Convert Polar Equations To Rectangular Form and vice versa - Pre-Calculus

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Convert to polar coordinates.

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Write the Cartesian to polar conversion formulas.

$\theta=tan^$-^1$ \left($\frac{y}{x}$\right)$

Substitute the coordinate point to the equations and solve for $(r,\theta)$ .

$\theta=tan^$-^1$ \left($\frac{y}{x}$\right)=tan^$-^1$ \left($\frac{2}{0}\right)=\frac{\pi}{2}$$

Since is located in between the first and second quadrant, this is the correct angle.

Therefore, the answer is $\left(2,$\frac{\pi}{2}\right)$ .

Convert to polar form.

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Write the Cartesian to polar conversion formulas.

$\theta=tan^$-^1$ \left($\frac{y}{x}$\right)$

Substitute the coordinate point to the equations to find $(r,\theta)$ .

$\theta=tan^$-^1$ \left($\frac{y}{x}$\right)=tan^$-^1$ \left($\frac{-3}{-3}$\right)=\frac{\pi}{4}$$

Since is not located in between the first quadrant, this is not the correct angle. The correct location of this coordinate is in the third quadrant. Add $\pi$ radians to get the correct angle.

$\theta=\frac{\pi}{4}$+\pi=\frac{5\pi}{4}$$

Therefore, the answer is $\left(3\sqrt2,$\frac{5\pi}{4}\right)$ .

Convert from polar form to rectangular form:

$r=4\cos\theta$

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Start by multiplying both sides by .

$r=4\cos\theta$

Keep in mind that

Remember that $x=r\cos\theta$

So then,

Now, complete the square.

Convert the polar equation to rectangular form:

$r=8\sin\theta$

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Start by multiplying both sides by .

$r=8\sin\theta$

Remember that

Keep in mind that $y=r\sin\theta$

So then,

Now, complete the square.

This is a graph of a circle with a radius of and a center at

Convert the polar equation into rectangular form:

$r=8\sec\theta$

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Remember that $\sec\theta=\frac{1}{\cos\theta}$$

So then $r=8\sec\theta$ becomes

$r=\frac{8}{\cos\theta}$$

Now, multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=8$

Since $x=r\cos\theta,$ the rectangular form of this equation is

Convert the polar equation into rectangular form:

$r=16\csc\theta$

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Recall that $\csc\theta=\frac{1}{\sin\theta}$$

Now, substitute in that value into the given equation.

$r=\frac{16}{\sin\theta}$$

Multiply both sides by $\sin\theta$ to get rid of the fraction.

$r\sin\theta=16$

Remember that $y=r\sin\theta$

The rectangular form of this equation is then

Convert the polar equation into rectangular form:

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Start by using the double angle formula for $\sin$ .

$\sin(2\theta)=2\sin\theta\cos\theta$

Substitute that into the equation gives the following:

Because we need $r\cos\theta$ and $r\sin\theta$ to get and respectively, multiply both sides by .

Now, recall that .

Convert to polar form:

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Convert by making the substitutions $x = r \cos \theta$ and $y = r \sin \theta$ :

$r \sin \theta = 2 r \cos \theta + 1$ subtract $2 r \cos \theta$ from both sides

$r \sin \theta - 2 r \cos \theta = 1$ factor out r

$r (\sin \theta - 2 \cos \theta ) = 1$ divide

$r = $\frac{ 1 }{ \sin \theta - 2 \cos \theta}$$

Convert the polar equation to rectangular form:

$\theta=\frac{5\pi}{6}$$

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Start by taking the tangent.

$\tan\theta=\tan$\frac{5\pi}{6}$$

$\tan\theta=-$\frac{\sqrt3}{3}$$

Recall that $\tan\theta=\frac{y}{x}$$

$\frac{y}{x}$=-$\frac{\sqrt3}{3}$

$y=-$\frac{\sqrt3}{3}$x$

Convert the polar equation to rectangular form:

$\theta=\frac{5\pi}{4}$$

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Start by taking the tangent.

$\tan\theta=\tan$\frac{5\pi}{4}$$

$\tan\theta=1$

Recall that $\tan\theta=\frac{y}{x}$$

$$\frac{y}{x}$=1$

Convert the polar equation to rectangular form:

$r=4\sec\theta\tan\theta$

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Recall that $\sec\theta=\frac{1}{\cos\theta}$$

Substituting this into the given equation gives

$r=4$\frac{1}{\cos\theta}$\tan\theta$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=4\tan\theta$

Recall that $x=r\cos\theta$ and that $\tan\theta=\frac{y}{x}$$

$x=4$\frac{y}{x}$$

Convert the polar equation into rectangular form:

$r=2\csc\theta\cot\theta$

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Recall that $\csc\theta=\frac{1}{sin\theta}$$ and $\cot\theta=\frac{1}{tan\theta}$$

$r=2$\frac{1}{\sin\theta}$$\frac{1}{\tan\theta}$$

Multiply both sides by $\sin\theta$

$r\sin\theta=2($\frac{1}{\tan\theta}$)$

Recall that $\tan\theta=\frac{y}{x}$$ and $r\sin\theta=y$

$y=2($\frac{1}{\frac{y}${x}})$

$y=2($\frac{x}{y}$)$

$y=$\sqrt{2x}$$

Convert the polar equation into rectangular form:

$r=-3\sec\theta$

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Recall that $\sec\theta=\frac{1}{\cos\theta}$$

Plugging this into the equation gives us

$r=-3$\frac{1}{\cos\theta}$$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=-3$

Recall that $r\cos\theta=x$

So then the rectangular form of the equation is

Convert the polar equation into rectangular form.

$6r=\csc\theta$

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Recall that $\csc\theta=\frac{1}{\sin\theta}$$

$6r=\frac{1}{\sin\theta}$$

Multiply both sides by $\sin\theta$ to get rid of the fraction.

$6r\sin\theta=1$

Recall that $y=r\sin\theta$ .

$y=\frac{1}{6}$$

Convert the polar equation into rectangular form.

$r=2+\cos\theta$

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$r=2+\cos\theta$

Start by multiplying both sides by .

Now, isolate the to one side.

Square both sides.

Recall that and that $x=r\cos\theta$ .

Convert the rectangular equation to polar form:

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Because $x=r\cos\theta$ and $y=r\sin\theta$ , substitute those values in the rectangular form.

Now, expand the equation.

Subtract from both sides.

Recall the trigonometric identity

Factor the equation.

$r(r+8\cos\theta)=0$

Convert the rectangular equation to polar form:

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Recall that $x=r\cos\theta$

Substitute that into the equation.

$r\cos\theta=9$

Recall that,

$$\frac{1}{\cos\theta}$=\sec\theta$

$r=\frac{9}{\cos\theta}$=9\sec\theta$

Convert the rectangular equation to polar form:

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Recall that $x=r\cos\theta$ .

Substitute that into the equation.

$r\sin\theta=-5$

Now, isolate the on one side.

$r=\frac{-5}{\sin\theta}$$

Recall that,

$$\frac{1}{\cos\theta}$=\sec\theta$

$r=-5\sec\theta$

Convert the rectangular equation into polar form.

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Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=2$

Recall that,

$$\frac{1}{\sin\theta}$=\csc\theta$

$r=\frac{2}{\sin\theta}$=2\csc\theta$

Convert the rectangular equation into polar form:

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Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=-4$

Recall that,

$$\frac{1}{\sin\theta}$=\csc\theta$

$r=\frac{-4}{\sin\theta}$=-4\csc\theta$