Chapter 11.4, Problem 11.91P

The motion of a particle is defined by the equations $\frac{x=\left(4\cos \pi t-1\right)}{\left(2-\cos \pi t\right)}$ and $\frac{y=3\sin \pi t}{\left(2-\cos \pi t\right)}$ , where x and y are expressed in feet and t is expressed in seconds. Show that the path of the particle is part of the ellipse shown, and determine the velocity when (a) $t=0$ , (b) $t=1/3$ s, (c) $t=1$ s.

  

To determine

(a)

To show:

The path of the particle is the part of the ellipse 0 $\frac{x^2}{4}+\frac{y^2}{3}=1$.

The velocity when the $t=0$.

Answer to Problem 11.91P

The velocity of the particle when $t=0$ is $2.356\text{units}$

Explanation of Solution

Given:

The motion of particle is governed by the equations $x=\frac{\left(4\text{cos}\pi \text{t-2}\right)}{\left(2-\cos \pi t\right)}$ and $y=\frac{3\text{sin}\pi \text{t}}{\left(2-\cos \pi t\right)}$.

Equation of the ellipse is:

$\frac{x^2}{4}+\frac{y^2}{3}=1$ ...... (1)

Concept used:

Write the formula for velocity as derivative of displacement in $x-$ axis.

$v_x=\frac{dx}{dt}$ ...... (2)

Write the formula for velocity as derivative of displacement in $y-$ axis.

$v_y=\frac{dy}{dt}$ ...... (3)

Substitute $\frac{\left(4\text{cos}\pi \text{t-2}\right)}{\left(2-\cos \pi t\right)}$ for $x$ in the equation 2.

$\begin{array}{c}{v}_x=\frac{d}{dt}\left(\frac{\left(4\text{cos}\pi \text{t-2}\right)}{\left(2-\cos \pi t\right)}\right)\\ =\frac{-6\pi \sin \pi t}{{\left(2-\cos \pi t\right)}^2}\end{array}$

Therefore $v_x=\frac{-6\pi \sin \pi t}{{\left(2-\cos \pi t\right)}^2}$ ...... (4)

Substitute $\frac{3\text{sin}\pi \text{t}}{\left(2-\cos \pi t\right)}$ for $y$ in the equation 2.

$\begin{array}{c}{v}_y=\frac{d}{dt}\left(\frac{3\text{sin}\pi \text{t}}{\left(2-\cos \pi t\right)}\right)\\ =\frac{6\pi \text{cos}\pi t-3\pi }{{\left(2-\cos \pi t\right)}^2}\end{array}$

Therefore $v_y=\frac{6\pi \text{cos}\pi t-3\pi }{{\left(2-\cos \pi t\right)}^2}$ ...... (5)

Write the formula for velocity.

$v=\sqrt{v_x{}^2+v_y{}^2}$ ......(6)

Calculation:

Substitute $\frac{\left(4\text{cos}\pi \text{t-2}\right)}{\left(2-\cos \pi t\right)}$ for $x$ and $\frac{3\text{sin}\pi \text{t}}{\left(2-\cos \pi t\right)}$ for $y$ in the left-side of equation 1.

$\begin{array}{c}\frac{{\left(4\text{cos}\pi \text{t-2}\right)}^2}{4{\left(2-\cos \pi t\right)}^2}+\frac{{\left(3\text{sin}\pi \text{t}\right)}^2}{3{\left(2-\cos \pi t\right)}^2}=\frac{{\left(\text{2cos}\pi \text{t-1}\right)}^2}{{\left(2-\cos \pi t\right)}^2}+\frac{3{\left(\text{sin}\pi \text{t}\right)}^2}{{\left(2-\cos \pi t\right)}^2}\\ =\frac{4{\text{cos}}^2\pi t-4\text{cos}\pi t+1+3{\sin }^2\pi }{\left(4+{\cos }^2\pi t-4\cos \pi t\right)}\\ =\frac{4+{\cos }^2\pi t-4\cos \pi t}{\left(4+{\cos }^2\pi t-4\cos \pi t\right)}\\ =1\end{array}$

Therefore, the right- side after solving the above equation is equal to right side of equation 1.

Substitute $0\text{seconds }$ for $t$ in equation 4.

$\begin{array}{c}{v}_x=\frac{-6\pi \sin \pi t}{{\left(2-\cos \pi t\right)}^2}\\ =\frac{-6\pi \sin \pi \left(0\right)}{{\left(2-\cos \pi \left(0\right)\right)}^2}\\ =0\text{units}\end{array}$

Substitute $0\text{seconds }$ for $t$ in equation 5.

$\begin{array}{c}{v}_y=\frac{6\pi \text{cos}\pi \left(0\right)-3\pi }{{\left(2-\cos \pi \left(0\right)\right)}^2}\\ =\frac{6\pi -3\pi }{4}\\ =2.356\text{units}\end{array}$

Substitute $0$ for $v_x$ and $2.356\text{units}$ for $v_y$ in equation 6.

$\begin{array}{c}v=\sqrt{0^2+{\left(2.356\right)}^2}\\ =2.356\text{units}\end{array}$

Conclusion:

Thus,the path of the particle is the part of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$

The velocity of the particle when $t=0$ is $2.356\text{units}$.

To determine

(b)

The velocity of the particle.

Answer to Problem 11.91P

The velocity of the particle when $t=\frac{1}{3}\text{s}$ is $7.255\text{units}$.

Explanation of Solution

Calculation:

Substitute $\frac{1}{3}\text{seconds }$ for $t$ in equation 4.

$\begin{array}{c}{v}_x=\frac{-6\pi \sin \pi \left(\frac{1}{3}\right)}{{\left(2-\cos \pi \left(\frac{1}{3}\right)\right)}^2}\\ =\frac{-6\pi \left(0.866\right)}{{\left(2-0.5\right)}^2}\\ =-7.255\text{units}\end{array}$

Substitute $\frac{1}{3}\text{seconds }$ for $t$ in equation 5.

$\begin{array}{c}{v}_y=\frac{6\pi \text{cos}\pi \left(\frac{1}{3}\right)-3\pi }{{\left(2-\cos \pi \left(\frac{1}{3}\right)\right)}^2}\\ =\frac{3\pi -3\pi }{{1.5}^2}\\ =0\text{units}\end{array}$

Substitute $-7.255\text{units}$ for $v_x$ and $0$ for $v_y$ in equation 6.

$\begin{array}{c}v=\sqrt{{\left(-7.255\right)}^2+{\left(0\right)}^2}\\ =7.255\text{units}\end{array}$

Conclusion:

Thus,the velocity of the particle when $t=\frac{1}{3}\text{s}$ is $7.255\text{units}$.

To determine

(c)

The velocity of the particle.

Answer to Problem 11.91P

The velocity of the particle when $t=1\text{s}$ is $3.14\text{units}$.

Explanation of Solution

Calculation:

Substitute $1\text{seconds }$ for $t$ in equation 4.

$\begin{array}{c}{v}_x=\frac{-6\pi \sin \pi }{{\left(2-\cos \pi \right)}^2}\\ =\frac{0}{{\left(2-\cos \pi \right)}^2}\\ =0\text{units}\end{array}$

Substitute $1\text{seconds }$ for $t$ in equation 5.

$\begin{array}{c}{v}_y=\frac{6\pi \text{cos}\pi -3\pi }{{\left(2-\cos \pi \right)}^2}\\ =\frac{-6\pi -3\pi }{3^2}\\ =-\pi \text{units}\end{array}$

Substitute $0\text{units}$ for $v_x$ and $-\pi \text{units}$ for $v_y$ in equation 6.

$\begin{array}{c}v=\sqrt{{\left(0\right)}^2+{\left(-\pi \right)}^2}\\ =3.14\text{units}\end{array}$

Conclusion:

Thus,the velocity of the particle when $t=1\text{s}$ is $3.14\text{units}$.