Q. 23 - Calculus

Question

In Exercises 17-23 the polar coordinates for several sets of points are given. Find the rectangular coordinates for each of the points, and then plot and label the points in the same polar coordinate system.

$\left(1,-\frac{\pi }{2}\right),\left(2,-\frac{\pi }{2}\right),\left(3,-\frac{\pi }{2}\right)$ and $\left(4,-\frac{\pi }{2}\right)$

Step-by-Step Solution

Answer

The rectangular coordinates of all the points are$(0,-1),(0,-2),(0,-3),(0,-4)$.

The graphical representation is as follows,

Step 1: Given information

The points in the same polar coordinate system $\left(1,-\frac{\pi }{2}\right),\left(2,-\frac{\pi }{2}\right),\left(3,-\frac{\pi }{2}\right)$ and $\left(4,-\frac{\pi }{2}\right)$

Step 2: Calculation

Consider the polar coordinates $\left(1,-\frac{\pi }{2}\right),\left(2,-\frac{\pi }{2}\right),\left(3,-\frac{\pi }{2}\right)$and $\left(4,-\frac{\pi }{2}\right)$.

The objective is to convert the polar coordinates to the rectangular coordinates.

In the point $\left(1,-\frac{\pi }{2}\right)$, then$r=1,\theta =-\frac{\pi }{2}$.

Use the equations $x=r\cos \theta ,y=r\sin \theta$

Take $x=r\cos \theta$ and substitute width="101" style="max-width: none; vertical-align: -15px;" $r=1,\theta =-\frac{\pi }{2}$ then,

$x=1\times \cos \left(-\frac{\pi }{2}\right)$

$x=1\times 0\left[since\cos \frac{\pi }{2}=0\right]$

Then the value is x=0.

Now take $y=r\sin \theta$ and substitute $r=1,\theta =-\frac{\pi }{2}$ then,

$$\begin{gathered} y=1·\sin \left(-\frac{\pi }{2}\right) \\ y=1·-1\left[\text{ since }\sin \left(-\frac{\pi }{2}\right)=-1\right] \\ y=-1 \end{gathered}$$

The rectangular coordinate $(x, y)=(0,-1)$.

Therefore, the rectangular coordinates are $(x, y)=(0,-1)$.

Step 3: Simplification

For the point $\left(2,-\frac{\pi }{2}\right)$ then $r=2,\theta =-\frac{\pi }{2}$.

Use the equations $x=r\cos \theta ,y=r\sin \theta$,

Take $x=r\cos \theta$ and substitute $r=2,\theta =-\frac{\pi }{2}$ then

$$\begin{gathered} x=2\times \cos \left(-\frac{\pi }{2}\right) \\ x=2\times 0\left[\text{ since }\cos \frac{\pi }{2}=0\right] \end{gathered}$$

Then the value is x=0.

Now take $y=r\sin \theta$and substitute $r=2,\theta =-\frac{\pi }{2}$ then

$$\begin{gathered} y=2\times \sin \left(-\frac{\pi }{2}\right) \\ y=2\times -1\left[\text{ since }\sin \left(-\frac{\pi }{2}\right)=-1\right] \end{gathered}$$

Thus, $y=-2$.

The rectangular coordinate $(x, y)=(0,-2)$.

Therefore, the rectangular coordinates are $(x, y)=(0,-2)$.

Step 4: Further calculation

For the point $\left(3,-\frac{\pi }{2}\right)$ then$r=3,\theta =-\frac{\pi }{2}$.

Use the equations$x=r\cos \theta ,y=r\sin \theta$,

Take $x=r\cos \theta$ and substitute $r=3,\theta =-\frac{\pi }{2}$ then

$$\begin{gathered} x=3\times \cos \left(-\frac{\pi }{2}\right) \\ x=3\times 0\left[since\cos \frac{\pi }{2}=0\right] \end{gathered}$$

Then the value is x=0.

Now take $y=r\sin \theta$ and substitute $r=3,\theta =-\frac{\pi }{2}$ then

$y=3·\sin \left(-\frac{\pi }{2}\right)$

$y=3\times -1\left[since\sin \frac{\pi }{2}=1\right]$

Thus, y = -3 .

The rectangular coordinate $(x, y)=(0,-3)$.

Therefore, the rectangular coordinates are $(x, y)=(0,-3)$.

For the point $\left(4,-\frac{\pi }{2}\right)$ then $r=4,\theta =-\frac{\pi }{2}$.

Use the equations $x=r\cos \theta ,y=r\sin \theta$.

Take $x=r\cos \theta$ and substitute $r=4,\theta =-\frac{\pi }{2}$ then

$$\begin{gathered} x=4\times \cos \left(-\frac{\pi }{2}\right) \\ x=4\times 0 \left[\text{ since }\cos \frac{\pi }{2}=0\right] \end{gathered}$$

Then the value is x=0.

Now take $y=r\sin \theta$ and substitute $r=4,\theta =-\frac{\pi }{2}$ then

$$\begin{gathered} y=4\times \sin \left(-\frac{\pi }{2}\right) \\ y=4\times -1\left[\text{ since }\sin \frac{\pi }{2}=1\right] \end{gathered}$$

Thus, y=-4

The rectangular coordinate is $(x, y)=(0,-4)$.

Therefore, the rectangular coordinates are $(x, y)=(0,-4)$.

The rectangular coordinates of all the points are $(0,-1),(0,-2),(0,-3),(0,-4)$.

Therefore, the answer is $(0,-1),(0,-2),(0,-3),(0,-4)$.

The graphical representation is as follows,

 

Hence the solution.