Chapter 10, Problem 47RE

(a)

To determine

The length of the velocity at $t=3$

Answer to Problem 47RE

The length of the velocity at $t=3$ is $3.238$

Explanation of Solution

Given:

The position of particle in coordinates are given as:

  $x=3\cos \left(\frac{\pi }{4}t\right),y=5\sin \left(\frac{\pi }{4}t\right)$

Calculation:

The length of the velocity can be find out as:

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

As given, $x=3\cos \left(\frac{\pi }{4}t\right),y=5\sin \left(\frac{\pi }{4}t\right)$

Now,

  $\begin{array}{l}\frac{dx}{dt}=3.\left(-\sin \left(\frac{\pi t}{4}\right)\right).\frac{\pi }{4}\\ \frac{dx}{dt}=-\frac{3\pi }{4}.\sin \left(\frac{\pi t}{4}\right)\end{array}$

  $\begin{array}{l}\frac{dy}{dt}=5.\left(\cos \left(\frac{\pi t}{4}\right)\right).\frac{\pi }{4}\\ \frac{dy}{dt}=\frac{5\pi }{4}.cos\left(\frac{\pi t}{4}\right)\end{array}$

Now, $\text{At t=3}$

  $\begin{array}{l}{\left.\frac{dx}{dt}\right|}_{t=3}=-\frac{3\pi }{4}.\sin \left(\frac{\pi .3}{4}\right)\\ {\left.\frac{dx}{dt}\right|}_{t=3}=-\frac{3\pi }{4}.sin\left(\frac{3\pi }{4}\right)\\ {\left.\frac{dx}{dt}\right|}_{t=3}=-\frac{3\pi }{4}.\left(\frac{\sqrt{2}}{2}\right)\\ {\left.\frac{dx}{dt}\right|}_{t=3}=-\frac{3\sqrt{2}\pi }{8}\end{array}$

  $\begin{array}{l}{\left.\frac{dy}{dt}\right|}_{t=3}=\frac{5\pi }{4}.cos\left(\frac{\pi .3}{4}\right)\\ {\left.\frac{dy}{dt}\right|}_{t=3}=\frac{5\pi }{4}.cos\left(\frac{3\pi }{4}\right)\\ {\left.\frac{dy}{dt}\right|}_{t=3}=\frac{5\pi }{4}.\left(-\frac{\sqrt{2}}{2}\right)\\ {\left.\frac{dy}{dt}\right|}_{t=3}=-\frac{5\sqrt{2}\pi }{8}\end{array}$

Substitute the values to find the velocity.

  $\begin{array}{l}v\left(3\right)=〈-\frac{3\sqrt{2}\pi }{8},-\frac{5\sqrt{2}\pi }{8}〉\\ \left|v\left(3\right)\right|=\sqrt{{\left(-\frac{3\sqrt{2}\pi }{8}\right)}^2+{\left(-\frac{5\sqrt{2}\pi }{8}\right)}^2}\\ \left|v\left(3\right)\right|=\sqrt{\frac{9.2{\pi }^2}{64}+\frac{25.2{\pi }^2}{64}}\\ \left|v\left(3\right)\right|=\sqrt{\frac{68{\pi }^2}{64}}\\ \left|v\left(3\right)\right|=\sqrt{\frac{\mathrm{2.2.17}{\pi }^2}{64}}\\ \left|v\left(3\right)\right|=\frac{\sqrt{17}\pi }{4}\\ \left|v\left(3\right)\right|=3.238\end{array}$

Conclusion:

The length of the velocity at $t=3$ is $3.238$

(b)

To determine

The x and y components of the acceleration.

Answer to Problem 47RE

The x and y components are $\frac{3\sqrt{2}{\pi }^2}{32}$ and $\frac{5\sqrt{2}{\pi }^2}{32}$

Explanation of Solution

Given:

The position of particle in coordinates are given as:

  $x=3\cos \left(\frac{\pi }{4}t\right),y=5\sin \left(\frac{\pi }{4}t\right)$

Calculation:

Acceleration can be find out as:

  $a(t)=〈\frac{d^{2}x}{d{t}^2},\frac{d^{2}y}{d{t}^2}〉$

Now,

  $\begin{array}{l}\frac{dy}{dt}=\frac{5\pi }{4}.cos\left(\frac{\pi t}{4}\right)\\ \frac{d^{2}y}{d{t}^2}=\frac{d\left(\frac{dy}{dt}\right)}{dt}\\ \frac{d^{2}y}{d{t}^2}=\frac{5\pi }{4}\left(-\sin \left(\frac{\pi t}{4}\right)\right).\frac{\pi }{4}\\ \frac{d^{2}y}{d{t}^2}=-\frac{5{\pi }^2}{16}\sin \left(\frac{\pi t}{4}\right)\end{array}$

  $\text{At t=3}$

  $\begin{array}{l}{\left.\frac{d^{2}x}{d{t}^2}\right|}_{t=3}=-\frac{3{\pi }^2}{16}\cos \left(\frac{\pi .3}{4}\right)\\ {\left.\frac{d^{2}x}{d{t}^2}\right|}_{t=3}=-\frac{3{\pi }^2}{16}\left(-\frac{\sqrt{2}}{2}\right)\\ {\left.\frac{d^{2}x}{d{t}^2}\right|}_{t=3}=\frac{3\sqrt{2}{\pi }^2}{32}\\ {\left.\frac{d^{2}x}{d{t}^2}\right|}_{t=3}=\frac{3{\pi }^2}{16\sqrt{2}}\end{array}$

  $\begin{array}{l}\frac{dy}{dt}=\frac{5\pi }{4}.cos\left(\frac{\pi t}{4}\right)\\ {\left.\frac{d^{2}y}{d{t}^2}\right|}_{t=3}=-\frac{5{\pi }^2}{16}\sin \left(\frac{\pi .3}{4}\right)\\ {\left.\frac{d^{2}y}{d{t}^2}\right|}_{t=3}=-\frac{5{\pi }^2}{16}\left(\frac{\sqrt{2}}{2}\right)\\ {\left.\frac{d^{2}y}{d{t}^2}\right|}_{t=3}=-\frac{5\sqrt{2}{\pi }^2}{32}\\ {\left.\frac{d^{2}y}{d{t}^2}\right|}_{t=3}=-\frac{5{\pi }^2}{16\sqrt{2}}\end{array}$

  $a(3)=〈\frac{3\sqrt{2}{\pi }^2}{32},\frac{5\sqrt{2}{\pi }^2}{32}〉$

So, the x- component is $\frac{3\sqrt{2}{\pi }^2}{32}$ and y-component is $\frac{5\sqrt{2}{\pi }^2}{32}$

Conclusion:

The x and y components are $\frac{3\sqrt{2}{\pi }^2}{32}$ and $\frac{5\sqrt{2}{\pi }^2}{32}$

(c)

To determine

A single equation in x and y.

Answer to Problem 47RE

The equation is $\frac{x^2}{9}+\frac{y^2}{25}=1$

Explanation of Solution

Given:

The position of particle in coordinates are given as:

  $x=3\cos \left(\frac{\pi }{4}t\right),y=5\sin \left(\frac{\pi }{4}t\right)$

Calculation:

  $\begin{array}{l}x=3\cos \left(\frac{\pi t}{4}\right);\frac{x}{3}=\cos \left(\frac{\pi t}{4}\right)\\ y=5\sin \left(\frac{\pi t}{4}\right);\frac{y}{5}=\sin \left(\frac{\pi t}{4}\right)\\ \text{As},{\sin }^2\theta +{\cos }^2\theta =1\\ {\left(\frac{x}{3}\right)}^2+{\left(\frac{y}{5}\right)}^2={\cos }^2\left(\frac{\pi t}{4}\right)+{\sin }^2\left(\frac{\pi t}{4}\right)\\ \frac{x^2}{9}+\frac{y^2}{25}=1\end{array}$

Conclusion:

The equation is $\frac{x^2}{9}+\frac{y^2}{25}=1$