AP Calculus AB 8.2 Connecting Position Velocity and Acceleration of Functions Using Integrals Questions

D Particle J is at position x=7 at time t=0. Find the position of particle J at time t=2. Show the

work that leads to your answer.

Scoring Notes for Part D

- To earn P7, a response must present an indefinite or definite integral with an integrand of $v_{J}(t)$ or

$2 t\left(t^{2}-1\right)^{3}$. (See below for notes on how to handle a missing differential d t.)

- P8 is earned for an antiderivative of the form $k\left(t^{2}-1\right)^{4}$ or equivalent, for k>0. If $k \neq \frac{1}{4}$, then

the response is not eligible to earn P9.

- A response of $7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right)$ or equivalent banks P9 (i.e., subsequent errors in simplification

will not be considered in scoring for P9).

Note: An ambiguous response, such as $7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right.$, does not bank P9 and therefore must go

on to resolve the ambiguity with a correct final answer (e.g., $7+\frac{1}{4}(80)$ or 27) to earn P9.

- If the differential d t is missing:

○ Writing $\int_{0}^{2} v_{J}(t)$ earns P7 and is eligible to earn P8 and P9.

○ Writing $7+\int_{0}^{2} v_{J}(t)$ earns P7 and is eligible to earn P8 and P9.

○ Writing $\int_{0}^{2} v_{J}(t)+7$ introduces an ambiguity for the intended integrand.

- $\int_{0}^{2} v_{J}(t)+7=\left[\frac{1}{4}\left(t^{2}-1\right)^{4}\right]_{0}^{2}+7$ resolves the ambiguity.

Therefore, this earns P7 and P8 and is eligible for P9.

- $\int_{0}^{2} v_{J}(t)+7=\left[\frac{1}{4}\left(t^{2}-1\right)^{4}+7 t\right]_{0}^{2}$ confirms that an incorrect integrand was used.

Therefore, this does not earn P7, earns P8, and is not eligible for P9.

- If the ambiguity is not resolved, this does not earn P7, P8, or P9.

- Alternate solution using u-substitution:

$$\begin{aligned} & \text { Let } u=t^{2}-1, \text { then } d u=2 t d t. \\ & t=0 \Rightarrow u=-1 \\ & t=2 \Rightarrow u=3 \\ & x_{J}(2)=x_{J}(0)+\int_{0}^{2} v_{J}(t) d t=7+\int_{0}^{2} 2 t\left(t^{2}-1\right)^{3} d t \\ & =7+\int_{-1}^{3} u^{3} d u=7+\left[\frac{1}{4} u^{4}\right]_{-1}^{3} \\ & =7+\frac{1}{4}\left((3)^{4}-(-1)^{4}\right)=7+\frac{1}{4}(80)=27 \end{aligned}$$

- Alternate solution using indefinite integral:

$$\begin{aligned} & \int 2 t\left(t^{2}-1\right)^{3} d t=\frac{1}{4}\left(t^{2}-1\right)^{4}+C \\ & x_{J}(0)=7=\frac{1}{4}\left(0^{2}-1\right)^{4}+C \Rightarrow C=\frac{27}{4} \\ & x_{J}(t)=\frac{1}{4}\left(t^{2}-1\right)^{4}+\frac{27}{4} \\ & x_{J}(2)=\frac{1}{4}\left(2^{2}-1\right)^{4}+\frac{27}{4}=\frac{108}{4}=27 \end{aligned}$$

Part B (AB): Graphing calculator not allowed