Convert Polar Equations To Rectangular Form and vice versa

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

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1

Convert to polar form.

$\left(3\sqrt2,\frac{5\pi}{4}\right)$

$\left(3\sqrt2,\frac{\pi}{4}\right)$

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

$\left(3,\frac{5\pi}{4}\right)$

$\left(-3,-\frac{\pi}{4}\right)$

Explanation

Write the Cartesian to polar conversion formulas.

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

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

$r=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt2$

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

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

2

Write the equation in polar form

$r = 2+3 \sin \theta$

$r = 3 \sin \theta$

$r = 4 + 3 \sin \theta$

$r = 4 - 3 \sin \theta$

$r = 2 - 3 \sin \theta$

Explanation

First re-arrange the original equation so that the 4 is factored out on the right side, and put and next to each other:

Make the substitutions and $y = r \sin \theta$ :

$(r^2 - 3 r \sin \theta )^2 = 4r^2$ take the square root of both sides

$r^2 - 3 r \sin \theta = 2 r$ divide both sides by r

$r - 3 \cos \theta = 2$ add $3 \cos \theta$ to both sides

$r = 2 + 3 \cos \theta$

3

Convert from polar form to rectangular form:

$r=4\cos\theta$

Explanation

Start by multiplying both sides by .

$r=4\cos\theta$

$r^2=4r\cos\theta$

Keep in mind that

$x^2+y^2=4r\cos\theta$

Remember that $x=r\cos\theta$

So then,

Now, complete the square.

4

Convert the polar equation to rectangular form:

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

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

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

Explanation

Start by taking the tangent.

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

$\tan\theta=1$

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

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

5

Convert to polar coordinates.

$\left(2,\frac{\pi}{2}\right)$

$(\sqrt2,\pi)$

$\left(\sqrt2,\frac{\pi}{2}\right)$

Explanation

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 $(r,\theta)$ .

$r=\sqrt{0^2+2^2}=2$

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

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

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

6

Convert the polar equation to rectangular form:

$r=8\sin\theta$

Explanation

Start by multiplying both sides by .

$r=8\sin\theta$

$r^2=8r\sin\theta$

Remember that

$x^2+y^2=8r\sin\theta$

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

7

Write the equation for $r = \frac{\cos \theta - \frac{1}{2}}{ \cos ^2 \theta }$ in rectangular form

$y = \sqrt{4x^4 - 8 x^3 + 3x^2 }$

$y = x^2 - x + \frac{1}{2}$

$y = \sqrt{2x^4 - 4x^3 + 1.5x^2 }$

$y = \sqrt{\frac{3}{4}x^2 - x^4}$

$y = \sqrt{x^4 - 2x^3 + \frac{3}{4} x }$

Explanation

Multiply both sides by the right denominator:

$r \cos ^2 \theta = \cos \theta - \frac{1}{2}$ multiply both sides by r

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

Now we can substitute in $r = \sqrt{x^2 + y^2 }$ and $x = r \cos \theta$ to start converting to rectangular form:

$x^2 = x - \frac{1}{2} \sqrt{x^2 + y^2 }$ subtract x from both sides

$x^2 - x = -\frac{ 1}{2} \sqrt{x^2 + y^2 }$ square both sides

$x^4 -2x^3 + x^2 = \frac{1}{4} (x^ 2 + y^2 )$ multiply both sides by 4

subtract x squared from both sides

take the square root of both sides

$y = \sqrt{4x^4 - 8x^3 + 3x^2 }$

8

Convert the polar equation into rectangular form:

$r=8\sec\theta$

Explanation

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

9

Convert the polar equation into rectangular form:

$r=16\csc\theta$

Explanation

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

10

Convert $r=\frac{\sin \theta + 1 }{2 \cos ^2 \theta }$ to rectangular form

$y = x^2 - \frac{1}{4}$

$y = x^2 + \frac{1}{4}$

$y = \frac{x^2}{2}$

$y = 1 - \frac{x^2 }{4}$

Explanation

First, multiply both sides by the denominator:

$r \cdot 2 \cos ^2 \theta = \sin \theta + 1$ multiply both sides by r

$2r^2 \cos ^2 \theta = r \sin \theta +r$

Now we can make the substitutions $r = \sqrt{x^2 + y^2 }$ $x=r \cos \theta$ and $y=r \sin \theta$ :

$2x^2 = y + \sqrt{x^2 + y^2 }$ subtract y from both sides

$2x^2 - y = \sqrt{x^2 + y^2}$ square both sides

subtract y squared from both sides

we are trying to get this in the form of y=, so subtract from both sides

divide both sides by

$y = \frac{x^2} {-4x^2 } - \frac{4 x^4 }{-4x^2 }$ simplify

$y = \frac{-1}{4} + x^2$ or $y=x^2 - \frac{1}{4}$

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