AP Calculus AB Unit 4: Contextual Applications

B During what open intervals of time t, for 0<t<5, are particles H and J moving in opposite directions?

Give a reason for your answer.

- To earn P3, a response can do one of the following:

○ Set $x_{H}{ }^{\prime}(t)=0, v_{H}(t)=0$, or $(2 t-4) e^{t^{2}-4 t}=0$

○ Set $v_{J}(t)=0$ or $2 t\left(t^{2}-1\right)^{3}=0$

○ Identify t=2 for particle H and no other values in the interval 0<t<5

○ Identify t=1 for particle J and no other values in the interval 0<t<5

○ Identify the interval 1<t<2

- To earn P4, a response can provide an analysis of signs of velocity or direction of motion on the

interval 0<t<5 for either particle H or particle J.

- To be eligible for P5, a response must provide correct analyses of signs of velocity or direction of

motion on the interval 0<t<5 for both particles.

- Only analysis within the interval 0<t<5 will be considered in scoring.

C It can be shown that $v_{J}^{\prime}(2)>0$. Is the speed of particle J increasing, decreasing, or neither at time

t=2 ? Give a reason for your answer.

Answer with reason

Point 6 (P6)

Because $v_{J}(2)$ and $v_{J}{ }^{\prime}(2)$ have the same sign, the speed of

particle J is increasing at t=2.

Scoring Notes for Part C

- An evaluation of $v_{J}(2)$ is not necessary, but if a value is presented, it must be correct. The correct

value is $v_{J}(2)=108$.

- An evaluation of $v_{J}{ }^{\prime}(2)$ is not necessary, but if a value is presented, it must be correct. The correct

value is $v_{J}{ }^{\prime}(2)=486$.

- A response can either import the analysis for the sign of $v_{J}(2)$ from part B or restart.

- A response that stated " $v_{J}(t)>0$ for 1<t<5 " in part B does not need to restate $v_{J}(2)>0$ and

earns P6 for " $v_{J}(2)$ and $v_{J}{ }^{\prime}(2)$ have the same sign, so the speed is increasing."

$f^{\prime}(7)=-8.120$ (or -8.119)

After the store has been open 7 hours, the rate at which bananas are

being removed from the display table is decreasing by 8.120 (or 8.119)

pounds per hour per hour.

g(5)-f(5)=-2.263103<0

Because g(5)-f(5)<0, the number of pounds of bananas on the

display table is decreasing at time t=5.

B There is a point P on the curve G near (2,-1) with x-coordinate 1.6. Use the line tangent to the curve at

(2,-1) to approximate the y-coordinate of point P.

Scoring Notes for Part B

- A response can earn $\mathbf{P 3}$ with $\left.\frac{d y}{d x}\right|_{(x, y)=(2,-1)}=-\frac{1}{4}, \frac{d y}{d x}=-\frac{1}{4}$, "slope is $-\frac{1}{4}$," or equivalent.

- A response that presents a linear approximation with a slope of $-\frac{1}{4}$ also earns $\mathbf{P 3}$.

- A response that declares $\left.\frac{d y}{d x}\right|_{(x, y)=(2,-1)}$ (or the slope) equal to any nonzero value $k \neq-\frac{1}{4}$ does not

earn P3. Such a response earns P4 for a presented approximation mathematically equivalent to

-1+k(-0.4).

- P4 cannot be earned with a linear approximation using a slope other than $-\frac{1}{4}$ if that slope has not

been declared to be the value of $\left.\frac{d y}{d x}\right|_{(x, y)=(2,-1)}$.

- A response does not have to present the tangent line equation but must clearly demonstrate its use at

x=1.6 in finding the requested approximation to be eligible for P4.

- A response of $-1-\frac{1}{4}(-0.4)$ earns both P3 and P4.

- A response of $-1-\frac{1}{4}(1.6-2)$ or equivalent banks P4 (i.e., subsequent errors in simplification will

not be considered in scoring for P4).

Note: An ambiguous response, such as $-1-\frac{1}{4}(1.6-2$, does not bank P4 and therefore must go on

to resolve the ambiguity with a correct final answer (e.g., -0.9 ) to earn P4.

D A particle moves along the curve H defined by the equation $2 x y+\ln y=8$. At the instant when the

particle is at the point $(4,1), \frac{d x}{d t}=3$. Find $\frac{d y}{d t}$ at that instant. Show the work that leads to your answer.

Scoring Notes for Part D

- P7 is earned for implicitly differentiating $2 x y+\ln y=8$ with respect to t with at most one error.

- P8 is earned for an equation equivalent to $2 \frac{d x}{d t} y+2 x \frac{d y}{d t}+\frac{1}{y} \frac{d y}{d t}=0$.

- To be eligible for P9, a response must have earned P7 and P8, with no errors in implicit

differentiation.

- P9 is earned only for the value of $-\frac{2}{3}$.

- Alternate solution:

$$\begin{aligned} & \frac{d y}{d t}=\frac{d y}{d x} \cdot \frac{d x}{d t} \\ & \frac{d}{d x}(2 x y+\ln y)=\frac{d}{d x}(8) \Rightarrow 2 y+2 x \frac{d y}{d x}+\frac{1}{y} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-2 y}{2 x+\frac{1}{y}} \\ & \left.\frac{d y}{d x}\right|_{(x, y)=(4,1)}=\frac{-2(1)}{2(4)+\frac{1}{1}}=-\frac{2}{9} \\ & \left.\frac{d y}{d t}\right|_{(x, y)=(4,1)}=\left.\frac{d y}{d x} \cdot \frac{d x}{d t}\right|_{(x, y)=(4,1)}=-\frac{2}{9} \cdot 3=-\frac{2}{3} \end{aligned}$$

○ P7 is earned for an implicit differentiation with respect to x with at most one error, as long as

$\frac{d y}{d x}$ is eventually correctly linked to $\frac{d x}{d t}$.

○ P8 is earned for finding $\frac{d y}{d x}$ and multiplying the result by $\frac{d x}{d t}=3$.

○ P8 can be earned for a stated incorrect $\frac{d y}{d x}$, as long as it is multiplied by 3.

○ P9 is only earned for a correct value of $-\frac{2}{3}$ or equivalent.

○ Stating $\frac{d y}{d t}=\frac{d y}{d x} \cdot \frac{d x}{d t}$ alone does not earn any points.