Chapter 11, Problem 48RE

(a)

To determine

To find: The slope of the path of the particle at time $t=\pi$ .

Answer to Problem 48RE

The slope of the path of the particle at time $t=\pi$ is $1$ .

Explanation of Solution

Given:

The parametric equations are $x=e^t\cos t$ and $y=e^t\sin t$ at time $t\le 0\le 4$ .

Calculation:

The slope of the particle is given by, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ .

Differentiate the equation $x=e^t\cos t$ with respect to $t$ .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(e^t\cos t\right)\\ =e^t\cos t-e^t\sin t\end{array}$

Differentiate the equation $y=e^t\sin t$ with respect to $t$ .

  $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(e^t\sin t\right)\\ =e^t\sin t+e^t\cos t\end{array}$

Find the slope of the path of the particle at time $t=\pi$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{e^t\sin t+e^t\cos t}{\left(e^t\cos t-e^t\sin t\right)}\\ =\frac{\left(\sin t+\cos t\right)}{\left(\cos t-\sin t\right)}\\ =\frac{\sin \pi +\cos \pi }{\cos \pi -\sin \pi }\\ =1\end{array}$

Therefore, the slope of the path of the particle at time $t=\pi$ is $1$ .

(b)

To determine

To find: The speed of the particle when $t=3$ .

Answer to Problem 48RE

The speed of the particle when $t=3$ is $\sqrt{2}{e}^3$ .

Explanation of Solution

Given:

The parametric equations are $x=e^t\cos t$ and $y=e^t\sin t$ at time $t\le 0\le 4$ .

Calculation:

Find the velocity vector of the particle as follows.

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left(r\left(t\right)\right)\\ =\frac{d}{dt}\langle e^t\cos t,e^t\sin t\rangle \\ =\langle e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t\rangle \end{array}$

Find the magnitude of the velocity vector as follows.

  $\begin{array}{c}\left|v\left(t\right)\right|=\sqrt{{\left(e^t\cos t-e^t\sin t\right)}^2+{\left(e^t\sin t+e^t\cos t\right)}^2}\\ =\sqrt{e^{2t}\left({\sin }^{2}t+2\sin t\cos t+{\cos }^{2}t\right)+e^{2t}\left({\cos }^{2}t-2\cos t\sin t+{\sin }^{2}t\right)}\\ =\sqrt{e^{2t}\left(1+2\sin t\cos t\right)+e^{2t}\left(1-2\cos t\sin t\right)}\\ =\sqrt{2e^{2t}}\end{array}$

Find the speed of the particle when $t=3$ .

  $\begin{array}{c}\left|v\left(t\right)\right|=\sqrt{2}{e}^t\\ {\left|v\left(t\right)\right|}_{t=3}=\sqrt{2}{e}^3\end{array}$

Therefore, the speed of the particle when $t=3$ is $\sqrt{2}{e}^3$ .

(c)

To determine

To find: The distance traveled by the particle along the path from $t=0$ to $t=3$ .

Answer to Problem 48RE

The distance traveled by the particle along the path from $t=0$ to $t=3$ is $\sqrt{2}\left(e^3-1\right)$ .

Explanation of Solution

Given:

The parametric equations are $x=e^t\cos t$ and $y=e^t\sin t$ at time $t\le 0\le 4$ .

Calculation:

Find distance traveled by the particle along the path from $t=0$ to $t=3$ as follows.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{3}{\int }}\left|v\left(t\right)\right|}={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{2}{e}^{t}dt}\\ =\sqrt{2}{\left[e^t\right]}_0^3\\ =\sqrt{2}\left(e^3-e^0\right)\\ =\sqrt{2}\left(e^3-1\right)\end{array}$

Therefore, the distance traveled by the particle along the path from $t=0$ to $t=3$ is $\sqrt{2}\left(e^3-1\right)$ .