AP Calculus AB Cha 3 D Calculate Related Rates in Applied Contexts Questions

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.