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Precalculus : Polar Coordinates and Complex Numbers

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Example Questions

Example Question #11 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form.

$r=2+\cos\theta$

Possible Answers:

(x^2+y^2-x)^2=4

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

(y^2-x^2)=4x^2

x^2+y^2=4

Correct answer:

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

Explanation:

$r=2+\cos\theta$

Start by multiplying both sides by .

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

Now, isolate the to one side.

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

Square both sides.

$4r^2=(r^2-r\cos\theta)^2$

Recall that r^2=x^2+y^2 and that $x=r\cos\theta$ .

4(x^2+y^2)=(x^2+y^2-x)^2

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

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Example Question #12 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r=16\csc\theta$

Possible Answers:

x^2+y^2=16

y=16

x^2+(y+8)^2=16

x=16

Correct answer:

y=16

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 y=16

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Example Question #13 : Convert Polar Equations To Rectangular Form And Vice Versa

What would be the rectangular equation form for the polar equation $r = \cos^2 \theta$ ?

Possible Answers:

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

$y=\sqrt[6]{x^4-x^6}$

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

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

$y=\sqrt[3]{x^2}$

Correct answer:

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

Explanation:

To convert from polar coordinates to rectangular coordinates, know that r is the hypotenuse of a right triangle with legs x and y, so $r = \sqrt{x^2 + y^2}$ .

The cosine of theta is this triangle's adjacent side over the hypotenuse r, so $\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$ . Making these substitutions into $r = \cos ^2 \theta$ we get:

$\sqrt{x^2 + y^2 } = (\frac{x}{\sqrt{x^2+y^2}})^2$ square the right side to simplify

$\sqrt{x^2 + y^2 } = \frac{x^2}{x^2 + y^2}$ square both sides to remove the radical

$x^2 + y^2 = \frac{x^4}{(x^2+y^2)^2}$ multiply both sides by the right denominator

(x^2 + y^2 )^3 = x^4 take both sides to the $\frac{1}{3}$ power

$x^2 + y^2 = x^{\frac{4}{3}}$ subtract x^2 from both sides

$y^2 = x^{\frac{4}{3}} - x^2$ take the square root

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

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Example Question #14 : Convert Polar Equations To Rectangular Form And Vice Versa

Which is equivalent to $r = \frac{1}{2} \cos \theta \sin \theta$ in rectangular form?

Possible Answers:

$x^2 + y^2 = \frac{xy}{2(x^2 + y^2)}$

$\sqrt[3]{4} (x^2 + y^2 ) = \sqrt{x^2y^2}$

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

4(x^2 + y^2 )^3 = x^2 y^2

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

Correct answer:

4(x^2 + y^2 )^3 = x^2 y^2

Explanation:

To convert from polar form to rectangular form, substitute in $r = \sqrt{x^2 + y^2 }$ , $y = r \sin \theta$ , and $x = r\cos\theta$ . Equivalently, $\sin \theta = \frac{y }{r} = \frac{ y }{ \sqrt{x^2 + y^2 }}$ and $\cos \theta = \frac{ x }{ r } = \frac{ x }{\sqrt{x^2 + y^2}}$ :

Substituting these into the original polar equation, we get:

$\sqrt{x^2 +y^2} = \frac {1}{2} \cdot \frac{y}{ \sqrt{x^2 + y^2}} \cdot \frac{x}{\sqrt{x^2 + y^2}}$ multiply the second two fractions

$\sqrt{x^2 + y^2 } = \frac{1}{2} \cdot \frac{xy}{x^2 + y^2 }$ now multiply these fractions

$\sqrt{x^2 + y^2 } = \frac{xy}{2(x^2 + y^2 )}$ square both sides

$x^2 + y^2 = \frac{ x^2 y^2 }{4 (x^2 + y^2 )^2}$ multiply both sides by the denominator

4(x^2 + y^2 )^3 = x^2 y^2

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Example Question #15 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation $r= 2 \cos \theta$ in rectangular form

Possible Answers:

y=x(2-x)

y = 2x^2

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

$y = \sqrt{3x^2}$

y^2= x^2 - 2x

Correct answer:

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

Explanation:

To convert to rectangular form, it is easiest to first multiply both sides by r:

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

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

x^2 + y^2 = 2 x

We want to solve for y, so subtract x squared from both sides:

y^2 = 2x - x^2 now take the square root of both sides

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

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Example Question #16 : Convert Polar Equations To Rectangular Form And Vice Versa

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

Possible Answers:

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

y = 2x^2 -1

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

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

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

Correct answer:

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

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

4x^4 - 4 x^2 y + y^2 = x^2 + y^2 subtract y squared from both sides

4x^4 - 4x^2 y = x^2 we are trying to get this in the form of y=, so subtract 4x^2 from both sides

-4x^2 y = x^2 - 4x^4 divide both sides by -4x^2

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

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

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Example Question #11 : Polar Coordinates

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

Possible Answers:

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

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

y = - x^2 + 1

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

y=x^2 - 1

Correct answer:

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

Explanation:

First, multiply both sides by the denominator:

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

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

To convert, make the substitutions $r = \sqrt{x^2 +y^2}$ , $y=r \sin \theta$ , and $x = r \cos \theta$

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

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

x^4 - 2x^2 y + y^2 = x^2 + y^2 subtract y squared from both sides

x^4 - 2x^2 y = x^2 we want to get y by itself, so subtract x^4 from both sides

-2x^2 y = x^2 - x^4

divide both sides by -2x^2

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

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Example Question #18 : Convert Polar Equations To Rectangular Form And Vice Versa

Which rectangular equation is equivalent to $r = \frac{\cos \theta }{\sin ^2 \theta + 2 }$ ?

Possible Answers:

$9y^2 + 2(x - \frac{1}{2})^2 = 1$

$3y^2 - 2(x-\frac{1}{2})^2 = 1$

3y^2 -2(x-1)^2 = 1

$9 y^2 = -2(x + \frac{1}{4} ) ^2$

$24y^2 +16(x-\frac{1}{4})^2 = 1$

Correct answer:

$24y^2 +16(x-\frac{1}{4})^2 = 1$

Explanation:

First multiply both sides by the denominator:

$r (\sin ^2 \theta + 2 ) = \cos \theta$ multiply both sides by r

$r^2 (\sin ^2 \theta + 2 ) = r \cos \theta$ distribute

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

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

y^2 + 2(x^2 +y^2 ) = x distribute

y^2 +2x^2 + 2y^2 = x combine like terms

3y^2 + 2x^2 = x subtract 2 x squared from both sides

3y^2 = -2x^2 + x We want to complete the square on the right, so factor our the -2:

$3y^2 = -2( x^2 -\frac{1}{2}x \enspace \enspace )$ to complete the square, add $(\frac{1}{2}\cdot \frac{1}{2})^2 = \frac{1}{16}$ inside the parentheses. This multiplied by the -2 outside the parentheses is $-\frac{1}{8}$ , so this means we're actually subtracting $\frac{1}{8}$ from both sides:

$3y^3 - \frac{1}{8} = -2(x - \frac{1}{4}) ^2$ add $\frac{1}{8}$ and $2(x-\frac{1}{4})^2$ to both sides:

$3y^3 + 2(x - \frac{1}{4} ) ^2 = \frac{1}{8 }$ multiply both sides by 8

$24y^2 + 16(x-\frac{1}{4})^2 = 1$

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Example Question #19 : Convert Polar Equations To Rectangular Form And Vice Versa

Which is the rectangular form for $r = \frac{\sin \theta } { 2 \cos ^2 \theta + 1 }$ ?

Possible Answers:

$y = \frac{-1 + \sqrt{1-12x^2}}{2}$

$y = \frac{1 + \sqrt{1 + 12x^2}}{2}$

$y = \frac{1 + \sqrt{1 - 12x^2}}{2}$

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

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

Correct answer:

$y = \frac{1 + \sqrt{1 - 12x^2}}{2}$

Explanation:

First multiply both sides by the right denominator:

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

$r^2 (2\cos ^2 \theta + 1) = r \sin \theta$ distribute

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

Now we can start to convet to rectangular by making the substitutions $y = r \sin \theta$ , $x = r \cos \theta$ , and r^2 = x^2 + y^2 :

2x^2 + x^2 + y^2 = y combine like terms:

3x^2 + y^2 = y subtract y from both sides, and re-order this in decending order of powers of y:

y^2 - y + 3x^2 = 0 this is a quadratic, so we can use the quadratic equation to get y by itself:

$y = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(3x^2)}}{2(1)} = \frac{1 \pm \sqrt{1 - 12x^2 }}{2 }$

The answer choice that works is

$y = \frac{1 + \sqrt{1 - 12x^2}}{2}$

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Example Question #20 : Convert Polar Equations To Rectangular Form And Vice Versa

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

4x^4 - 8 x ^2 + 4 x ^2 = x^2 + y^2 subtract x squared from both sides

4x^4 - 8x^3 + 3x^2 = y^2 take the square root of both sides

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

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Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #12 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #13 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #14 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #15 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #16 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #11 : Polar Coordinates

Example Question #18 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #19 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #20 : Convert Polar Equations To Rectangular Form And Vice Versa

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