AP Calculus BC 9.4 Defining and Differentiating Vector- Valued Functions - FRQs - Exam Style Questions
Calc-Ok Question
For \(0\le t\le \pi\), a particle moves along a curve so that its position at time \(t\) is \((x(t),y(t))\).
The \(x(t)\) component is not explicitly given and \(y(t)=2\sin t\). It is known that \(\displaystyle \frac{dx}{dt}=e^{\cos t}\). At \(t=0\) the particle is at \((1,0)\).
(a) Find the acceleration vector of the particle at time \(t=1\). Show the setup for your calculations.
(b) For \(0\le t\le \pi\), find the first time \(t\) at which the speed of the particle is \(1.5\). Show the work that leads to your answer.
(c) Find the slope of the line tangent to the path at time \(t=1\). Find the \(x\)-coordinate of the position of the particle at time \(t=1\). Show the work that leads to your answers.
(d) Find the total distance traveled by the particle over \(0\le t\le \pi\). Show the setup for your calculations.
Most-appropriate topic codes:
• TOPIC 9.4: Defining & Differentiating Vector-Valued Functions — part (a)
• TOPIC 9.6: Solving Motion Problems Using Parametric & Vector-Valued Functions — part (b)
• TOPIC 9.1: Defining & Differentiating Parametric Equations — part (c, slope)
• TOPIC 9.5: Integrating Vector-Valued Functions — part (c, find \(x(1)\) from \(x'(t)\)+IV)
• TOPIC 9.3: Finding Arc Lengths of Curves Given by Parametric Equations — part (d)
• TOPIC 9.6: Solving Motion Problems Using Parametric & Vector-Valued Functions — part (b)
• TOPIC 9.1: Defining & Differentiating Parametric Equations — part (c, slope)
• TOPIC 9.5: Integrating Vector-Valued Functions — part (c, find \(x(1)\) from \(x'(t)\)+IV)
• TOPIC 9.3: Finding Arc Lengths of Curves Given by Parametric Equations — part (d)
▶️ Answer/Explanation
(a) Acceleration vector at \(t=1\)
\(x'(t)=e^{\cos t}\Rightarrow x”(t)=\dfrac{d}{dt}\!\big(e^{\cos t}\big)=e^{\cos t}(-\sin t)=-e^{\cos t}\sin t.\)\(y(t)=2\sin t\Rightarrow y'(t)=2\cos t,\; y”(t)=-2\sin t.\)
Evaluate at \(t=1\): \(x”(1)=-e^{\cos 1}\sin 1\approx-1.444\), \(y”(1)=-2\sin 1\approx-1.683\).
\(\displaystyle \boxed{\mathbf a(1)=\langle x”(1),\,y”(1)\rangle\approx\langle-1.444,\,-1.683\rangle}\).
(b) First time when speed \(=1.5\)
Speed \(=\sqrt{\left(\dfrac{dx}{dt}\right)^{2}+\left(\dfrac{dy}{dt}\right)^{2}} =\sqrt{\big(e^{\cos t}\big)^{2}+\big(2\cos t\big)^{2}} =\sqrt{e^{2\cos t}+4\cos^{2}t}.\)Solve for \(0\le t\le\pi\): \[ \sqrt{e^{2\cos t}+4\cos^{2}t}=1.5. \] Numerically (unique solution in \((0,\tfrac{\pi}{2})\)): \(\boxed{t\approx 1.254}\) (radians).
(c) Slope at \(t=1\) and \(x(1)\)
\(\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2\cos t}{e^{\cos t}}.\) At \(t=1\): \(\displaystyle \left.\frac{dy}{dx}\right|_{t=1}= \frac{2\cos 1}{e^{\cos 1}}\approx \boxed{0.630}.\)To find \(x(1)\), use the initial value \(x(0)=1\) and accumulate change: \[ x(1)=x(0)+\int_{0}^{1}x'(t)\,dt =1+\int_{0}^{1}e^{\cos t}\,dt \approx \boxed{3.342}. \]
(d) Total distance on \(0\le t\le \pi\)
Distance \(=\displaystyle\int_{0}^{\pi}\text{speed}\;dt =\int_{0}^{\pi}\sqrt{e^{2\cos t}+4\cos^{2}t}\;dt \approx \boxed{6.035\ \text{units}}. \)