Question
The velocity vector of an object in motion in the plane for $0 \leq t \leq 6$ is given by its components $\frac{d x}{d t}=3 t \cdot \sin \left(t^2\right)$ and $\frac{d y}{d t}=2 e^{\cos (t)}$. At time $t=5$, the position of the object is $(3,-2)$.
(a) Find the speed of the object at time $t=4$, and find the acceleration vector of the object at time $t=4$.
(b) Find the position of the object at time $t=4$.
(c) Find the total distance traveled by the object from $t=0$ to $t=4$.
▶️Answer/Explanation
(a) Note: $\frac{d x}{d t}$ and $x^{\prime}(t)$ are equivalent notations for the $x$-component for velocity. Likewise, $\frac{d y}{d t}$ and $y^{\prime}(t)$ are equivalent notations for the $y$-component for velocity.
$
\begin{aligned}
& \text { Speed }=\sqrt{\left(x^{\prime}(4)\right)^2+\left(y^{\prime}(4)\right)^2}=3.608 \\
& \text { Acceleration }=\left\langle x^{\prime \prime}(4), y^{\prime \prime}(4)\right\rangle=\langle-92.799,0.787\rangle \\
& \qquad x(4)=x(5)+\int_5^4 x^{\prime}(t) d t=3+\int_5^4 x^{\prime}(t) d t=5.923
\end{aligned}
$
(b)
$
y(4)=y(5)+\int_5^4 y^{\prime}(t) d t=-2+\int_5^4 y^{\prime}(t) d t=-3.697
$
At time $t=4$, the object is at position $(5.923,-3.697)$
(c) Total Distance $=\int_0^4 \sqrt{\left(x^{\prime}(t)\right)^2+\left(y^{\prime}(t)\right)^2} d t=20.261$
2. See the solution for $\mathrm{AB} / \mathrm{BC}$ 2, page 523.