In Exercises 16–23 sketch the parametric curve by plotting points$\mathrm{x}={\mathrm{t}}^3-\mathrm{t},\mathrm{y}={\mathrm{t}}^3+\mathrm{t},\mathrm{t}\in \mathrm{ℝ}$

The given plotting points are$\mathrm{x}={\mathrm{t}}^3-\mathrm{t},\mathrm{y}={\mathrm{t}}^3+\mathrm{t},\mathrm{t}\in \mathrm{ℝ}$

Take the polar coordinates into consideration $\left(1,\frac{\mathrm{\pi }}{4}\right),\left(2,\frac{\mathrm{\pi }}{4}\right),\left(3,\frac{\mathrm{\pi }}{4}\right),\left(4,\frac{\mathrm{\pi }}{4}\right)$

The goal is to transform polar coordinates into rectangle coordinates

In this case $\left(1,\frac{\mathrm{\pi }}{4}\right)$, then $\mathrm{r}=1,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

Apply the formulas. $\mathrm{x}=\mathrm{rcos}\mathrm{\theta },\mathrm{y}=\mathrm{rsin}\mathrm{\theta }$

using, width="65">x=rcos⁡θand put width="83">r=1,θ=π4, then

localid="1651900923789" width="232">x=1⋅cos⁡π4x=1⋅12sincecos⁡π4=12

The value is then, $\mathrm{x}=\frac{1}{\sqrt{2}}$

Now, using $\mathrm{y}=\mathrm{rsin}\mathrm{\theta }$and put $\mathrm{r}=1,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

$\begin{array}{r}\mathrm{y}=1\cdot \sin \frac{\mathrm{\pi }}{4}\\ \mathrm{y}=1\cdot \frac{1}{\sqrt{2}}\left[\text{since}\sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\right]\\ \mathrm{y}=\frac{1}{\sqrt{2}}\end{array}$

rectangle position are$(\mathrm{x},\mathrm{y})=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

As a result, rectangular coordinates are $(\mathrm{x},\mathrm{y})=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$

To illustrate the point $\left(2,\frac{\mathrm{\pi }}{4}\right)$, then $\mathrm{r}=2,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

Use the formulas,$\mathrm{x}=\mathrm{rcos}\mathrm{\theta },\mathrm{y}=\mathrm{rsin}\mathrm{\theta }$

Taken,$\mathrm{x}=\mathrm{rcos}\mathrm{\theta }$and put $\mathrm{r}=2,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$, then

$\begin{array}{r}\mathrm{x}=2\cdot \cos \left(\frac{\mathrm{\pi }}{4}\right)\\ \mathrm{x}=2\cdot \frac{1}{\sqrt{2}}\left[\text{since}\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\right]\end{array}$

finally , the value is $\mathrm{x}=\frac{2}{\sqrt{2}}$

We can taking $\mathrm{y}=\mathrm{rsin}\mathrm{\theta }$and put $\mathrm{r}=2,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

$\begin{array}{r}\mathrm{y}=2\cdot \sin \frac{\mathrm{\pi }}{4}\\ \mathrm{y}=2\cdot \frac{1}{\sqrt{2}}\left[\text{since}\sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\right]\end{array}$

That is, $\mathrm{y}=\frac{2}{\sqrt{2}}$

rectangle position are role="math" localid="1651903134558" $(\mathrm{x},\mathrm{y})=\left(\frac{2}{\sqrt{2}},\frac{2}{\sqrt{2}}\right)$

As a result, rectangular coordinates are $(\mathrm{x},\mathrm{y})=\left(\frac{2}{\sqrt{2}},\frac{2}{\sqrt{2}}\right)$

To illustrate the point$\left(3,\frac{\mathrm{\pi }}{4}\right)$, then $\mathrm{r}=3,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

Use the formulas,

Taken,and put , then

finally , the value is

We can takingand put