AP Calculus BC 9.4 Defining and Differentiating Vector- Valued Functions - Exam Style Questions - MCQs - New Syllabus
Question
If at any time \(t\) the position vector of a particle moving in the \(xy\)-plane is \(\langle \sin(\pi t),\,\cos(\pi t)\rangle\), what is the acceleration vector of the particle at time \(t=\tfrac12\)?
A. \(\langle -\pi^2,\,0\rangle\)
B. \(\langle 0,\,-\pi^2\rangle\)
C. \(\langle -1,\,0\rangle\)
D. \(\langle 0,\,1\rangle\)
B. \(\langle 0,\,-\pi^2\rangle\)
C. \(\langle -1,\,0\rangle\)
D. \(\langle 0,\,1\rangle\)
▶️ Answer/Explanation
Detailed solution
Velocity: \( \mathbf{v}(t)=\langle \pi\cos(\pi t),\, -\pi\sin(\pi t)\rangle \).
Acceleration: \( \mathbf{a}(t)=\mathbf{v}'(t)=\langle -\pi^2\sin(\pi t),\, -\pi^2\cos(\pi t)\rangle \).
At \(t=\tfrac12\): \(\sin(\pi/2)=1,\ \cos(\pi/2)=0\Rightarrow \mathbf{a}\!\left(\tfrac12\right)=\langle -\pi^2,\,0\rangle\).
✅ Correct: A