Q. 21 - Calculus

Question

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

Step-by-Step Solution

Answer

The Resultant Answer is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}},\frac{2}{\sqrt{2}}\right)\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)\left(\frac{4}{\sqrt{2}},\frac{4}{\sqrt{2}}\right)$

Step 1: Given information

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

Step 2: Simplications

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" style="max-width: none; vertical-align: -4px;" $\mathrm{x}=\mathrm{rcos}\mathrm{\theta }$ and put width="83" style="max-width: none; vertical-align: -15px;" $\mathrm{r}=1,\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$, then

width="232" style="max-width: none; vertical-align: -46px;" $$\begin{gathered} \mathrm{x}=1\cdot \cos \frac{\mathrm{\pi }}{4} \\ \mathrm{x}=1\cdot \frac{1}{\sqrt{2}}\left[\text{ since }\cos \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\right] \end{gathered}$$

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

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 taking  and put