AP Calculus BC 9.2 Parametric Second Derivatives Overview

Review parametric second derivatives by differentiating dy over dx with respect to the parameter and dividing again to describe curve concavity.

Covered syllabus points

CHA-3.G—Calculate derivatives of parametric functions—Topic 9.2

[Maximum number: 2]

For $0 \leq t \leq \pi$ , a particle is moving along the curve shown so that its position at time t is (x(t), y(t)), where x(t) is not explicitly given and $y(t)=2 \sin t$ . It is known that $\frac{d x}{d t}=e^{\cos t}$ . At time t=0, the particle is at position (1, 0).

Find the acceleration vector of the particle at time t=1. Show the setup for your calculations.

Since
$\frac{dx}{dt}=e^{\cos t},$ $\frac{d^2x}{dt^2}=-e^{\cos t}\sin t, \qquad x''(1)=-e^{\cos1}\sin1=-1.444407.$
Also,
$y(t)=2\sin t \quad\Longrightarrow\quad \frac{d^2y}{dt^2}=-2\sin t, \qquad y''(1)=-2\sin1=-1.682942.$
Therefore,
$\mathbf a(1)=\langle-1.444407,-1.682942\rangle.$

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Covered syllabus points

9.2 Second Derivatives of Parametric Equations question 1