AP Calculus BC Cha 3 G Calculate Derivatives of Parametric Functions Topic 9 2 QuestionsPractice AP Calculus BC questions on differentiating parametric rates again to analyze curvature or acceleration components.SyllabusEffective Fall 2025CourseAP Calculus BC
AP Calculus BC Cha 3 G Calculate Derivatives of Parametric Functions Topic 9 2 Questions question 1[Maximum number: 2]For 0≤t≤π0 \leq t \leq \pi0≤t≤π, 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)=2sinty(t)=2 \sin ty(t)=2sint. It is known that dxdt=ecost\frac{d x}{d t}=e^{\cos t}dtdx=ecost. 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.Show AnswerSincedxdt=ecost,\frac{dx}{dt}=e^{\cos t},dtdx=ecost,d2xdt2=−ecostsint,x′′(1)=−ecos1sin1=−1.444407.\frac{d^2x}{dt^2}=-e^{\cos t}\sin t, \qquad x''(1)=-e^{\cos1}\sin1=-1.444407.dt2d2x=−ecostsint,x′′(1)=−ecos1sin1=−1.444407.Also,y(t)=2sint⟹d2ydt2=−2sint,y′′(1)=−2sin1=−1.682942.y(t)=2\sin t \quad\Longrightarrow\quad \frac{d^2y}{dt^2}=-2\sin t, \qquad y''(1)=-2\sin1=-1.682942.y(t)=2sint⟹dt2d2y=−2sint,y′′(1)=−2sin1=−1.682942.Therefore,a(1)=⟨−1.444407,−1.682942⟩.\mathbf a(1)=\langle-1.444407,-1.682942\rangle.a(1)=⟨−1.444407,−1.682942⟩.Add to Test