91影视

The given plotting points are$\mathrm{x}={\mathrm{t}}^3鈭�\mathrm{t},\mathrm{y}={\mathrm{t}}^3+\mathrm{t},\mathrm{t}鈭�\mathrm{鈩�}$

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

The goal is to transform polar coordinates into rectangle coordinates

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

Apply the formulas. $\mathrm{x}=\mathrm{rcos}鈦�\mathrm{胃},\mathrm{y}=\mathrm{rsin}鈦�\mathrm{胃}$

using, width="65">x=rcos鈦�胃and put width="83">r=1,胃=蟺4, then

localid="1651900923789" width="232">x=1鈰�cos鈦�蟺4x=1鈰�12sincecos鈦�蟺4=12

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

Now, using $\mathrm{y}=\mathrm{rsin}鈦�\mathrm{胃}$and put $\mathrm{r}=1,\mathrm{胃}=\frac{\mathrm{蟺}}{4}$

$\begin{array}{r}\mathrm{y}=1鈰�\sin 鈦�\frac{\mathrm{蟺}}{4}\\ \mathrm{y}=1鈰�\frac{1}{\sqrt{2}}\left[\text{since}\sin 鈦�\frac{\mathrm{蟺}}{4}=\frac{1}{\sqrt{2}}\right]\\ \mathrm{y}=\frac{1}{\sqrt{2}}\end{array}$

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 $\left(2,\frac{\mathrm{蟺}}{4}\right)$, then $\mathrm{r}=2,\mathrm{胃}=\frac{\mathrm{蟺}}{4}$

Use the formulas,$\mathrm{x}=\mathrm{rcos}鈦�\mathrm{胃},\mathrm{y}=\mathrm{rsin}鈦�\mathrm{胃}$

Taken,$\mathrm{x}=\mathrm{rcos}鈦�\mathrm{胃}$and put $\mathrm{r}=2,\mathrm{胃}=\frac{\mathrm{蟺}}{4}$, then

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

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

We can taking $\mathrm{y}=\mathrm{rsin}鈦�\mathrm{胃}$and put $\mathrm{r}=2,\mathrm{胃}=\frac{\mathrm{蟺}}{4}$

$\begin{array}{r}\mathrm{y}=2鈰�\sin 鈦�\frac{\mathrm{蟺}}{4}\\ \mathrm{y}=2鈰�\frac{1}{\sqrt{2}}\left[\text{since}\sin 鈦�\frac{\mathrm{蟺}}{4}=\frac{1}{\sqrt{2}}\right]\end{array}$

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$\left(3,\frac{\mathrm{蟺}}{4}\right)$, then $\mathrm{r}=3,\mathrm{胃}=\frac{\mathrm{蟺}}{4}$

Use the formulas,

Taken,and put , then

finally , the value is

We can takingand put