linear-algebra/source/02-EV/01.ptx
@@ -509,7 +509,7 @@ in \IR^3.
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/02-EV/02.ptx b/source/linear-algebra/source/02-EV/02.ptx
index 19e982773..1303f8821 100644
--- a/source/linear-algebra/source/02-EV/02.ptx
+++ b/source/linear-algebra/source/02-EV/02.ptx
@@ -471,7 +471,7 @@ Given no additional information about the vectors \vec{v}_1,\dots, \vec{v}_m<
-
Exercises available at .
+
Exercises available at .
diff --git a/source/linear-algebra/source/02-EV/03.ptx b/source/linear-algebra/source/02-EV/03.ptx
index aaf957664..f5c746e9b 100644
--- a/source/linear-algebra/source/02-EV/03.ptx
+++ b/source/linear-algebra/source/02-EV/03.ptx
@@ -808,8 +808,10 @@ Let W be a subspace of a vector space V. How are \vspan WIndividual Practice
+
Recall that in we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals.
In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
+
@@ -892,7 +894,7 @@ Let W be a subspace of a vector space V. How are \vspan W
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/02-EV/04.ptx b/source/linear-algebra/source/02-EV/04.ptx
index 33754f55c..12100dff6 100644
--- a/source/linear-algebra/source/02-EV/04.ptx
+++ b/source/linear-algebra/source/02-EV/04.ptx
@@ -489,8 +489,10 @@ vectors that can form a linearly independent set?
Individual Practice
+
Recall that in we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals.
In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
+
@@ -548,7 +550,7 @@ vectors that can form a linearly independent set?
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/02-EV/05.ptx b/source/linear-algebra/source/02-EV/05.ptx
index 29a249868..636f68584 100644
--- a/source/linear-algebra/source/02-EV/05.ptx
+++ b/source/linear-algebra/source/02-EV/05.ptx
@@ -13,8 +13,10 @@
Warm Up
- Recall that in we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals.
- In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
+
+Recall that in we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals.
+In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
+
@@ -243,8 +245,9 @@ In particular, the standard basis for \mathbb R^3 is \{\vec e_1,\vec e
- Take the RREF of an appropriate matrix to determine
+ Use technology to find the RREF of an appropriate matrix and determine
if each of the following sets is a basis for \IR^4.
+ (Don't forget to use format rat to nicely format Octave's output.)
@@ -545,7 +548,7 @@ Not a basis, because not only does it fail to span \IR^4, it's also linea
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/02-EV/06.ptx b/source/linear-algebra/source/02-EV/06.ptx
index f626848cd..62cead82e 100644
--- a/source/linear-algebra/source/02-EV/06.ptx
+++ b/source/linear-algebra/source/02-EV/06.ptx
@@ -442,7 +442,7 @@ Thus the basis for a subspace is not unique in general.
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/02-EV/07.ptx b/source/linear-algebra/source/02-EV/07.ptx
index 1d58746f0..68b95d2a1 100644
--- a/source/linear-algebra/source/02-EV/07.ptx
+++ b/source/linear-algebra/source/02-EV/07.ptx
@@ -127,15 +127,81 @@ Rewrite this solution space in the form \setBuilder{ a \left[\begin{array}{c
+
-Rewrite this solution space in the form \vspan\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}.
+Which of these choices best describes the set of two vectors
+\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}
+used in this solution space?
+
+
+
+ The set is linearly dependent.
+
+
+
+
+ The set is linearly independent.
+
+
+
+
+ The set spans the solution space.
+
+
+
+
+ The set is a basis of the solution space.
+
+
+
+
+
+
+
+ D.
+
+
+
+
+
+
+
+
+
+
+
+
+
+Consider the homogeneous system of equations
+
+
+2x_1&\,+\,&4x_2&\,+\,&2x_3 &\,-\,&3 x_4 &\,+\,&31x_5&\,+\,&2x_6&\,-\,&16x_7&=& 0
+
+
+-1x_1&\,-\,&2x_2&\,+\,&4x_3 &\,-\,&x_4 &\,+\,&2x_5&\,+\,&9x_6&\,+\,&3x_7&=& 0
+
+
+x_1&\,+\,&2x_2&\,+\,&x_3 &\,+\,& x_4 &\,+\,&3x_5&\,+\,&6x_7&\,+\,&7x_7&=& 0
+
+
+
+
+
+
+Find its solution set (a subspace of \IR^7).
+Rewrite this solution space in the form \setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+c \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+d \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] }{a,b,c,d \in \IR}.
+
+
+
+
+
Which of these choices best describes the set of two vectors
-\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}
-used in this span?
+\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]\right\}
+used in this solution space?
@@ -149,22 +215,32 @@ used in this span?
- The set spans all of \IR^4.
+ The set spans the solution space.
-
+
- The set fails to span the solution space.
+ The set is a basis for the solution space.
+
+
@@ -298,7 +338,7 @@ at the point (x,y,z)\in\mathbb R^3 onto the pixel located at
-What homoegeneous linear system describes the positions (x,y,z)
+What homogeneous linear system describes the positions (x,y,z)
within the original scene that would be aligned with the
pixel (0,0) on the screen?
@@ -387,7 +427,7 @@ Solve this system to describe these locations.
-
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/03-AT/02.ptx b/source/linear-algebra/source/03-AT/02.ptx
index c313a069b..1e26ed98c 100644
--- a/source/linear-algebra/source/03-AT/02.ptx
+++ b/source/linear-algebra/source/03-AT/02.ptx
@@ -557,7 +557,7 @@ Write the standard matrix [T(\vec e_1) \,\cdots\, T(\vec e_n)] for T
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
@@ -568,7 +568,7 @@ Write the standard matrix [T(\vec e_1) \,\cdots\, T(\vec e_n)] for T
- For each of the following properties, determine if it is held by the dot product. Either provide a proof it the property holds, or provide a counter-example if it does not.
+ For each of the following properties, determine if it is held by the dot product. Either provide a proof that the property holds, or provide a counter-example if it does not.
diff --git a/source/linear-algebra/source/03-AT/03.ptx b/source/linear-algebra/source/03-AT/03.ptx
index 7d0fd5e31..32fe7e3a5 100644
--- a/source/linear-algebra/source/03-AT/03.ptx
+++ b/source/linear-algebra/source/03-AT/03.ptx
@@ -641,7 +641,7 @@ The dimension of the image is called the rank of T (or A<
- What picture, or other study strategy would be helpful to you in conceptualizing how these defintions fit together?
+ What picture, or other study strategy would be helpful to you in conceptualizing how these definitions fit together?
@@ -654,7 +654,8 @@ The dimension of the image is called the rank of T (or A<
-
Video: The kernel and image of a linear transformation
+
Video: The kernel and image of a linear transformation.
+ Note that there is a typo: if you're following along, you should find that T\left(\left[\begin{array}{c}2\\1\\3\\0\end{array}\right]\right)=\left[\begin{array}{c}14\\-5\\9\end{array}\right].
diff --git a/source/linear-algebra/source/03-AT/05.ptx b/source/linear-algebra/source/03-AT/05.ptx
index d0346d8ea..d42d4f794 100644
--- a/source/linear-algebra/source/03-AT/05.ptx
+++ b/source/linear-algebra/source/03-AT/05.ptx
@@ -70,7 +70,7 @@ Consider the following applications of properties of the real numbers
-Which of the following properites of \IR^2 Euclidean vectors is NOT true?
+Which of the following properties of \IR^2 Euclidean vectors is NOT true?
\left[\begin{array}{c} x_1\\x_2\end{array}\right]
@@ -123,7 +123,7 @@ Which of the following properites of \IR^2 Euclidean vectors is NOT true?
- Consider the following applications of properites of the real numbers
+ Consider the following applications of properties of the real numbers
\mathbb R:
@@ -159,7 +159,7 @@ Which of the following properites of \IR^2 Euclidean vectors is NOT true?
-Which of the following properites of \IR^2 Euclidean vectors is NOT true?
+Which of the following properties of \IR^2 Euclidean vectors is NOT true?
@@ -337,7 +337,7 @@ The space of m \times nmatrices
Consider the
- set \IC of complex numbers with the usual defintion for addition:
+ set \IC of complex numbers with the usual definition for addition:
(a+b\mathbf i)\oplus(c+d\mathbf i)=(a+c)+(b+d)\mathbf i.
Let
@@ -687,14 +687,14 @@ for allc\in \IR,\, (x_1,y_1),(x_2,y_2) \in V.
What are some objects that are important to you personally, academically, or otherwise that appear vector-like to you?
- What makes them feel vector-like? Which axiom for vector spaces does not hold for these objects, if any.
+ What makes them feel vector-like? Which axiom for vector spaces does not hold for these objects, if any?
- Our vector space axioms have eight properties. While these eight properties are enough to capture vectors, the objects that we study in the real-world often have additional structures not captured by these axioms.
+ Our vector space axioms have eight properties. While these eight properties are enough to capture vectors, the objects that we study in the real world often have additional structures not captured by these axioms.
What are some structures that you have encountered in other classes, or in previous experiences, that are not captured by these eight axioms?
@@ -721,7 +721,7 @@ for allc\in \IR,\, (x_1,y_1),(x_2,y_2) \in V.
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/03-AT/06.ptx b/source/linear-algebra/source/03-AT/06.ptx
index e6da3f68f..aa1cc50ae 100644
--- a/source/linear-algebra/source/03-AT/06.ptx
+++ b/source/linear-algebra/source/03-AT/06.ptx
@@ -448,12 +448,12 @@ as a linear combination of polynomials from the set
-
Describe the vector space involved in this problem, and an isomorphic Euclidean space and relevant Eucldean vectors that can be used to solve this problem.
+
Describe the vector space involved in this problem, and an isomorphic Euclidean space, and relevant Euclidean vectors that can be used to solve this problem.
-
Show how to construct an appropriate Euclidean vector from an approriate set of Euclidean vectors.
+
Show how to construct an appropriate Euclidean vector from an appropriate set of Euclidean vectors.
@@ -545,7 +545,7 @@ as a linear combination of polynomials from the set
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/04-MX/01.ptx b/source/linear-algebra/source/04-MX/01.ptx
index f8306cd23..3e0c6ce25 100644
--- a/source/linear-algebra/source/04-MX/01.ptx
+++ b/source/linear-algebra/source/04-MX/01.ptx
@@ -330,7 +330,7 @@ in terms of matrix multiplication.
-
diff --git a/source/linear-algebra/source/04-MX/04.ptx b/source/linear-algebra/source/04-MX/04.ptx
index 39d4cc67a..877059d88 100644
--- a/source/linear-algebra/source/04-MX/04.ptx
+++ b/source/linear-algebra/source/04-MX/04.ptx
@@ -364,7 +364,7 @@ If you and a teammate were to do this independently, would you necessarily come
-
Exercises available at .
+
Exercises available at .
diff --git a/source/linear-algebra/source/05-GT/01.ptx b/source/linear-algebra/source/05-GT/01.ptx
index 8529f77fe..8f7c4459c 100644
--- a/source/linear-algebra/source/05-GT/01.ptx
+++ b/source/linear-algebra/source/05-GT/01.ptx
@@ -1042,7 +1042,7 @@ Or we may use a formula:
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/05-GT/02.ptx b/source/linear-algebra/source/05-GT/02.ptx
index f04fe517f..64ca9a869 100644
--- a/source/linear-algebra/source/05-GT/02.ptx
+++ b/source/linear-algebra/source/05-GT/02.ptx
@@ -452,7 +452,7 @@ Based on the previous activities, which technique is easier for computing determ
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/05-GT/03.ptx b/source/linear-algebra/source/05-GT/03.ptx
index 7cd53d3fe..071f88c6b 100644
--- a/source/linear-algebra/source/05-GT/03.ptx
+++ b/source/linear-algebra/source/05-GT/03.ptx
@@ -359,7 +359,7 @@ Set this characteristic polynomial equal to zero and factor to determine the eig
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/05-GT/04.ptx b/source/linear-algebra/source/05-GT/04.ptx
index 98e2f0f1a..c7aa37066 100644
--- a/source/linear-algebra/source/05-GT/04.ptx
+++ b/source/linear-algebra/source/05-GT/04.ptx
@@ -160,7 +160,7 @@ associated with the eigenvalue 2.
-
Exercises available at .
+
Exercises available at .
Mathematical Writing Explorations
diff --git a/source/linear-algebra/source/05-GT/05.ptx b/source/linear-algebra/source/05-GT/05.ptx
new file mode 100644
index 000000000..b2a39f227
--- /dev/null
+++ b/source/linear-algebra/source/05-GT/05.ptx
@@ -0,0 +1,238 @@
+
+
+ Change of Basis (GT5)
+
+
+
+
+ Calculate the change-of-basis matrix between two bases of \IR^n.
+
+
+
+
+
+
+ Warm Up
+
+
+
+Class Activities
+
+
+ So far, when working with the Euclidean vector space \IR^n, we have primarily worked with the standard basis \mathcal{E}=\setList{\vec{e}_1,\dots, \vec{e}_n}.
+ We can explore alternative perspectives more easily if we expand our toolkit to analyze different bases.
+
+
+
+
+
+ Let \mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}=\setList{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}1\\-1\\1\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}}.
+
+
+
+
+ Is \cal{B} a basis of \IR^3?
+
+
+
+
+
+ Yes.
+
+
+
+
+ No.
+
+
+
+
+
+
+
+ Since \cal{B} is a basis, we know that if \vec{v}\in \IR^3, the following vector equation will have a unique solution:
+ x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\vec{v}
+ Given this, we define a map C_{\mathcal{B}}\colon\IR^3\to\IR^3 via the rule that C_{\mathcal{B}}(\vec{v}) is equal to the unique solution to the above vector equation.
+ The map C_{\mathcal{B}} is a linear map.
+
+ Compute C_\mathcal{B}(\vec{e}_1),C_\mathcal{B}(\vec{e}_2), C_\mathcal{B}(\vec{e}_3) and, in doing so, write down the standard matrix M_\mathcal{B} of C_\mathcal{B}.
+
+
+
+
+
+
+
+
+
+
+
+
+
+ Note that one way to compute M_{\mathcal{B}} is calculate the RREF of the following matrix:
+ \left[\begin{array}{ccc|ccc}1&1&0&1&0&0\\
+ 0&-1&1&0&1&0\\
+ 1&1&1&0&0&1\end{array}\right]\sim\left[\begin{array}{ccc|ccc}1&0&0&2&1&-1\\
+ 0&1&0&-1&-1&1\\
+ 0&0&1&-1&0&1\end{array}\right]
+ Thus, the matrix M_{\mathcal{B}} is the inverse of the matrix [\vec{v}_1\ \vec{v}_2\ \vec{v}_3].
+ That is:
+ M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \vec{v}_2\ \vec{v}_3].
+
+
+
+
+
+ Given a basis \cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n} of \IR^n, the change of basis/coordinate transformation from the standard basis to\mathcal{B} is the transformation C_\mathcal{B}\colon\IR^n\to\IR^n defined by the property that, for any vector \vec{v}\in\IR^n, the vector C_\mathcal{B}(\vec{v}) is the unique solution to the vector equation:
+ x_1\vec{v}_1+\dots+x_n\vec{v}_n=\vec{v}.
+ Its standard matrix is called the change-of-basis matrix from the standard basis to \mathcal{B} and is denoted by M_{\mathcal{B}}.
+ It satisfies the following:
+ M_{\mathcal{B}}^{-1}=[\vec{v}_1\ \cdots\ \vec{v}_n].
+
+
+
+
+
+
+ The vector C_\mathcal{B}(\vec{v}) is the \mathcal{B}-coordinates of \vec{v}.
+ If you work with standard coordinates, and I work with \mathcal{B}-coordinates, then to build the vector that you call \vec{v}=\begin{bmatrix}a_1\\\vdots\\a_n\end{bmatrix}=a_1\vec{e}_1+\cdots+a_n\vec{e}_n, I would first compute C_\mathcal{B}(\vec{v})=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} and then build \vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n.
+
+
+ In particular, notation as above, we would have:
+ a_1\vec{e}_1+\cdots a_n\vec{e}_n=\vec{v}=x_1\vec{v}_1+\cdots+x_n\vec{v}_n.
+
+
+
+
+
+
+
+ Let \vec{v}_1=\begin{bmatrix}1\\-2\\1\end{bmatrix},\ \vec{v}_2=\begin{bmatrix}-1\\0\\3\end{bmatrix},\ \vec{v}_3=\begin{bmatrix}0\\1\\-1\end{bmatrix}, and \mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}
+
+
+
+
+ Calculate M_{\mathcal{B}} using technology.
+
+
+
+
+
+
+
+
+
+
+ Use your result to calculate C_\mathcal{B}\left(\begin{bmatrix}1\\1\\1\end{bmatrix}\right) and express the vector \begin{bmatrix}1\\1\\1\end{bmatrix} as a linear combination of \vec{v}_1,\vec{v}_2,\vec{v}_3.
+
+
+
+
+
+ Let T\colon\IR^n\to\IR^n be a linear transformation and let A denote its standard matrix.
+ If \cal{B}=\setList{\vec{v}_1,\dots, \vec{v}_n} is some other basis, then we have:
+
+ M_\mathcal{B}AM_{\mathcal{B}}^{-1} \amp= M_\mathcal{B}A[\vec{v_1}\cdots\vec{v}_n]
+ \amp= M_\mathcal{B}[T(\vec{v}_1)\cdots T(\vec{v}_n)]
+ \amp= [C_\mathcal{B}(T(\vec{v}_1))\cdots C_\mathcal{B}(T(\vec{v}_n))]
+
+ In other words, the matrix M_{\mathcal{B}}AM_{\mathcal{B}}^{-1} is the matrix whose columns consist of \mathcal{B}-coordinate vectors of the image vectors T(\vec{v}_i).
+ The matrix M_{\mathcal{B}}AM_{\mathcal{B}}^{-1} is called the matrix of T with respect to \mathcal{B}-coordinates.
+
+
+
+
+
+
+ Let \mathcal{B}=\setList{\vec{v}_1,\vec{v}_2,\vec{v}_3}=\setList{\begin{bmatrix}1\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\0\\3\end{bmatrix},\begin{bmatrix}0\\1\\-1\end{bmatrix}} be basis from the previous Activity.
+ Let T denote the linear transformation whose standard matrix is given by:
+ A=\begin{bmatrix}9&4&4\\6&9&2\\-18&-16&-9\end{bmatrix}.
+
+
+
+
+ Calculate the matrix M_\mathcal{B}AM_{\mathcal{B}}^{-1}.
+
+
+
+
+
+
+
+
+
+
+ The matrix A describes how T transforms the standard basis of \IR^3.
+ The matrix M_\mathcal{B}AM_{\mathcal{B}}^{-1} describes how T transforms the basis \mathcal{B} (in \mathcal{B}-coordinates).
+
+
+ Which of these two descriptions of T is most helpful to you in describing/understanding/visualizing the transformation T and why?
+
+
+
+
+
+
+
+
+ Sample Problem and Solution
+
+
+
+ Let \mathcal{B}=\setList{\begin{bmatrix}-2\\-2\\1\end{bmatrix},\begin{bmatrix}-1\\-2\\-1\end{bmatrix},\begin{bmatrix}1\\3\\2\end{bmatrix}}, and \vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}.
+
+
+
+ Explain and demonstrate how to verify that \mathcal{B} is a basis of \IR^3 and how to calculate M_\mathcal{B}, the change-of-basis matrix from the standard basis of \IR^3 to \mathcal{B}.
+
+
+
+
+ Explain and demonstrate how to use M_\mathcal{B} to express \vec{v} in terms of \mathcal{B}-basis vectors.
+
+
+
+
+
+
+ We can accomplish both tasks by calculating the RREF of the following matrix:
+ \RREF\left[\begin{array}{ccc|ccc}-2&-1&1&1&0&0\\
+ -2&-2&3&0&1&0\\1&-1&2&0&0&1\end{array}\right]
+ =
+ \left[\begin{array}{ccc|ccc}1&0&0&1&-1&1\\
+ 0&1&0&-7&5&-4\\
+ 0&0&1&-4& 3&-2\end{array}\right].
+ The fact that the matrix to the left of the vertical bar is the identity matrix tells that \mathcal{B} is a basis.
+ The matrix on the right hand side of the bar is equal to the change-of-basis matrix:
+ M_\mathcal{B}=\left[\begin{array}{ccc}1&-1&1\\
+ -7&5&-4\\
+ -4& 3&-2\end{array}\right].
+
+
+
+
+ By definition of the change of basis matrix, if \vec{v}=\begin{bmatrix}1\\2\\3\end{bmatrix}, then the coordinates of \vec{v} with respect to \mathcal{B} are given by:
+ M_\mathcal{B}\vec{v}=M_\mathcal{B}=\left[\begin{array}{ccc}1&-1&1\\
+ -7&5&-4\\
+ -4& 3&-2\end{array}\right]\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}2\\-9\\-4\end{bmatrix}.
+ It follows that:
+ \begin{bmatrix}1\\2\\3\end{bmatrix}=2\begin{bmatrix}-2\\-2\\1\end{bmatrix} -9\begin{bmatrix}-1\\-2\\-1\end{bmatrix} -4\begin{bmatrix}1\\3\\2\end{bmatrix}.
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/source/linear-algebra/source/05-GT/main.ptx b/source/linear-algebra/source/05-GT/main.ptx
index 4cf21668a..299537a5f 100644
--- a/source/linear-algebra/source/05-GT/main.ptx
+++ b/source/linear-algebra/source/05-GT/main.ptx
@@ -7,5 +7,5 @@
-
+
diff --git a/source/linear-algebra/source/applications/pagerank.ptx b/source/linear-algebra/source/applications/pagerank.ptx
index d6865fdeb..d4262fe8a 100644
--- a/source/linear-algebra/source/applications/pagerank.ptx
+++ b/source/linear-algebra/source/applications/pagerank.ptx
@@ -6,7 +6,7 @@
-The $978,000,000,000 Problem
+The $2,110,000,000,000 Problem
@@ -58,7 +58,7 @@ most important to least important.
-The $978,000,000,000 Idea
+The $2,110,000,000,000 Idea
@@ -187,7 +187,7 @@ the matrix equation A\vec{x}=1\vec{x}.
-Thus, our $978,000,000,000 problem is what kind of problem?
+Thus, our $2,110,000,000,000 problem is what kind of problem?
\left[\begin{array}{ccc}0&1&0\\\frac{1}{2}&0&\frac{1}{2}\\\frac{1}{2}&0&0\end{array}\right]
\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]
@@ -395,7 +395,7 @@ it's reasonable to consider page 2 as the most important page.
-Based upon this page rank vector,
+Based on this page rank vector,
here is a complete ranking of all seven pages from most important to least important:
2, 4, 1, 3, 7, 5, 6
diff --git a/source/linear-algebra/source/applications/truss.ptx b/source/linear-algebra/source/applications/truss.ptx
index 65a6692d5..defcb3e50 100644
--- a/source/linear-algebra/source/applications/truss.ptx
+++ b/source/linear-algebra/source/applications/truss.ptx
@@ -82,7 +82,7 @@ determine many of the forces at play.
-For example, at the bottom left node there are 3 forces acting.
+For example, at the bottom left node, 3 forces are acting.
@@ -287,7 +287,7 @@ one variable may be used to represent each.
-Since the angle of the normal forces for each anchor point are unknown,
+Since the angle of the normal forces for each anchor point is unknown,
two variables may be used to represent each.
The full augmented matrix given by the ten equations in this linear system
-is given below, where the eleven columns correspond to x_1,\dots,x_7,y_1,y_2,z_1,z_2,
+is shown below, where the eleven columns correspond to x_1,\dots,x_7,y_1,y_2,z_1,z_2,
and the ten rows correspond to the horizontal and vertical components of the
forces acting at A,\dots,E.
@@ -455,7 +455,7 @@ y_2=z_2&=5000
In particular, the negative x_1,x_2,x_5 represent tension (forces pointing into the nodes),
-and the postive x_3,x_4 represent compression (forces pointing out of the nodes).
+and the positive x_3,x_4 represent compression (forces pointing out of the nodes).
The vertical normal forces y_2+z_2 counteract the 10000 load.
diff --git a/source/linear-algebra/source/future-ON/02.ptx b/source/linear-algebra/source/future-ON/02.ptx.bak
similarity index 100%
rename from source/linear-algebra/source/future-ON/02.ptx
rename to source/linear-algebra/source/future-ON/02.ptx.bak
diff --git a/source/linear-algebra/source/future-ON/03.ptx b/source/linear-algebra/source/future-ON/03.ptx.bak
similarity index 100%
rename from source/linear-algebra/source/future-ON/03.ptx
rename to source/linear-algebra/source/future-ON/03.ptx.bak
diff --git a/source/linear-algebra/source/future-ON/05.ptx b/source/linear-algebra/source/future-ON/05.ptx.bak
similarity index 100%
rename from source/linear-algebra/source/future-ON/05.ptx
rename to source/linear-algebra/source/future-ON/05.ptx.bak
diff --git a/source/linear-algebra/source/meta/frontmatter.ptx b/source/linear-algebra/source/meta/frontmatter.ptx
index 3952ca371..67d93ad7d 100644
--- a/source/linear-algebra/source/meta/frontmatter.ptx
+++ b/source/linear-algebra/source/meta/frontmatter.ptx
@@ -1,6 +1,6 @@
-
+
@@ -43,11 +43,15 @@
-
-
-
+
+
+
+
+
+
+ Attribution and Acknowledgements
This work includes materials used under license from the following works:
@@ -76,7 +80,7 @@ This work includes materials used under license from the following works:
-
+
diff --git a/source/precalculus/exercises/outcomes/EL/EL4/generator.sage b/source/precalculus/exercises/outcomes/EL/EL4/generator.sage
index 264551a88..5dc27581b 100644
--- a/source/precalculus/exercises/outcomes/EL/EL4/generator.sage
+++ b/source/precalculus/exercises/outcomes/EL/EL4/generator.sage
@@ -1,3 +1,4 @@
+from sage.symbolic.function_factory import function_factory
class Generator(BaseGenerator):
def data(self):
a=1
@@ -22,10 +23,8 @@ class Generator(BaseGenerator):
d = choice(range(1,6))*choice([-1,1])
f=a*log(k*x-c)/log(b)+d
- if a == 1:
- f_string = f"\\log _{{ {b} }} \\left({k*x-c}\\right)+{d}"
- else:
- f_string = f"-\\log _{{ {b} }} \\left({k*x-c}\\right)+{d}"
+ logb = function_factory("logb", 1, f"\\log_{{ {b} }}")
+ f_string = latex(a*logb(k*x-c)+d)
return {
"f": f,
diff --git a/source/precalculus/exercises/outcomes/FN/FN1/generator.sage b/source/precalculus/exercises/outcomes/FN/FN1/generator.sage
index de62eab6b..e044f14b9 100644
--- a/source/precalculus/exercises/outcomes/FN/FN1/generator.sage
+++ b/source/precalculus/exercises/outcomes/FN/FN1/generator.sage
@@ -118,4 +118,6 @@ class Generator(BaseGenerator):
"plot0": plot(q[0]),
"plot1": plot(q[1]),
"plot2": plot(q[2]),
- }
\ No newline at end of file
+ }
+
+# trigger preview build...
diff --git a/source/precalculus/exercises/outcomes/LF/LF4/generator.sage b/source/precalculus/exercises/outcomes/LF/LF4/generator.sage
index f289357d2..c2896c004 100644
--- a/source/precalculus/exercises/outcomes/LF/LF4/generator.sage
+++ b/source/precalculus/exercises/outcomes/LF/LF4/generator.sage
@@ -41,9 +41,9 @@ class Generator(BaseGenerator):
point5=(point5[0],point5[1]+choice([-4..-1,1..4]))
slope = -1*line.lhs().coefficients(x)[1][0]/line.lhs().coefficients(y)[1][0]
- if orientation1=="parallel":
+ if orientation2=="parallel":
slope2=slope
- elif orientation1=="perpendicular":
+ elif orientation2=="perpendicular":
slope2=-1/slope
new_line = y-point5[1]==(x-point5[0]).mul(slope2,hold=True)
diff --git a/source/precalculus/source/01-EQ/01.ptx b/source/precalculus/source/01-EQ/01.ptx
index 93d44e55d..3ccc1c7c2 100644
--- a/source/precalculus/source/01-EQ/01.ptx
+++ b/source/precalculus/source/01-EQ/01.ptx
@@ -13,7 +13,9 @@
Activities
+
Recall that when solving a linear equation, you use addition, subtraction, multiplication and division to isolate the variable.
+
@@ -156,7 +158,9 @@
+
A linear equation with one unique solution is a conditional equation. A linear equation that is true for all values of the variable is an identity equation. A linear equation with no solutions is an inconsistent equation.
+
@@ -390,8 +394,9 @@
-
+
You can treat solving linear inequalities, just like solving an equation. The one exception is when you multiply or divide by a negative value, reverse the inequality symbol.
+
@@ -630,7 +635,9 @@
+
Solving a compound linear inequality, uses the same methods as a single linear inequality ensuring that you perform the same operations on all three parts. Alternatively, you can break the compound inquality up into two and solve separately.
+
@@ -693,7 +700,7 @@
-
Exercises available at .
+
Exercises available at .
-
\ No newline at end of file
+
diff --git a/source/precalculus/source/03-LF/07.ptx b/source/precalculus/source/03-LF/07.ptx
index 02e488efa..c850cd3c6 100644
--- a/source/precalculus/source/03-LF/07.ptx
+++ b/source/precalculus/source/03-LF/07.ptx
@@ -13,7 +13,9 @@
Activities
+
Now that we have explored multiple methods for solving systems of linear equations, let's put those in to practice using some real-world application problems.
+
@@ -231,7 +233,7 @@
-
Exercises available at .
+
Exercises available at .
-
\ No newline at end of file
+
diff --git a/source/precalculus/source/05-EL/02.ptx b/source/precalculus/source/05-EL/02.ptx
index eaac32408..c698db5fb 100644
--- a/source/precalculus/source/05-EL/02.ptx
+++ b/source/precalculus/source/05-EL/02.ptx
@@ -1076,7 +1076,7 @@
-
+
+
+
+ p=plot(csc(x),xmin=-2*pi,xmax=2*pi,ymin=-2*pi,ymax=2*pi,thickness=3,aspect_ratio=1,detect_poles=True,gridlines=True, ticks=[pi/2,1],tick_formatter=[pi,1])
+ for i in [-2..2]:
+ p+=line([(i*pi,-2*pi),(i*pi,2*pi)],thickness=3,linestyle='dashed',color='red')
+ p
+
+
+
+
+
+ p=plot(sec(x),xmin=-2*pi,xmax=2*pi,ymin=-2*pi,ymax=2*pi,thickness=3,aspect_ratio=1,detect_poles=True,gridlines=True, ticks=[pi/2,1],tick_formatter=[pi,1])
+ for i in [-2..1]:
+ p+=line([(i*pi+pi/2,-2*pi),(i*pi+pi/2,2*pi)],thickness=3,linestyle='dashed',color='red')
+ p
+
+
Graph of y=\sec x
+
+
+
+
+
+ p=plot(tan(x),xmin=-2*pi,xmax=2*pi,ymin=-2*pi,ymax=2*pi,thickness=3,aspect_ratio=1,detect_poles=True,gridlines=True, ticks=[pi/2,1],tick_formatter=[pi,1])
+ for i in [-2..1]:
+ p+=line([(i*pi+pi/2,-2*pi),(i*pi+pi/2,2*pi)],thickness=3,linestyle='dashed',color='red')
+ p
+
+
Graph of y=\tan x
+
+
+
+ p=plot(cot(x),xmin=-2*pi,xmax=2*pi,ymin=-2*pi,ymax=2*pi,thickness=3,aspect_ratio=1,detect_poles=True,gridlines=True, ticks=[pi/2,1],tick_formatter=[pi,1])
+ for i in [-2..2]:
+ p+=line([(i*pi,-2*pi),(i*pi,2*pi)],thickness=3,linestyle='dashed',color='red')
+ p
+
+
+ \sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}} or \sin\frac{\theta}{2}=-\sqrt{\frac{1-\cos\theta}{2}}, depending on which quadrant \theta is in.
+
+
+
+
+ \cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}} or \cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos\theta}{2}}, depending on which quadrant \theta is in.
+
+
+
+
+ \tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta} or \tan\frac{\theta}{2}=-\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}, depending on which quadrant \theta is in.
+