Chapter 10, Problem 48RE

(a)

To determine

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

Answer to Problem 48RE

The slope is $1$

Explanation of Solution

Given:

The position of particle in parametric equations are given as:

  $x=e^t\cos t,y=e^t\sin t$

Calculation:

To find the slope, $\frac{dy}{dx}$ needs to be calculated.

To calculate $\frac{dy}{dx}$ chain rule can be used such as;

  $\begin{array}{l}\frac{dy}{dt}=\frac{dy}{dx}.\frac{dx}{dt}\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{array}$

Now, find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

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

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

Hence, the slope can be find out as:

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

At $t=\pi$

  $\begin{array}{l}{\left[\frac{dy}{dx}\right]}_{t=\pi }=\frac{\sin \pi +\cos \pi }{\cos \pi -\sin \pi }\\ {\left[\frac{dy}{dx}\right]}_{t=\pi }=\frac{0+\left(-1\right)}{-1-0}\\ {\left[\frac{dy}{dx}\right]}_{t=\pi }=1\end{array}$

Conclusion:

The slope is $1$

(b)

To determine

The speed of the particle when $t=3$

Answer to Problem 48RE

The speed of the particle is approximately $28.405$

Explanation of Solution

Given:

The position of particle in parametric equations are given as:

  $x=e^t\cos t,y=e^t\sin t$

Calculation:

As, the speed of a particle at a time t is equal to the magnitude of velocity vector.

So, it can be written as:

  $v(t)=〈\frac{dx}{dt},\frac{dy}{dt}〉$

Also,

  $\begin{array}{l}v=〈v_1,v_2〉\\ \left|v\right|=\sqrt{v_1^2+v_2^2}\end{array}$

Substitute the value of $\frac{dx}{dt}$ and $\frac{dy}{dt}$ in the above formula.

  $v(t)=〈e^t.\left(\cos t-\sin t\right),e^t.\left(\sin t+\cos t\right)〉$

  $\begin{array}{l}\left|v\left(t\right)\right|=\sqrt{{\left(e^t\left(\cos t-\sin t\right)\right)}^2+{\left(e^t\left(\sin t+\cos t\right)\right)}^2}\\ \left|v\left(t\right)\right|=\sqrt{{\left(e^t\right)}^2{\left(\cos t-\sin t\right)}^2+{\left(e^t\right)}^2{\left(\sin t+\cos t\right)}^2}\\ \left|v\left(t\right)\right|=\sqrt{e^{2t}\left({\cos }^{2}t-2\cos t\sin t+{\sin }^{2}t\right)+e^{2t}\left({\sin }^{2}t+2\sin t\cos t+{\cos }^{2}t\right)}\\ \left|v\left(t\right)\right|=\sqrt{e^{2t}\left(1-2\cos t\sin t\right)+e^{2t}\left(1+2\sin t\cos t\right)}\\ \left|v\left(t\right)\right|=\sqrt{e^{2t}\left(1-2\cos t\sin t+1+2\cos t\sin t\right)}\\ \left|v\left(t\right)\right|=\sqrt{e^{2t}.2}\\ \left|v\left(t\right)\right|=\sqrt{e^{2t}}.\sqrt{2}\\ \left|v\left(t\right)\right|=\sqrt{2}{e}^t\\ \text{At t = 3}\\ \left|v\left(3\right)\right|=\sqrt{2}{e}^3\\ \left|v\left(3\right)\right|=28.405\end{array}$

Conclusion:

The speed of the particle is approximately $28.405$

(c)

To determine

The distance travelled by the particle from $t=0\text{ to t = 3}$

Answer to Problem 48RE

The distance travelled by the particle from $t=0\text{ to t = 3}$ is approximately $26.991$

Explanation of Solution

Given:

The position of particle in parametric equations are given as:

  $x=e^t\cos t,y=e^t\sin t$

Calculation:

  $\begin{array}{l}\underset{a}{\overset{b}{{\displaystyle \int }}}\left|v\left(t\right)\right|dt=\underset{a}{\overset{b}{{\displaystyle \int }}}\sqrt{v_1^2(t)+v_2^2(t)dt}\hfill \\ v_1\left(t\right)=e^t.\left(\cos t-\sin t\right)\hfill \\ v_2(t)=e^t.\left(\sin t+\cos t\right)\hfill \end{array}$

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

  $\left|v(t)\right|=\sqrt{e^{2t}.\left({\cos }^{2}t-2\cos t\sin t+{\sin }^{2}t\right)+e^{2t}\left({\sin }^{2}t+2\sin t\cos t+{\cos }^{2}t\right)}$

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

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

  $\begin{array}{l}\underset{0}{\overset{3}{{\displaystyle \int }}}\left|v(t)\right|dt=\sqrt{2}\left(e^3-1\right)\hfill \\ \underset{0}{\overset{3}{{\displaystyle \int }}}\left|v(t)\right|dt=26.991\hfill \end{array}$

Conclusion:

The distance travelled by the particle from $t=0\text{ to t = 3}$ is approximately $26.991$