AP Calculus BC 9.3 Finding Arc Lengths of Curves Given by Parametric Equations - FRQs - Exam Style Questions
Calc-Ok Question
A particle moving along a curve in the xy-plane has position \((x(t),y(t))\) at time \(t\) seconds, where \(x(t)\) and \(y(t)\) are measured in centimeters. It is known that \[ x'(t)=8t-t^{2}\quad\text{and}\quad y'(t)=-t+\sqrt{t^{1.2}+20}. \] At time \(t=2\) seconds, the particle is at the point \((3,6)\).
(a) Find the speed of the particle at time \(t=2\) seconds. Show the setup for your calculations.
(b) Find the total distance traveled by the particle over the time interval \(0\le t\le 2\). Show the setup for your calculations.
(c) Find the \(y\)-coordinate of the position of the particle at the time \(t=0\). Show the setup for your calculations.
(d) For \(2\le t\le 8\), the particle remains in the first quadrant. Find all times \(t\) in the interval \(2\le t\le 8\) when the particle is moving toward the \(x\)-axis. Give a reason for your answer.
Most-appropriate topic codes:
• TOPIC 9.3: Derivatives of Parametric Equations (parts a, d) | • TOPIC 9.5: Arc Length of Parametric Curves (distance via \(\int \sqrt{(x’)^{2}+(y’)^{2}}\,dt\)) (part b) | • TOPIC 8.3: Using Accumulation Functions in Applied Contexts (recovering \(y(0)\) from \(y(2)\) and \(\int y'(t)\,dt\)) (part c)
▶️ Answer/Explanation
(a) Speed at \(t=2\)
\[ \text{speed}(t)=\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2},\quad x'(2)=12,\quad y'(2)=-2+\sqrt{2^{1.2}+20}\approx 2.7231. \] \[ \text{speed}(2)=\sqrt{12^{2}+2.7231^{2}}\approx \boxed{12.305\ \text{cm/s}}. \](b) Total distance on \(0\le t\le 2\)
\[ \text{Distance}=\int_{0}^{2}\sqrt{\big(8t-t^{2}\big)^{2}+\Big(-t+\sqrt{t^{1.2}+20}\Big)^{2}}\,dt \approx \boxed{15.902\ \text{cm}}. \](c) \(y\)-coordinate at \(t=0\)
\[ y(0)=y(2)+\int_{2}^{0}y'(t)\,dt =6-\int_{0}^{2}\!\Big(-t+\sqrt{t^{1.2}+20}\Big)\,dt \approx 6-7.173613=\boxed{-1.174}. \](d) Times moving toward the \(x\)-axis on \(2\le t\le 8\)
Require \(y(t)>0\) (first quadrant) and \(y'(t)<0\): \[ -t+\sqrt{t^{1.2}+20}<0 \;\Longleftrightarrow\; t^{2}>t^{1.2}+20. \] The root of \(t^{2}-t^{1.2}-20=0\) in \([2,8]\) is \(t\approx 5.222\). Hence \[ \boxed{5.222<t<8}. \]