Question
(a) Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(b) Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(c) Topic-6.7 The Fundamental Theorem of Calculus and Definite Integrals
(d) Topic-8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled
2. For \(0\leq t\leq \pi \), a particle is moving along the curve shown so that its position at time t is ( x(t) , ( t)y), where x( t) is not explicitly given and y(t ) = 2 sin t. It is known that \(\frac{dx}{dt}=e^{cos\\\ t}\). At time t = 0, the particle is at position (1, 0)
(a) Find the acceleration vector of the particle at time t = 1. Show the setup for your calculations.
(b) For \(0\leq t\leq \pi \), 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 of the particle 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 the time interval \(0\leq t\leq \pi \). Show the setup for your calculations.
▶️Answer/Explanation
2(a) Find the acceleration vector of the particle at time t = 1. Show the setup for your calculations.
\(x”(1)=\frac{d}{dt}(e^{cos t})|_{t=1}=-1.444407\)
\(y(t) =2sint\Rightarrow y'(t)=2cos t\)
\(y” (1)=\frac{d}{dt}(2cos t)|_{t=1}=-1.682942\)
The acceleration vector at time t = 1 is
\(a(1)=\left< -1.444,-1.683(or-1.682)\right>\)
2(b) For 0 ≤ t≤ π, find the first time t at which the speed of the particle is 1.5. Show the work that leads to your answer.
Speed =\(\sqrt{\left ( \frac{dx}{dt}\right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}=\sqrt{(e^{cos t})+(2cos t)^{2}}\)
\(0\leq t\leq \pi and \sqrt{(e^{cos t})+(2cos t)^{2}}=1.5\)
\(\Rightarrow t=1.254472,t=2.358077\)
The first time at which the speed of the particle is 1.5 is t = 1.254.
2(c) Find the slope of the line tangent to the path of the particle 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.
\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2cos t}{e^{cos t}}\)
\(\frac{dy}{dx}|_{t=1}=\frac{2 cos 1}{e^{cos 1}} =0.629530\)
The slope of the line tangent to the curve at t = 1 is 0.630 (or 0.629 ).
\(x(1)=x(0)+\int_{0}^{1}\frac{dx}{dt}dt=1+\int_{0}^{1}e^{cos t}dt=3.341575\)
The x -coordinate of the position at t = 1 is 3.342 (or 3.341).
2(d) Find the total distance traveled by the particle over the time interval 0 ≤ t≤ π. Show the setup for your calculations.
\(\int_{0}^{\pi }\sqrt{(e^{cos t})^{2}+(2 cos t)^{2}} dt\)
= 6.034611
The total distance traveled by the particle over 0 ≤ t≤ π is 6.035 (or 6.034 ).