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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #51 : Polar Coordinates

Write the equation (x^2 -3y +y^2 ) ^2 = 4x^2 + 4y^2 in polar form

Possible Answers:

$r = 2+3 \sin \theta$

$r = 4 + 3 \sin \theta$

$r = 3 \sin \theta$

$r = 2 - 3 \sin \theta$

$r = 4 - 3 \sin \theta$

Correct answer:

$r = 2+3 \sin \theta$

Explanation:

First re-arrange the original equation so that the 4 is factored out on the right side, and put x^2 and y^2 next to each other:

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

Make the substitutions r^2 = x^2 + y^2 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

$r = 2 + 3 \cos \theta$

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Example Question #52 : Polar Coordinates

Which is equivalent to $y = \frac{1}{2}x^2$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$r \sin \theta = \frac{ 1}{2} r^2 \cos ^2 \theta$ divide both sides by r

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

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

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

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Example Question #53 : Polar Coordinates

Convert to polar form: y = 2x + 1

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #51 : Polar Coordinates

Example Question #52 : Polar Coordinates

Example Question #53 : Polar Coordinates

All Precalculus Resources

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