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Precalculus : Polar Coordinates

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Example Questions

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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Example Question #21 : Convert Polar Equations To Rectangular Form And Vice Versa

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First multiply both sides by cosine

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

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

$x = \frac{y}{\sqrt{x^2 + y^2 }} - 2$ add 2 to both sides

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

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

(x+2)^2 (x^2 + y^2 ) = y^2 distribute on the left side

x^2 (x+2)^2 + y^2 (x+2)^2 = y^2 subtract from both sides

x^2 (x+2)^2 = y^2 - y^2(x+2)^2 factor out y squared

x^2(x+2)^2 = y^2 [1 - (x+2)^2] divide both sides by 1-(x+2)^2

$\frac{x^2 (x+2)^2}{ 1 - (x+2)^2} = y^2$ take the square root

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

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Example Question #21 : Convert Polar Equations To Rectangular Form And Vice Versa

Which is equivalent to $r=\sin^2 \theta$ in rectangular form?

Possible Answers:

x^2 + y^2 = y^3

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

y^6 - y^4 + x^6 = 0

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

x = y^2

Correct answer:

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

Explanation:

Multiply both sides by r squared

$r^3 = r^2 \sin ^2 \theta$

Now we can substitute $r= \sqrt{x^2 + y^2 }$ and $y = r \sin \theta$

$(\sqrt{x^2+y^2 })^3 = y^2$ square both sides

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

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Example Question #23 : Convert Polar Equations To Rectangular Form And Vice Versa

Which is equivalent to $r = 4 \sin \theta -1$ in rectangular form?

Possible Answers:

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

0 = 3y^2 -x^2 -x^4 -y^4

2x^2 + 2y^2 -4y = 0

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

2x^2 - 4y = 0

Correct answer:

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

Explanation:

First multiply both sides by r:

$r^2 = 4 r \sin \theta - r$

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

$x^2 + y^2 = 4y - \sqrt{x^2 + y^2 }$ add $\sqrt{x^2 + y^2 }$ to both sides

$\sqrt{x^2 + y^2 } +x^2 + y^2 = 4y$ subtract x squared and y squared from both sides

$\sqrt{x^2 + y^2 } = 4y - x^2 - y^2$ square both sides

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

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Example Question #21 : Polar Coordinates

Write $r = 2 \cos \theta + \sin \theta$ in rectangular form

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

multiply both sides by r:

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

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

x^2 + y^2 = 2x + y

If we subtract the terms on the right from both sides, we can complete the square twice and make this into a regular circle equation:

$x^2 - 2x \enspace \enspace + y^2 - y \enspace \enspace = 0$

To complete the square for x, add 1, since $(\frac{1}{2}(-2))^2= (-1)^ 2 =1$ .

To complete the square for y, add $\frac{1}{4}$ since $(\frac{1}{2}(-1))^2 = (\frac{1}{2})^2 = \frac{1}{4}$ .

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

re-write the left side as binomials squared:

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

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Example Question #22 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation for $r = 8\sin \theta - 2 \cos \theta$ in rectangular form.

Possible Answers:

(x-1)^2 + (y-4)^2 = 17

(x+1)^2 + (y-4)^2 = 17

(x-2)^2 + (y+8)^2 = 68

(y-1)^2 + (x + 4)^2 = 17

(y+4)^2 + (x+1)^2 = 17

Correct answer:

(x+1)^2 + (y-4)^2 = 17

Explanation:

Multiply both sides by r:

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

Now we can substitute in r^2 = x^2 + y^2 , $x= r \cos \theta$ , and $y = r \sin \theta$ .

x^2 + y^2 = 8y-2x if we subtract both terms on the right from both sides, we can complete the square twice to get this into the normal form for a circle.

$x^2 + 2x \enspace \enspace + y^2 -8y \enspace \enspace =0$

To complete the square for x, add 1 to both sides, since $(\frac{1}{2}(2))^2 = (1)^2 = 1$

To complete the square for y, add 16 to both sides, since $(\frac{1}{2}(-8))^2 = (-4)^2 = 16$ :

(x^2 + 2x + 1 ) + (y^2 - 8y + 16 ) = +1 + 16 Re-write the left side as two binomials squared

(x+1)^2 + (y-4)^2 = 17

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Example Question #23 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation for $r = \tan \theta$ in rectangular form

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

$x^2 + y^2 = \frac{y^2}{x^2}$ multiply both sides by x^2

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

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

x^4 + x^2 y^2 - y^2 = 0 subtract x^4 from both sides

x^2 y^2 - y^2 = -x^4 factor our y^2

y^2 (x^2 - 1 ) = -x^4 divide both sides by x^2 - 1

$y ^2 = \frac{ -x^4}{ x^2 - 1 }$ take the square root of both sides

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

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Example Question #22 : Polar Coordinates

Convert the equation $r = \frac{ 8 \cos \theta }{1 + 7 \sin ^2 \theta }$ to rectangular form

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Multiply both sides by $1 + 7 \sin ^2 \theta$ :

$r(1 + 7 \sin ^2 \theta ) = 8 \cos \theta$ distribute

$r + 7 r \sin ^2 \theta = 8 \cos \theta$ multiply both sides by r

$r^2 + 7 r^2 \sin^2 \theta = 8 r \cos \theta$

convert to rectangular by making the substitutions x^2 + y^2 = r^2 , $y = r \sin \theta$ , and $x = r \cos \theta$ :

x^2 + y^2 + 7 y^2 = 8 x subtract 8x from both sides and combine like terms

x^2 - 8x + 8y^2 = 0 complete the square for x by adding 16 to both sides

x^2 - 8x + 16 + 8y^2 = 16

(x-4)^2 + 8y^2 = 16 divide both sides by 16

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

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Example Question #91 : Polar Coordinates And Complex Numbers

Convert (0,2) to polar coordinates.

Possible Answers:

(0,2)

$(\sqrt2,\pi)$

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

(2,0)

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

Correct answer:

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

Explanation:

Write the Cartesian to polar conversion formulas.

r^2=x^2+y^2

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

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

r^2=x^2+y^2

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

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

Since (0,2) is located in between the first and second quadrant, this is the correct angle.

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

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Example Question #92 : Polar Coordinates And Complex Numbers

Convert (-3,-3) to polar form.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Write the Cartesian to polar conversion formulas.

r^2=x^2+y^2

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

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

r^2=x^2+y^2

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

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

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

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Precalculus : Polar Coordinates

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #21 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #21 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #23 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #21 : Polar Coordinates

Example Question #22 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #23 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #22 : Polar Coordinates

Example Question #91 : Polar Coordinates And Complex Numbers

Example Question #92 : Polar Coordinates And Complex Numbers

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