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AP Calculus BC : Polar Form

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Polar Form

Rewrite in polar form:

$x^{2} = 3xy + 1$ $\displaystyle x^{2} = 3xy + 1$

Possible Answers:

$r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }$ $\displaystyle r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta + 3 \cos \theta\; \sin \theta }}$ $\displaystyle r =\sqrt{ \frac{1}{\cos ^{2} \theta + 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }$ $\displaystyle r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$ $\displaystyle r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

$r = \sqrt{\frac{1}{\sin ^{2} \theta - 3 \cos \theta\; \sin \theta }}$ $\displaystyle r = \sqrt{\frac{1}{\sin ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Correct answer:

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$ $\displaystyle r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Explanation:

$x^{2} = 3xy + 1$ $\displaystyle x^{2} = 3xy + 1$

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$ $\displaystyle \left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$ $\displaystyle r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$ $\displaystyle r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$ $\displaystyle r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$ $\displaystyle r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$ $\displaystyle r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

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Example Question #11 : Functions, Graphs, And Limits

What is the following coordinate in polar form?

(-2,-5) $\displaystyle (-2,-5)$

Provide the angle in degrees.

Possible Answers:

$(\sqrt{29},248)$ $\displaystyle (\sqrt{29},248)$

$\displaystyle (10,248)$

$(\sqrt{29},68)$ $\displaystyle (\sqrt{29},68)$

$\displaystyle (16,112)$

$(\sqrt{29},202)$ $\displaystyle (\sqrt{29},202)$

Correct answer:

$(\sqrt{29},248)$ $\displaystyle (\sqrt{29},248)$

Explanation:

To calculate the polar coordinate, use

$r=\sqrt{x^2+y^2}$ $\displaystyle r=\sqrt{x^2+y^2}$

$r=\sqrt{29}$ $\displaystyle r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$ $\displaystyle \theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ $\displaystyle \theta=248$.

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Example Question #2 : Polar Form

What is the equation $y=2x^{2}$ $\displaystyle y=2x^{2}$ in polar form?

Possible Answers:

$r=\frac{1}{2}\cos \theta}{\tan \theta$ $\displaystyle r=\frac{1}{2}\cos \theta}{\tan \theta$

$r=\frac{1}{2}\tan \theta}{\sec \theta$ $\displaystyle r=\frac{1}{2}\tan \theta}{\sec \theta$

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

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

$r=\frac{1}{2}\sin \theta}{\sec \theta$ $\displaystyle r=\frac{1}{2}\sin \theta}{\sec \theta$

Correct answer:

$r=\frac{1}{2}\tan \theta}{\sec \theta$ $\displaystyle r=\frac{1}{2}\tan \theta}{\sec \theta$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ $\displaystyle y=r\sin \theta$ and $x=r\cos \theta$ $\displaystyle x=r\cos \theta$. Given $y=2x^{2}$ $\displaystyle y=2x^{2}$, then $r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ $\displaystyle r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$.

$r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ $\displaystyle r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$. Dividing both sides by $r\cos \theta$ $\displaystyle r\cos \theta$,

$\tan \theta=2r\cos \theta$ $\displaystyle \tan \theta=2r\cos \theta$

$\frac{\tan \theta}{\cos \theta}=2r$ $\displaystyle \frac{\tan \theta}{\cos \theta}=2r$

$\tan \theta}{\sec \theta=2r$ $\displaystyle \tan \theta}{\sec \theta=2r$

$\frac{1}{2}\tan \theta}{\sec \theta=r$ $\displaystyle \frac{1}{2}\tan \theta}{\sec \theta=r$

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Example Question #202 : Ap Calculus Bc

What is the equation $y=\frac{6}{x}$ $\displaystyle y=\frac{6}{x}$ in polar form?

Possible Answers:

$r=\sqrt\frac{6}{\sin \theta cos \theta}$ $\displaystyle r=\sqrt\frac{6}{\sin \theta cos \theta}$

$r=\sqrt\frac{6}{\cos \theta}$ $\displaystyle r=\sqrt\frac{6}{\cos \theta}$

$r=\sqrt\frac{6}{\sin \theta}$ $\displaystyle r=\sqrt\frac{6}{\sin \theta}$

None of the above

$r=\sqrt\frac{6}{\tan \theta}$ $\displaystyle r=\sqrt\frac{6}{\tan \theta}$

Correct answer:

$r=\sqrt\frac{6}{\sin \theta cos \theta}$ $\displaystyle r=\sqrt\frac{6}{\sin \theta cos \theta}$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ $\displaystyle y=r\sin \theta$ and $x=r\cos \theta$ $\displaystyle x=r\cos \theta$. Given $y=\frac{6}{x}$ $\displaystyle y=\frac{6}{x}$, then $r\sin \theta=\frac{6}{r\cos \theta}$ $\displaystyle r\sin \theta=\frac{6}{r\cos \theta}$. Multiplying both sides by $r\cos \theta$ $\displaystyle r\cos \theta$,

$r^{2}\sin \theta cos \theta=6$ $\displaystyle r^{2}\sin \theta cos \theta=6$

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

$r=\sqrt\frac{6}{\sin \theta cos \theta}$ $\displaystyle r=\sqrt\frac{6}{\sin \theta cos \theta}$

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Example Question #1 : Polar Form

Convert the following function into polar form:

$\displaystyle f(x)=2x^2+5x+2+2y^2$

Possible Answers:

$r^2+1+5r\cos(\theta)$ $\displaystyle r^2+1+5r\cos(\theta)$

$r^2+\sin(\theta)$ $\displaystyle r^2+\sin(\theta)$

$2r^2+5r\sin(\theta)+2$ $\displaystyle 2r^2+5r\sin(\theta)+2$

$2r^2+5r\cos(\theta )+2$ $\displaystyle 2r^2+5r\cos(\theta )+2$

Correct answer:

$2r^2+5r\cos(\theta )+2$ $\displaystyle 2r^2+5r\cos(\theta )+2$

Explanation:

The following formulas were used to convert the function from polar to Cartestian coordinates:

$x=r\cos(\theta), y=r\sin(\theta), x^2+y^2=r^2$ $\displaystyle x=r\cos(\theta), y=r\sin(\theta), x^2+y^2=r^2$

Note that the last formula is a manipulation of a trignometric identity.

Simply replace these with x and y in the original function.

$\displaystyle f(x)=2x^2+5x+2+2y^2$

$\displaystyle f(x)=(x^2+y^2)2+5x+2$

$f(x)=2r^2+5rcos(\theta)+2$ $\displaystyle f(x)=2r^2+5rcos(\theta)+2$

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Example Question #12 : Functions, Graphs, And Limits

What is the equation $y=-\frac{1}{2}x^{2}$ $\displaystyle y=-\frac{1}{2}x^{2}$ in polar form?

Possible Answers:

$r=\frac{1}{2}\tan \theta\sec\theta$ $\displaystyle r=\frac{1}{2}\tan \theta\sec\theta$

$r=-2\tan \theta\sec\theta$ $\displaystyle r=-2\tan \theta\sec\theta$

$r=-\tan 2\theta\sec\theta$ $\displaystyle r=-\tan 2\theta\sec\theta$

$r=2\tan \theta\sec\theta$ $\displaystyle r=2\tan \theta\sec\theta$

$r=-\frac{1}{2}\tan \theta\sec\theta$ $\displaystyle r=-\frac{1}{2}\tan \theta\sec\theta$

Correct answer:

$r=-2\tan \theta\sec\theta$ $\displaystyle r=-2\tan \theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ $\displaystyle y=r\sin\theta$ and $x=r\cos\theta$ $\displaystyle x=r\cos\theta$. Given $y=-\frac{1}{2}x^{2}$ $\displaystyle y=-\frac{1}{2}x^{2}$, then:

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

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

Dividing both sides by $r\cos\theta$ $\displaystyle r\cos\theta$, we get:

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

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

$\tan \theta\sec\theta=-\frac{1}{2}r$ $\displaystyle \tan \theta\sec\theta=-\frac{1}{2}r$

$r=-2\tan \theta\sec\theta$ $\displaystyle r=-2\tan \theta\sec\theta$

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Example Question #3 : Polar Form

What is the polar form of $y=-7x^{2}$ $\displaystyle y=-7x^{2}$?

Possible Answers:

$r=-\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta$

$r=\tan7\theta\sec\theta$ $\displaystyle r=\tan7\theta\sec\theta$

$r=7\tan\theta\sec\theta$ $\displaystyle r=7\tan\theta\sec\theta$

$r=\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{7}\tan\theta\sec\theta$

$r=-7\tan\theta\sec\theta$ $\displaystyle r=-7\tan\theta\sec\theta$

Correct answer:

$r=-\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=-7x^{2}$ $\displaystyle y=-7x^{2}$, then:

$r\sin\theta=-7(r\cos\theta)^{2}$ $\displaystyle r\sin\theta=-7(r\cos\theta)^{2}$

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

Dividing both sides by $r\cos\theta$ , we get:

$\tan\theta=-7r\cos\theta$ $\displaystyle \tan\theta=-7r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=-7r$ $\displaystyle \frac{\tan\theta}{\cos\theta}=-7r$

$\tan\theta\sec\theta=-7r$ $\displaystyle \tan\theta\sec\theta=-7r$

$r=-\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta$

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Example Question #11 : Functions, Graphs, And Limits

What is the polar form of y=3-4x $\displaystyle y=3-4x$?

Possible Answers:

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

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

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

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

None of the above

Correct answer:

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

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given y=3-4x $\displaystyle y=3-4x$, then:

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

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

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

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

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Example Question #14 : Functions, Graphs, And Limits

What is the polar form of $y=-x^{2}$ $\displaystyle y=-x^{2}$?

Possible Answers:

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

None of the above

$r=-\tan\theta\sec\theta$ $\displaystyle r=-\tan\theta\sec\theta$

$r=-\frac{\tan\theta}{\sec\theta}$ $\displaystyle r=-\frac{\tan\theta}{\sec\theta}$

$r=\tan\theta\sec\theta$ $\displaystyle r=\tan\theta\sec\theta$

Correct answer:

$r=-\tan\theta\sec\theta$ $\displaystyle r=-\tan\theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=-x^{2}$ $\displaystyle y=-x^{2}$, then:

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

Dividing both sides by $r\cos\theta$ , we get:

$\tan\theta=-r\cos\theta$ $\displaystyle \tan\theta=-r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=-r$ $\displaystyle \frac{\tan\theta}{\cos\theta}=-r$

$\tan\theta\sec\theta=-r$ $\displaystyle \tan\theta\sec\theta=-r$

$r=-\tan\theta\sec\theta$ $\displaystyle r=-\tan\theta\sec\theta$

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Example Question #1 : Polar Form

What is the polar form of $\displaystyle y=-10x+5$?

Possible Answers:

$r=\frac{5}{\sin\theta-10\cos\theta}$ $\displaystyle r=\frac{5}{\sin\theta-10\cos\theta}$

$r=-\frac{5}{\sin\theta+10\cos\theta}$ $\displaystyle r=-\frac{5}{\sin\theta+10\cos\theta}$

$r=-\frac{5}{\sin\theta-10\cos\theta}$ $\displaystyle r=-\frac{5}{\sin\theta-10\cos\theta}$

$r=\frac{5}{\sin\theta+10\cos\theta}$ $\displaystyle r=\frac{5}{\sin\theta+10\cos\theta}$

None of the above

Correct answer:

$r=\frac{5}{\sin\theta+10\cos\theta}$ $\displaystyle r=\frac{5}{\sin\theta+10\cos\theta}$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $\displaystyle y=-10x+5$, then:

$r\sin\theta=-10(r\cos\theta)+5$ $\displaystyle r\sin\theta=-10(r\cos\theta)+5$

$r\sin\theta+10(r\cos\theta)=5$ $\displaystyle r\sin\theta+10(r\cos\theta)=5$

$r(\sin\theta+10\cos\theta)=5$ $\displaystyle r(\sin\theta+10\cos\theta)=5$

$r=\frac{5}{\sin\theta+10\cos\theta}$ $\displaystyle r=\frac{5}{\sin\theta+10\cos\theta}$

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AP Calculus BC : Polar Form

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Polar Form

Example Question #11 : Functions, Graphs, And Limits

Example Question #2 : Polar Form

Example Question #202 : Ap Calculus Bc

Example Question #1 : Polar Form

Example Question #12 : Functions, Graphs, And Limits

Example Question #3 : Polar Form

Example Question #11 : Functions, Graphs, And Limits

Example Question #14 : Functions, Graphs, And Limits

Example Question #1 : Polar Form

All AP Calculus BC Resources

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