Chapter 11.2, Problem 49E

(a)

To determine

To find: The slope of the path.

Answer to Problem 49E

The slope of the path is $-1$ .

Explanation of Solution

Differentiate the component of the position vector,

  $\frac{d}{dt}\langle e^t\sin t,e^t\cos t\rangle =\langle e^t\sin t+e^t\cos t,e^t\cos t-e^t\sin t\rangle$

The slope of the path at $t=\frac{\pi }{2}$ is calculated as,

  $\begin{array}{c}m=\frac{e^t\cos t-e^t\sin t}{e^t\sin t+e^t\cos t}\\ =-1\end{array}$

Therefore, the slope is $-1$ .

(b)

To determine

To find: The speed of the particle.

Answer to Problem 49E

The speed of the particle is $3.844\text{m}/\text{s}$ .

Explanation of Solution

The velocity vector is calculated as,

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

The velocity at $t=1$ is,

  $\begin{array}{c}v\left(1\right)=\langle e^1\sin 1+e^1\cos 1,e^1\cos 1-e^1\sin 1\rangle \\ =\langle 3.756,-0.871\rangle \end{array}$

The magnitude of the velocity is calculated as,

  $\begin{array}{c}v=\sqrt{{3.756}^2+{\left(-0.817\right)}^2}\\ =3.844\end{array}$

Therefore, the speed of the particle is $3.844\text{m}/\text{s}$ .

(c)

To determine

To find: The distance travelled by the particle.

Answer to Problem 49E

The distance travelled by the particle is $2.430$ .

Explanation of Solution

The distance travelled by the particle is calculated as,

  $\begin{array}{c}d={\displaystyle {\int }_0^1\sqrt{{\left(v_1\left(t\right)\right)}^2+{\left(v_2\left(t\right)\right)}^2}dt}\\ ={\displaystyle {\int }_0^1\sqrt{{\left(e^t\sin t+e^t\cos t\right)}^2+{\left(e^t\cos t-e^t\sin t\right)}^2}}\\ =2.430\end{array}$

Therefore, the distance travelled by the particle is 2.430.