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Precalculus : Polar Coordinates and Complex Numbers

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Example Questions

Example Question #111 : Polar Coordinates And Complex Numbers

Convert the rectangular equation into polar form:

y=-4

Possible Answers:

$r=-4\sin\theta$

$r=-4\cos\theta$

$r=-4\sec\theta$

$r=-4\csc\theta$

Correct answer:

$r=-4\csc\theta$

Explanation:

Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=-4$

Recall that,

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

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

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Example Question #1632 : Pre Calculus

How could you write the equation y=x^2 in polar coordinates?

Possible Answers:

$r = cos\theta tan \theta$

$r = tan^2 \theta$

$r=\frac{sin\theta}{cos^2\theta}$

$r = cot \theta$

$r = tan\theta$

Correct answer:

$r=\frac{sin\theta}{cos^2\theta}$

Explanation:

To convert from rectangular to polar, use the equivalent forms $x = r*cos\theta$ and $y = r*sin\theta$ . Substituting these in, we get:

$r*sin\theta = (r*cos\theta)^2$

$r*sin\theta = r^2*cos^2 \theta$ divide both sides by r

$sin \theta = r*cos^2 \theta$ divide both sides by $cos^2 \theta$ to get this equation in terms of r=

$r = \frac{sin\theta}{cos^2\theta}$

Note that we could simplify this a little bit if we wanted to $r = \frac{tan\theta}{cos\theta}$ but that wasn't one of the choices.

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Example Question #1633 : Pre Calculus

How could you express the rectangular equation $y = \sqrt{x + \frac{1}{4}}$ in polar form?

Possible Answers:

$r = \frac{1}{4 \sin^2 \theta - \cos \theta}$

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

$r = \frac { \cos \theta}{2 \sin ^2 \theta }$

$r = \frac { \cos \theta + \frac {1}{4}}{ \sin ^2 \theta}$

$r = \frac{-\cos \theta \pm \sqrt { \cos 2 \theta }}{2 \sin ^2 \theta}$

Correct answer:

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

Explanation:

To convert from rectangular to polar, we can substitute in $y = r \sin \theta$ and $x = r \cos \theta$ . Our equation now becomes:

$r \sin \theta = \sqrt{ r \cos \theta + \frac {1}{4}}$ square both sides to remove the radical

$r^2 \sin^2 \theta = r \cos \theta + \frac{1}{4}$

Now we can see that in terms of r, this is a quadratic. We can solve using the quadratic formula if we subtract everything from the right side and get our equation equal to 0:

$r^2 \sin ^2 \theta - r \cos \theta - \frac{1}{4} = 0$

Put our coefficents a, b, and c into the quadratic formula:

$r = \frac{ \cos \theta \pm \sqrt{\cos ^2 \theta - 4* \sin ^2 \theta * -\frac{1}{4}}}{2 \sin ^2 \theta}$ multiplying $-4 * -\frac{1}{4}$ yields 1, so this now becomes:

$r = \frac{ \cos \theta \pm \sqrt{ \cos ^2 \theta + \sin ^2 \theta }}{2 \sin ^2 \theta}$ we can simplify this knowing the trigonometric identity that $\cos ^2 \theta + \sin ^2 \theta = 1$

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

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Example Question #42 : Polar Coordinates

Which equation is the polar equivalent of the rectangular quadratic $y = -\frac{1}{4} x^2 + 1$ ?

Possible Answers:

$r = \frac{ 4 \sin \theta - 2 \cos \theta}{ \cos ^ 2 \theta }$

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

$r = \frac {1}{ \sin \theta + \frac {1}{4} \cos \theta }$

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

$r = \frac{ 4}{ 4 \sin \theta + \cos \theta}$

Correct answer:

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

Explanation:

To convert from rectangular to polar, we can substitute $y = r \sin \theta$ and $x = r \cos \theta$ . That gives us:

$r \sin \theta = -\frac {1}{4} r^2 \cos ^2 \theta + 1$

We can see that this is a quadratic in terms of r, so to solve, just like any other quadratic, we want to subtract everything from the right side so that it is equal to 0.

$\frac { 1}{4} \cos ^2 \theta * r^2 + \sin \theta * r - 1 = 0$

Now we can use the quadratic formula to solve for r:

$r = \frac {- \sin \theta \pm \sqrt { \sin ^2 \theta - 4*\frac{1}{4}cos^2 \theta *(-1) }}{2*\frac{1}{4} \cos ^2 \theta}$

$r = \frac { - \sin \theta \pm \sqrt { \sin ^2 \theta + \cos ^2 \theta }}{ \frac {1}{2} \cos ^2 \theta}$

we can simplify using the trig identity $\sin ^2 \theta + \cos ^2 \theta = 1$

$r = \frac { - \sin \theta \pm 1 }{ \frac {1}{2} \cos ^ 2 \theta }$ to get rid of the fraction in the denominator, multiply top and bottom by 2

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

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Example Question #43 : Polar Coordinates

How would you write the equation x^2 + y^2 = 9 as a polar equation?

Possible Answers:

r=9

$r = \theta$

$r^2 = 9\theta$

$r=3\theta$

r=3

Correct answer:

r=3

Explanation:

This simple rectangular equation represents a circle centered at the origin with radius 3,

x^2+y^2=r^2

since 3^2 = 9 .

The way to write that in polar form is just r = 3 .

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Example Question #1641 : Pre Calculus

Write y = 2x + 3 in polar form.

Possible Answers:

$r = \frac{ \sin \theta - 2 \cos \theta }{3}$

$r = \frac{ 3}{ 2\cos \theta - \sin \theta }$

$r = \frac{3}{ \sin \theta +2 \cos \theta }$

$r = \frac{ 3}{ \sin \theta - 2 \cos \theta }$

$r = \frac{ \sin \theta + 2 \cos \theta }{3}$

Correct answer:

$r = \frac{ 3}{ \sin \theta - 2 \cos \theta }$

Explanation:

To convert from rectangular to polar, substitute $y = r \sin \theta$ and $x = r \cos \theta$ :

$r \sin \theta = 2(r \cos \theta ) + 3$

$r \sin \theta - 2 r \cos \theta = 3$ factor out r

$r(\sin \theta - 2 \cos \theta ) = 3$ divide

$r = \frac{ 3}{ \sin \theta - 2 \cos \theta }$

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Example Question #1641 : Pre Calculus

Write the equation y = 3x^2 in polar form.

Possible Answers:

$r = \frac{ \sin \theta }{ 3 \cos \theta }$

$r = \frac{ \sin \theta}{ 3 \cos ^ 2 \theta }$

$r = 3 \tan \theta$

$r = \frac{ 3\sin ^2 \theta}{ \cos \theta }$

$r = \frac{ 3\cos ^2 \theta }{ \sin \theta }$

Correct answer:

$r = \frac{ \sin \theta}{ 3 \cos ^ 2 \theta }$

Explanation:

To convert from rectangular to polar, substitute in $x = r \cos \theta$ and $y = r \sin \theta$ :

$r \sin \theta = 3 (r \cos \theta )^2$

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

$0 = r^2 \cdot 3 \cos ^2 \theta - r \cdot \sin \theta$ factor out r

$0 = r [ r \cdot 3 \cos ^2 \theta - \sin \theta ]$ this gives us a trivial answer of r = 0, and a second answer found by setting the second [more interesting] answer equal to zero:

$0 = r\cdot 3 \cos^2 \theta - \sin \theta$

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

$\frac{ \sin \theta } { 3 \cos ^2 \theta } = r$

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Example Question #113 : Polar Coordinates And Complex Numbers

Write $y = \frac{ 1}{2} x^2 + x$ in polar form.

Possible Answers:

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

$r= \frac{ 2 ( \sin \theta - \cos \theta )}{\cos ^2 \theta }$

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

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

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

Correct answer:

$r= \frac{ 2 ( \sin \theta - \cos \theta )}{\cos ^2 \theta }$

Explanation:

To convert, substitute $x = r \cos \theta$ and $y = r \sin \theta$

$r \sin \theta = \frac{1}{2} (r \cos \theta ) ^ 2 + r \cos \theta$

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

$0 =r^2 \cdot \frac{1}{2} \cos ^2 \theta + r \cos \theta - r \sin \theta$ factor out r

$0 = r [ r \cdot \frac{1}{2 } \cos ^2 \theta + \cos \theta - \sin \theta ]$

This gives us the trivial answer r = 0, but also another answer from setting the second factor equal to zero:

$0 = r \cdot \frac{1}{2} \cos ^2 \theta + \cos \theta - \sin \theta$

$\sin \theta - \cos \theta = r \cdot \frac{1}{2} \cos ^2 \theta$ multiply by 2

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

$\frac{ 2 (\sin \theta - \cos \theta )}{ \cos ^2 \theta } = r$

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Example Question #42 : Polar Coordinates

Write y = 4x^2 - 3x in polar form.

Possible Answers:

$r = \frac{ 4 \cos ^2 \theta }{ 3 \cos \theta - \sin \theta }$

$r = \frac{ 4 \cos ^2 \theta } { 3 \cos \theta + \sin \theta }$

$r = \frac{3 \cos \theta + \sin \theta }{ 4 \cos ^2 \theta }$

$r = \frac{ \sin \theta - 3 \cos \theta }{ 4 \cos ^2 \theta }$

$r = \frac{ 3 \cos \theta - \sin \theta }{ 4 \cos^2 \theta }$

Correct answer:

$r = \frac{3 \cos \theta + \sin \theta }{ 4 \cos ^2 \theta }$

Explanation:

To convert, substitute $x = r \cos \theta$ and $y = r \sin \theta$

$r \sin \theta = 4(r \cos \theta ) ^2 - 3 r \cos \theta$

$r \sin \theta = 4 \cos ^2 \theta \cdot r^2 - 3 r \cos \theta$ divide both sides by r

$\sin \theta = 4 \cos ^2 \theta \cdot r - 3 \cos \theta$

$\sin \theta + 3 \cos \theta = 4 \cos ^2 \theta \cdot r$

$\frac{ \sin \theta + 3 \cos \theta }{4 \cos ^2 \theta } = r$

The answer choice appears in a slightly different order,

$r = \frac{3 \cos \theta + \sin \theta }{ 4 \cos ^2 \theta }$ , but these are equivalent expressions.

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Example Question #1642 : Pre Calculus

Convert (x^2 + 2x + y^2 )^2 = 4x^2 +4y^2 to polar form

Possible Answers:

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

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

$r = 2 \cos ^2 \theta + \sqrt{ \cos ^4 \theta + \sin ^2 \theta }$

$r = 2 - 2 \cos \theta$

$r = -2 \cos \theta + 2 \sqrt{ \cos ^4 \theta - \cos ^2 \theta + 1 }$

Correct answer:

$r = 2 - 2 \cos \theta$

Explanation:

Re-arrange the left side so that x^2 is next to y^2 , and factor the 4 out on the right side :

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

Make the substitutions x^2 + y^2 = r^2 and $x = r \cos \theta$ :

$(r^2 + 2 r \cos \theta )^2 = 4r^2$ take the square root of both sides

$r^2 + 2 r \cos \theta = 2r$ divide both sides by r

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

$r = 2 - 2 \cos \theta$

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Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #111 : Polar Coordinates And Complex Numbers

Example Question #1632 : Pre Calculus

Example Question #1633 : Pre Calculus

Example Question #42 : Polar Coordinates

Example Question #43 : Polar Coordinates

Example Question #1641 : Pre Calculus

Example Question #1641 : Pre Calculus

Example Question #113 : Polar Coordinates And Complex Numbers

Example Question #42 : Polar Coordinates

Example Question #1642 : Pre Calculus

All Precalculus Resources

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