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Precalculus : Polar Coordinates

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Example Questions

Example Question #1622 : Pre Calculus

Convert the rectangular equation to polar form:

(x+4)^2+y^2=16

Possible Answers:

$r(r+8\cos\theta)=0$

$8r\cos\theta+r+16=0$

$r(r+8r\cos\theta)=0$

r^2+8=0

Correct answer:

$r(r+8\cos\theta)=0$

Explanation:

Because $x=r\cos\theta$ and $y=r\sin\theta$ , substitute those values in the rectangular form.

(x+4)^2+y^2=16

$(r\cos\theta+4)^2+(r\sin\theta)^2=16$

Now, expand the equation.

$r^2\cos^2\theta+8r\cos\theta+16+r^2\sin^2\theta=16$

Subtract from both sides.

$r^2\cos^2\theta+8r\cos\theta+r^2\sin^2\theta=0$

$r^2(\cos^2\theta+\sin^2\theta)+8r\cos\theta=0$

Recall the trigonometric identity $\sin^2\theta+\cos^2\theta=1$

$r^2+8r\cos\theta=0$

Factor the equation.

$r(r+8\cos\theta)=0$

Report an Error

Example Question #31 : Polar Coordinates

Convert the rectangular equation into polar form.

y=3x+2

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

Substitute these values into the equation.

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

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

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

Factor out the .

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

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

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

Report an Error

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

Conver the rectanglar equation into polar form.

y=-2x-5

Possible Answers:

$r=-5(\sin\theta+2\cos\theta)$

$r=-\frac{5}{\sin\theta+2\cos\theta}$

$r=-\frac{\sin\theta+2\cos\theta}{5}$

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

Correct answer:

$r=-\frac{5}{\sin\theta+2\cos\theta}$

Explanation:

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

Substitute these values into the equation.

$r\sin\theta=-2(r\cos\theta)-5$

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

$r\sin\theta+2r\cos\theta=-5$

Factor out the .

$r(\sin\theta+2\cos\theta)=-5$

Divide both sides by $\sin\theta+2\cos\theta$ .

$r=-\frac{5}{\sin\theta+2\cos\theta}$

Report an Error

Example Question #101 : Polar Coordinates And Complex Numbers

Convert the rectangular equation to polar form.

x=7y+2

Possible Answers:

$r=2(\cos\theta-7\sin\theta)$

$r=\frac{2}{\cos\theta-7\sin\theta}$

$r=\frac{\cos\theta-7\sin\theta}{2}$

$r=\frac{2}{\sin\theta-7\cos\theta}$

Correct answer:

$r=\frac{2}{\cos\theta-7\sin\theta}$

Explanation:

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

Substitute these values into the equation.

$r\cos\theta=7(r\sin\theta)+2$

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

$r\cos\theta-7r\sin\theta=2$

Factor out the .

$r(\cos\theta-7\sin\theta)=2$

Divide both sides by $\cos\theta-7\sin\theta$ .

$r=\frac{2}{\cos\theta-7\sin\theta}$

Report an Error

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 $\cos\theta+3\sin\theta$ .

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

Report an Error

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 .

$r=3\sin\theta$

Report an Error

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 .

$r=6\cos\theta$

Report an Error

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

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

Report an Error

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,

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

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

Report an Error

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,

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

$r=-5\sec\theta$

Report an Error

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Precalculus : Polar Coordinates

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1622 : Pre Calculus

Example Question #31 : Polar Coordinates

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

Example Question #101 : Polar Coordinates And Complex Numbers

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

All Precalculus Resources

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