AP Calculus BC 9.2 Second Derivatives of Parametric Equations - FRQs - Exam Style Questions
Calc-Ok Question
The motion satisfies \[ \frac{dx}{dt}=\sqrt{1+t^2}, \qquad \frac{dy}{dt}=\ln(2+t^2). \] At time \(t=4\), the particle is at the point \((1,5)\).
(a) Find the slope of the line tangent to the path at time \(t=4\).
(b) Find the speed of the particle at time \(t=4\).
Then find the acceleration vector of the particle at time \(t=4\).
(c) Find the \(y\)-coordinate of the particle’s position at time \(t=6\).
(d) Find the total distance the particle travels along the curve from \(t=4\) to \(t=6\).
Most-appropriate topic codes (CED):
• TOPIC 9.4: Defining & Differentiating Vector-Valued Functions (speed/acceleration) — part (b)
• TOPIC 8.3: Using Accumulation Functions in Applied Contexts (recover \(y\) from \(y’\) + initial value) — part (c)
• TOPIC 9.3: Arc Length of Parametric Curves (distance = integral of speed) — part (d)
▶️ Answer/Explanation
\[ \frac{dy}{dx}\Big|_{t=4}=\frac{y'(4)}{x'(4)} =\frac{\ln(2+4^2)}{\sqrt{1+4^2}} =\frac{\ln 18}{\sqrt{17}} \approx \boxed{0.701}. \]
Speed: \[ v(4)=\sqrt{(x'(4))^2+(y'(4))^2} =\sqrt{17+(\ln18)^2} \approx \boxed{5.035}. \] Acceleration vector \(a(t)=\langle x”(t),\,y”(t)\rangle\):
\(x”(t)=\dfrac{t}{\sqrt{1+t^2}}\Rightarrow x”(4)=\dfrac{4}{\sqrt{17}}\approx 0.970.\)
\(y”(t)=\dfrac{2t}{2+t^2}\Rightarrow y”(4)=\dfrac{8}{18}=\dfrac{4}{9}\approx 0.444.\)
\[ \boxed{a(4)=\langle 4/\sqrt{17},\,4/9\rangle \approx \langle 0.970,\;0.444\rangle }. \]
Use accumulation with \(y(4)=5\): \[ y(6)=y(4)+\int_{4}^{6}\!y'(t)\,dt =5+\int_{4}^{6}\!\ln(2+t^2)\,dt \approx \boxed{11.571}. \]
Distance equals the integral of speed: \[ \text{Dist}=\int_{4}^{6}\!\sqrt{(x'(t))^2+(y'(t))^2}\,dt =\int_{4}^{6}\!\sqrt{\,1+t^2+\big(\ln(2+t^2)\big)^2\,}\;dt \approx \boxed{12.136}. \]