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Calculus 2 : Polar

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Example Question #14 : Polar Form

What is the polar form of $y=\frac{1}{2}x^{2}$ ?

Possible Answers:

None of the above

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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Example Question #15 : Polar Form

What is the polar form of $y=4x^{2}$ ?

Possible Answers:

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

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

$r=\tan4\theta\sec\theta$

$r=-\frac{1}{4}\tan\theta\sec\theta$

Correct answer:

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

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 $y=4x^{2}$ , then:

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

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

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

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

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

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

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Example Question #1 : Polar Form

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

Possible Answers:

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

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

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

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

None of the above

Correct answer:

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

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 $y=-x^{2}$ , then:

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

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

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

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

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

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

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Example Question #16 : Polar Form

What is the polar form of $y=\frac{1}{6}x^{2}$ ?

Possible Answers:

$r=-\frac{1}{6}\tan\theta\sec\theta$

$r=\frac{1}{6}\tan\theta\sec\theta$

$r=6\tan\theta\sec\theta$

$r=\frac{\tan\theta}{6\sec\theta}$

$r=-6\tan\theta\sec\theta$

Correct answer:

$r=6\tan\theta\sec\theta$

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 $y=\frac{1}{6}x^{2}$ , then:

$r\sin\theta=\frac{1}{6}(r\cos\theta)^{2}$

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

$\tan\theta=\frac{1}{6}r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=\frac{1}{6}r$

$\tan\theta\sec\theta=\frac{1}{6}r$

$r=6\tan\theta\sec\theta$

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Example Question #17 : Polar Form

What is the polar form of $y=\frac{5}{3}x^{2}$ ?

Possible Answers:

$r=\frac{5}{3}\tan\theta\sec\theta$

None of the above

$r=-\frac{5}{3}\tan\theta\sec\theta$

$r=\frac{3}{5}\tan\theta\sec\theta$

$r=-\frac{3}{5}\tan\theta\sec\theta$

Correct answer:

$r=\frac{3}{5}\tan\theta\sec\theta$

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 $y=\frac{5}{3}x^{2}$ , then:

$r\sin\theta=\frac{5}{3}(r\cos\theta)^{2}$

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

$\tan\theta=\frac{5}{3}(r\cos\theta)$

$\frac{\tan\theta}{\cos\theta}=\frac{5}{3}r$

$\tan\theta\sec\theta=\frac{5}{3}r$

$r=\frac{3}{5}\tan\theta\sec\theta$

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Example Question #21 : Polar Form

What is the polar form of y=2x-1 ?

Possible Answers:

None of the above

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

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

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

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

Correct answer:

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

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=2x-1 , then:

$r\sin\theta=2(r\cos\theta)-1$

$r\sin\theta-2(r\cos\theta)=-1$

$r(\sin\theta-2\cos\theta)=-1$

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

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Example Question #8 : Polar Form

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

Possible Answers:

None of the above

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

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

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

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #10 : Polar Form

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

Possible Answers:

None of the above

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

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

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

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

Correct answer:

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

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 y=-10x+5 , then:

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

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

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

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

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Example Question #21 : Parametric, Polar, And Vector Functions

What is the polar form of $y=\frac{2}{3}x+9$ ?

Possible Answers:

$r=\frac{9}{\sin\theta+\frac{2}{3}\cos\theta}$

$r=-\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

$r=-\frac{9}{\sin\theta+\frac{2}{3}\cos\theta}$

None of the above

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

Correct answer:

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

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 $y=\frac{2}{3}x+9$ , then:

$r\sin\theta= \frac{2}{3}(r\cos\theta)+9$

$r\sin\theta-\frac{2}{3}(r\cos\theta)=9$

$r(\sin\theta-\frac{2}{3}\cos\theta)=9$

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

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Example Question #22 : Polar Form

What is the polar form of $y=\frac{1}{2}x+6$ ?

Possible Answers:

$r=-\frac{6}{\sin\theta+\frac{1}{2}\cos\theta}$

$r=\frac{6}{\sin\theta+\frac{1}{2}\cos\theta}$

$r=\frac{6}{\sin\theta-\frac{1}{2}\cos\theta}$

$r=-\frac{6}{\sin\theta-\frac{1}{2}\cos\theta}$

None of the above

Correct answer:

$r=\frac{6}{\sin\theta-\frac{1}{2}\cos\theta}$

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 $y=\frac{1}{2}x+6$ , then:

$r\sin\theta=\frac{1}{2}(r\cos\theta)+6$

$r\sin\theta-\frac{1}{2}(r\cos\theta)=6$

$r(\sin\theta-\frac{1}{2}\cos\theta)=6$

$r=\frac{6}{\sin\theta-\frac{1}{2}\cos\theta}$

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #14 : Polar Form

Example Question #15 : Polar Form

Example Question #1 : Polar Form

Example Question #16 : Polar Form

Example Question #17 : Polar Form

Example Question #21 : Polar Form

Example Question #8 : Polar Form

Example Question #10 : Polar Form

Example Question #21 : Parametric, Polar, And Vector Functions

Example Question #22 : Polar Form

All Calculus 2 Resources

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