Chapter 11.2, Problem 44E

To determine

To sketch the pat that the particle travels in exercise 40.

Explanation of Solution

Given:

The velocity $\mathrm{v}(\mathrm{t})$ of the particle is given as

  $\mathrm{v}(\mathrm{t})=\langle 2\mathrm{\pi }\cos 4\mathrm{\pi }\mathrm{t},4\mathrm{\pi }\sin 2\mathrm{\pi }\mathrm{t}\rangle ;(7,2)$

Calculation:

It is known that-

  $\begin{array}{l}\frac{\mathrm{d}\mathrm{x}(\mathrm{t})}{\mathrm{d}\mathrm{t}}=\mathrm{v}(\mathrm{t})=2\mathrm{\pi }\cos 4\mathrm{\pi }\mathrm{t}\\ \mathrm{x}(\mathrm{t})={\displaystyle \int 2\mathrm{\pi }\cos 4\mathrm{\pi }\mathrm{t}\mathrm{d}\mathrm{t}}\\ \mathrm{x}(\mathrm{t})=\frac{1}{2}\sin 4\mathrm{\pi }\mathrm{t}+\mathrm{C}\end{array}$

The particle is at $\mathrm{x}=7$ when $\mathrm{t}=0$

So by putting these values-

  $\begin{array}{l}7=0+\mathrm{C}\\ \mathrm{C}=7\end{array}$

Hence the position of the particle is given by

  $\mathrm{x}(\mathrm{t})=\frac{1}{2}\sin 4\mathrm{\pi }\mathrm{t}+7$

Similarly,

  $\begin{array}{l}\frac{\mathrm{d}\mathrm{y}(\mathrm{t})}{\mathrm{d}\mathrm{t}}=\mathrm{v}(\mathrm{t})=4\mathrm{\pi }\sin 2\mathrm{\pi }\mathrm{t}\\ \mathrm{y}(\mathrm{t})={\displaystyle \int 4\mathrm{\pi }\sin 2\mathrm{\pi }\mathrm{t}\mathrm{d}\mathrm{t}}\\ \mathrm{y}(\mathrm{t})=-2\cos 2\mathrm{\pi }\mathrm{t}+\mathrm{C}\end{array}$

The particle is at $\mathrm{x}=7$ when $\mathrm{t}=0$

So by putting these values-

  $\begin{array}{l}2=-2+\mathrm{C}\\ \mathrm{C}=4\end{array}$

Hence the position of the particle is given by

  $\mathrm{y}(\mathrm{t})=-2\cos 2\mathrm{\pi }\mathrm{t}+4$

Hence the position vector is $\mathrm{r}(\mathrm{t})=\langle \frac{1}{2}\sin 4\mathrm{\pi }\mathrm{t}+7,-2\cos 2\mathrm{\pi }\mathrm{t}+4\rangle$

Graphing the parametric form of the position vector and finding the path of particle as follows-