Free Printable WorksheetMathCalculus IIMath CurriculumAP Calculus BC Unit 7; FL MA.912.C.5.2-C.5.4; CCSS.HSM; CCSS.MP1, MP2, MP4, MP5, MP6; Integrated Calculus II Standards CAL2-ODE; TEKS 19 TAC section 111.55

CAL2.87 Logistic Solution and Inflection Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Calculus II Math worksheet provides structured practice on CAL2.87 Logistic Solution and Inflection. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

CAL2.87 Logistic Solution and Inflection - Choose the correct limiting behavior, fastest-growth point, or inflection conclusion for a logistic model. Standards: AP Calculus BC Unit 7; FL MA.912.C.5.2-C.5.4; CCSS.HSM; CCSS.MP1, MP2, MP4, MP5, MP6; Integrated Calculus II Standards CAL2-ODE; TEKS 19 TAC section 111.55.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Calculus II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

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Below are sample questions representative of the exercises included on this worksheet and related CAL2.87 Logistic Solution and Inflection skill assessments:

Problem 1Difficulty: Mastery
A logistic model has limiting value 1600 and fastest growth rate 80 units per day. Which differential equation matches those facts?
AdP/dt = 0.05P(1 - P/1600)
BdP/dt = 80P(1 - P/1600)
CdP/dt = 0.2P(1 - P/800)
DdP/dt = 0.2P(1 - P/1600)
Solution & Explanation:The limiting value gives K = 1600. Since maximum rate is rK/4, 80 = r(1600)/4 = 400r, so r = 0.2. Thus dP/dt = 0.2P(1 - P/1600).
Problem 2Difficulty: Mastery
A model P(t) = 1000/(1 + Ae^(-0.25t)) has inflection time t = 8. What is P(0)?
A1000/(1 + e^(-2))
B500
C1000/(1 + e2)
D1000/e2
Solution & Explanation:Inflection at t = 8 means Ae^(-0.25*8) = Ae^(-2) = 1, so A = e2. Therefore P(0) = 1000/(1 + A) = 1000/(1 + e2).
Problem 3Difficulty: Mastery
For P(t) = 720/(1 + 8e^(-0.6t)), compare concavity at t = 0 with concavity at the time when P = 600.
AConcave down at both times
BConcave up at t = 0; concave down when P = 600
CConcave up at both times
DConcave down at t = 0; concave up when P = 600
Solution & Explanation:P(0) = 720/9 = 80, which is below K/2 = 360, so the graph is concave up at t = 0. At P = 600, which is above 360, it is concave down.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Calculus II Math students. It aligns with educational standard AP Calculus BC Unit 7; FL MA.912.C.5.2-C.5.4; CCSS.HSM; CCSS.MP1, MP2, MP4, MP5, MP6; Integrated Calculus II Standards CAL2-ODE; TEKS 19 TAC section 111.55 (Choose the correct limiting behavior, fastest-growth point, or inflection conclusion for a logistic model. Standards: AP Calculus BC Unit 7; FL MA.912.C.5.2-C.5.4; CCSS.HSM; CCSS.MP1, MP2, MP4, MP5, MP6; Integrated Calculus II Standards CAL2-ODE; TEKS 19 TAC section 111.55.).

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