AP Calculus BC 9.2 Second Derivatives of Parametric Equations - MCQs - Exam Style Questions

Use \(\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dt}\!\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt}\).

\(\displaystyle \frac{dx}{dt}=5/(t+5).\)

Let \(S(t)=\pi\cos(\pi t)\,(t+5)\). Then
\(\displaystyle S'(t)=\pi\big[\cos(\pi t)\big]'(t+5)+\pi\cos(\pi t)\cdot 1 =\pi\!\left(-\pi\sin(\pi t)\right)(t+5)+\pi\cos(\pi t).\)

Thus \(\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{S'(t)}{dx/dt} =\frac{\pi\cos(\pi t)-\pi^{2}(t+5)\sin(\pi t)}{5/(t+5)} =\frac{t+5}{5}\Big(\pi\cos(\pi t)-\pi^{2}(t+5)\sin(\pi t)\Big).\)

At \(t=1\): \(\cos(\pi)=-1,\ \sin(\pi)=0,\ \frac{t+5}{5}=\frac{6}{5}\). Hence
\(\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{6}{5}\big(\pi(-1)-0\big)=-\frac{6\pi}{5}\approx-3.770.\)

✅ Answer: (B)