Chapter 8.2, Problem 73E

Solution

(a.)

The plot of $\left(x,y\right)$ relation for velocity versus position for $0\le t\le 2\pi$ .

The plot of $\left(x,y\right)$ relation for velocity versus position for $0\le t\le 2\pi$ has been provided.

Given:

As a pendulum swings toward and away from a motion detector, its distance (in meters) from the detector is given by the position function $x\left(t\right)=3+\cos \left(2t-5\right)$ , where $t$ represents time (in seconds). The velocity (in m/s) of the pendulum is given by $y\left(t\right)=-2\sin \left(2t-5\right)$ .

Concept used:

Parametric equations $x\left(t\right)$ and $y\left(t\right)$ can be plotted on the $xy$ plane for a given bound of $t$ .

Calculation:

The given parametric equations are $x\left(t\right)=3+\cos \left(2t-5\right)$ and $y\left(t\right)=-2\sin \left(2t-5\right)$ .

Plotting these parametric equations on the $xy$

plane as follows:

  

Conclusion:

The plot of $\left(x,y\right)$ relation for velocity versus position for $0\le t\le 2\pi$ has been provided.

(b.)

The equation of the resulting conic in standard form, in terms of $x$ and $y$ , and eliminating the parameter $t$ .

It has been determined that the equation of the resulting conic in standard form, in terms of $x$ and $y$ , and eliminating the parameter $t$ , is $\frac{{\left(x-3\right)}^2}{1^2}+\frac{y^2}{2^2}=1$ .

Given:

As a pendulum swings toward and away from a motion detector, its distance (in meters) from the detector is given by the position function $x\left(t\right)=3+\cos \left(2t-5\right)$ , where $t$ represents time (in seconds). The velocity (in m/s) of the pendulum is given by $y\left(t\right)=-2\sin \left(2t-5\right)$ .

Concept used:

Parametric equations $x\left(t\right)$ and $y\left(t\right)$ can be plotted on the $xy$ plane for a given bound of $t$ .

Calculation:

The given parametric equations are $x\left(t\right)=3+\cos \left(2t-5\right)$ and $y\left(t\right)=-2\sin \left(2t-5\right)$ .

Simplifying,

  $x-3=\cos \left(2t-5\right)$ ,

  $\frac{y}{2}=-\sin \left(2t-5\right)$

Squaring both sides of each equation,

  ${\left(x-3\right)}^2={\cos }^2\left(2t-5\right)$ ,

  $\frac{y^2}{4}={\sin }^2\left(2t-5\right)$

Adding both equations,

  ${\left(x-3\right)}^2+\frac{y^2}{4}={\cos }^2\left(2t-5\right)+{\sin }^2\left(2t-5\right)$

Using trigonometric identity; ${\sin }^2\theta +{\cos }^2\theta =1$ ,

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

Converting to standard form,

  $\frac{{\left(x-3\right)}^2}{1^2}+\frac{y^2}{2^2}=1$

This is the required equation of the resulting conic in standard form.

Conclusion:

It has been determined that the equation of the resulting conic in standard form, in terms of $x$ and $y$ , and eliminating the parameter $t$ , is $\frac{{\left(x-3\right)}^2}{1^2}+\frac{y^2}{2^2}=1$ .