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

Convert the rectangular equation to polar form.

x=-3y-4

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Recall that $x=r\cos\theta$ and $y=r\sin\theta$ .

Substitute these values into the equation.

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

Now, manipulate the equation so that the terms with are on the same side.

$r\cos\theta+3r\sin\theta=-4$

Factor out the .

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

Divide both sides by $\cos\theta+3\sin\theta$ .

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

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Example Question #36 : Polar Coordinates

Convert from rectangular form to polar.

y^2=3y-x^2

Possible Answers:

$r=3\sin\theta$

$r=3\cos^2\theta$

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

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

Correct answer:

$r=3\sin\theta$

Explanation:

Recall that $y=r\sin\theta$ and $x=r\cos\theta$ .

Substitute those into the equation.

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

Expand the equation.

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

Add $r^2\cos^2\theta$ to both sides.

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

Factor out r^2 .

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

Remember that $\sin^2\theta+\cos^2\theta=1$ .

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

Divide both sides by .

$r=3\sin\theta$

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Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the rectangular equation to polar form.

x^2-6x+y^2=0

Possible Answers:

$r=6\sin\theta$

$r=\frac{6}{\cos^2\theta}$

$r=6\cos\theta$

$r=6\cos^2\theta\sin\theta$

Correct answer:

$r=6\cos\theta$

Explanation:

Recall that $y=r\sin\theta$ and $x=r\cos\theta$ .

Substitute those into the equation.

$(r\cos\theta)^2-6(r\cos\theta)+(r\sin\theta)^2=0$

Expand this equation.

$r^2\cos^2\theta+r^2\sin^2\theta-6r\cos\theta=0$

Add $6r\cos\theta$ to both sides.

$r^2\cos^2\theta+r^2\sin^2\theta=6r\cos\theta$

Factor out r^2 on the left side of the equation.

$r^2(\cos^2\theta+\sin^2\theta)=6r\cos\theta$

Recall that $\sin^2\theta+\cos^2\theta=1$

$r^2=6r\cos\theta$

Divide both sides by .

$r=6\cos\theta$

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Example Question #31 : Polar Coordinates

Convert the rectangular equation into polar form.

y=4x^2

Possible Answers:

$r=\frac{\tan\theta\sec\theta}{4}$

$r=\frac{\cot\theta\sec\theta}{4}$

$r=\frac{4}{\tan\theta\sec\theta}$

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

Correct answer:

$r=\frac{\tan\theta\sec\theta}{4}$

Explanation:

Recall that $y=r\sin\theta$ and $x=r\cos\theta$ .

Substitute those into the equation.

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

Expand this equation.

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

Subtract both sides by $r\sin\theta$ .

$4r^2\cos^2\theta-r\sin\theta=0$

Factor out the .

$r(4r\cos^2\theta-\sin\theta)=0$

At this point, either r=0 or $4r\cos^2\theta-\sin\theta=0$ . Let's continue solving the latter equation to get a more meaningful answer.

$4r\cos^2\theta-\sin\theta=0$

Add $\sin\theta$ to both sides.

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

Divide both sides by $4\cos^2\theta$ to solve for .

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

Recall that $tan\theta=\frac{\sin\theta}{\cos\theta}$ and that $\frac{1}{\cos\theta}=\sec\theta$ .

$r=\frac{\tan\theta\sec\theta}{4}$

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Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the rectangular equation to polar form:

x=9

Possible Answers:

$r=9\cos\theta$

$r=9\sec\theta$

$r=9\sin\theta$

$r=9\csc\theta$

Correct answer:

$r=9\sec\theta$

Explanation:

Recall that $x=r\cos\theta$

Substitute that into the equation.

$r\cos\theta=9$

Recall that,

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

$r=\frac{9}{\cos\theta}=9\sec\theta$

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Example Question #40 : Polar Coordinates

Convert the rectangular equation to polar form:

x=-5

Possible Answers:

$r=-5\sec\theta$

$r=-5\cos\theta$

$r=-5\csc\theta$

$r=-5\sin\theta$

Correct answer:

$r=-5\sec\theta$

Explanation:

Recall that $x=r\cos\theta$ .

Substitute that into the equation.

$r\sin\theta=-5$

Now, isolate the on one side.

$r=\frac{-5}{\sin\theta}$

Recall that,

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

$r=-5\sec\theta$

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Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the rectangular equation into polar form.

y=2

Possible Answers:

$r=2\sin\theta$

$r=2\sec\theta$

$r=2\csc\theta$

$r=\cos\theta$

Correct answer:

$r=2\csc\theta$

Explanation:

Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=2$

Recall that,

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

$r=\frac{2}{\sin\theta}=2\csc\theta$

Report an Error

Example Question #41 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the rectangular equation into polar form:

y=-4

Possible Answers:

$r=-4\cos\theta$

$r=-4\csc\theta$

$r=-4\sec\theta$

$r=-4\sin\theta$

Correct answer:

$r=-4\csc\theta$

Explanation:

Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=-4$

Recall that,

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

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

Report an Error

Example Question #41 : Polar Coordinates

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

Possible Answers:

$r = tan\theta$

$r = tan^2 \theta$

$r = cot \theta$

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

$r = cos\theta tan \theta$

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #36 : Polar Coordinates

Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #31 : Polar Coordinates

Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #40 : Polar Coordinates

Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #41 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #41 : Polar Coordinates

Example Question #1633 : Pre Calculus

All Precalculus Resources

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