Q. 21 - 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.

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

Step-by-Step Solution

Answer

The tabular representation of the points is as follows,

\begin{tabular}{|l|l|l|l|l|l|}

\hline$t$ & $-2$ & $-1$ & 0 & 1 & 2 \\

\hline$x$ & $-6$ & $-2$ & 0 & 0 & 6 \\

\hline$y$ & $-10$ & $-2$ & 0 & 2 & 10 \\

\hline

\end{tabular}

The graphical representation is shown as below,

Step 1: Given information

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

Step 2: Calculation

Consider the parametric curves $x=t^{3}-t, y=t^{3}+t, t \in \mathbb{R}$.

The objective is to sketch the parametric curve.

To draw the graph for the parametric equations assume $t=-2,-1,0,1,2$.

Substitute different $t$ values in the parametric equations and find the values of $x, y$.

The point $(x, y)$ When $t=-2$ is,

$$

(x, y)=\left(t^{3}-t, t^{3}+t\right)

$$

$(x, y)=\left((-2)^{3}-(-2),(-2)^{3}+(-2)\right) \quad[$ since by substituting $t=-2]$

$$

\begin{aligned}

&(x, y)=(-8+2,-8-2) \\

&(x, y)=(-6,-10) \text { simplify }

\end{aligned}

$$

The point $(x, y)$ When $t=-1$ is,

$$

(x, y)=\left(t^{3}-t, t^{3}+t\right)

$$

$(x, y)=\left((-1)^{3}-1,(-1)^{3}+(-1)\right)[$ since by substituting $t=-1]$

$$

\begin{aligned}

&(x, y)=(-1-1,-1-1) \\

&(x, y)=(-2,-2) \text { simplify }

\end{aligned}

$$

The point $(x, y)$ When $t=0$ is,

$$

\begin{aligned}

&(x, y)=\left(t^{3}-t, t^{3}+t\right) \\

&\left.(x, y)=\left(0^{3}-0,0^{3}+0\right) \quad \text { [since by substituting } t=0\right] \\

&(x, y)=(0,0)

\end{aligned}

$$

The point $(x, y)$ When $t=1$ is,

$$

\begin{aligned}

&(x, y)=\left(t^{3}-t, t^{3}+t\right) \\

&\left.(x, y)=\left(1^{3}-1,1^{3}+1\right) \text { [since by substituting } t=1\right] \\

&(x, y)=(0,2) \text { simplify }

\end{aligned}

$$

The point $(x, y)$ When $t=2$ is,

$$

\begin{aligned}

&(x, y)=\left(t^{3}-t, t^{3}+t\right) \\

&\left.(x, y)=\left(2^{3}-2,2^{3}+2\right) \text { [since by substituting } t=2\right] \\

&(x, y)=(8-2,8+2) \\

&(x, y)=(6,10) \text { simplify }

\end{aligned}

$$

The tabular representation of the points is as follows,

\begin{tabular}{|l|l|l|l|l|l|}

\hline$t$ & $-2$ & $-1$ & 0 & 1 & 2 \\

\hline$x$ & $-6$ & $-2$ & 0 & 0 & 6 \\

\hline$y$ & $-10$ & $-2$ & 0 & 2 & 10 \\

\hline

\end{tabular}

The graphical representation is shown as below,

Therefore the solution is a required graph.