AP Calculus BC 9.2: Parametric Second Derivatives

Practice AP Calculus BC questions on differentiating parametric rates again to analyze curvature or acceleration components.

[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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CHA-3.G—Calculate derivatives of parametric functions—Topic 9.2 question 1