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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 #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r^2=4\sin(2\theta)$

Possible Answers:

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

(x^2-^2)^2=8xy

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

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

Correct answer:

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

Explanation:

Start by using the double angle formula for $\sin$ .

$\sin(2\theta)=2\sin\theta\cos\theta$

Substitute that into the equation gives the following:

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

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

Because we need $r\cos\theta$ and $r\sin\theta$ to get and respectively, multiply both sides by r^2 .

$(r^2)(r^2)=8(r\sin\theta)(r\cos\theta)$

Now, recall that r^2=x^2+y^2 .

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

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

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Example Question #2 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation to rectangular form:

$\theta=\frac{5\pi}{6}$

Possible Answers:

y=-3x

$y=-\frac{3}\sqrt3}x$

$y=-\frac{\sqrt3}{3}x$

$y=-\sqrt3x$

Correct answer:

$y=-\frac{\sqrt3}{3}x$

Explanation:

Start by taking the tangent.

$\tan\theta=\tan\frac{5\pi}{6}$

$\tan\theta=-\frac{\sqrt3}{3}$

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

$\frac{y}{x}=-\frac{\sqrt3}{3}$

$y=-\frac{\sqrt3}{3}x$

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Example Question #3 : Polar Coordinates

Convert the polar equation to rectangular form:

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

Possible Answers:

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

y=-x

y=x

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

Correct answer:

y=x

Explanation:

Start by taking the tangent.

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

$\tan\theta=1$

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

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

y=x

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation to rectangular form:

$r=4\sec\theta\tan\theta$

Possible Answers:

$y=2\sqrt x$

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

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

x=4y^2

Correct answer:

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

Explanation:

Recall that $\sec\theta=\frac{1}{\cos\theta}$

Substituting this into the given equation gives

$r=4\frac{1}{\cos\theta}\tan\theta$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=4\tan\theta$

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

$x=4\frac{y}{x}$

4y=x^2

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

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r=2\csc\theta\cot\theta$

Possible Answers:

y=2x^2

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

$y=\sqrt{2x}$

$y=-\sqrt{2x}$

Correct answer:

$y=\sqrt{2x}$

Explanation:

Recall that $\csc\theta=\frac{1}{sin\theta}$ and $\cot\theta=\frac{1}{tan\theta}$

$r=2\frac{1}{\sin\theta}\frac{1}{\tan\theta}$

Multiply both sides by $\sin\theta$

$r\sin\theta=2(\frac{1}{\tan\theta})$

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

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

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

2x=y^2

$y=\sqrt{2x}$

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Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r=-3\sec\theta$

Possible Answers:

x=-3

y=-3

(x-3)^2=4

x^2+y^2=-3

Correct answer:

x=-3

Explanation:

Recall that $\sec\theta=\frac{1}{\cos\theta}$

Plugging this into the equation gives us

$r=-3\frac{1}{\cos\theta}$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=-3$

Recall that $r\cos\theta=x$

So then the rectangular form of the equation is x=-3

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Example Question #3 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form.

$6r=\csc\theta$

Possible Answers:

$x=\frac{1}{6}$

y=6

$y=\frac{1}{6}$

y^2=6

Correct answer:

$y=\frac{1}{6}$

Explanation:

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

$6r=\frac{1}{\sin\theta}$

Multiply both sides by $\sin\theta$ to get rid of the fraction.

$6r\sin\theta=1$

Recall that $y=r\sin\theta$ .

6y=1

$y=\frac{1}{6}$

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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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #2 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #3 : Polar Coordinates

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #1 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #3 : Convert Polar Equations To Rectangular Form And Vice Versa

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

All Precalculus Resources

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