Question
For time t ≥ 0, a particle moves in the xy-plane with position (x(t), y(t)) , and velocity vector \(\left \langle (t-1)^{e^{t^{2}}}, sin \left ( t^{1.25} \right ) \right \rangle.\) At time t = 0 , the position of the particle is (−2, 5) .
(a) Find the speed of the particle at time t = 1.2. Find the acceleration vector of the particle at time t = 1.2.
(b) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 1.2.
(c) Find the coordinates of the point at which the particle is farthest to the left for t ≥ 0. Explain why there is no point at which the particle is farthest to the right for t ≥ 0.
Answer/Explanation
Ans:
(a)
\(Speed = \sqrt{(x'(t))^{2}+(y'(t))^{2}}\)
\(\sqrt{\left ( (1.2-1)e^{(1.2)^{2}} \right )^{2}+\left ( sin((1.2)^{1.25}) \right )^{2}}\)
Speed = 1.271 \(accel: \left \langle (t-1)(2+e^{t2})+(1)(e^{t2}),cos (t^{1.25}\cdot 1.25t^{.25}) \right \rangle\)
of t = 1.2
accel = < 6.247, .405>
(b)
\(\int_{0}^{1.2}\sqrt{\left ( (t-1)e^{t2} \right )^{2}+\left ( sin(t^{1.25}) \right )^{2}}dt\)
= 1.010
(c)
x'(t) is negative on (0, 1) and is positive on (1, 0). x'(1) = 0 Since x'(t) goes from negative to positive at t = 1 and x'(1) = 0, x(t) has a rel. minimum at t=1, since x(t) decreases until t=1 and always increases afterwards, x(1) is on abs. min / that is the location of further to the left, \(-2+\int_{0}^{1}(t-1)e^{t2}dt = x(1)\)
x(1) = -2.604
\(5+\int_{0}^{0}(sin (e^{1.25}))dt = y(1)\)
y (1) = 5.410
(-2.604, 5.410)
There is no point furthest to the right because x'(t) increases towards infinity on +20, meaning there is no abs max of x(t) (sint x(t) will thus be increasing towards ∝ on t > 1). This means the particle will continue to the right for t > 1, creating no furthest right point.