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Calculus 2 : Polar Form

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Example Questions

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Example Question #31 : Polar Form

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

Possible Answers:

$r=-\frac{2}{5}\tan\theta\sec\theta$

$r=\frac{5}{2}\tan\theta\sec\theta$

$r=-\frac{5}{2}\tan\theta\sec\theta$

$r=\frac{2}{5}\tan\theta\sec\theta$

$r=-\tan\frac{2}{5}\theta\sec\theta$

Correct answer:

$r=-\frac{5}{2}\tan\theta\sec\theta$

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

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

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

$\tan\theta=-\frac{2}{5}(r\cos \theta)$

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

$\tan\theta\sec\theta=-\frac{2}{5}r$

$r=-\frac{5}{2}\tan\theta\sec\theta$

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Example Question #32 : Polar Form

What is the polar form of $y=11-\frac{1}{2}x$ ?

Possible Answers:

$r=-\frac{11}{\sin\theta-\frac{1}{2}\cos\theta}$

$r=\frac{11}{\sin\theta+\frac{1}{2}\cos\theta}$

$r=\frac{11}{\sin\theta-\frac{1}{2}\cos\theta}$

$r=-\frac{11}{\sin\theta+\frac{1}{2}\cos\theta}$

None of the above

Correct answer:

$r=\frac{11}{\sin\theta+\frac{1}{2}\cos\theta}$

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 $y=11-\frac{1}{2}x$ , then:

$r\sin\theta=11-\frac{1}{2}(r\cos \theta)$

$r\sin\theta+\frac{1}{2}(r\cos \theta)=11$

$r(\sin\theta+\frac{1}{2}\cos\theta)=11$

$r=\frac{11}{\sin\theta+\frac{1}{2}\cos\theta}$

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Example Question #33 : Polar Form

What is the polar form of $y=-\frac{1}{4}x^{2}$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $y=-\frac{1}{4}x^{2}$ , then:

$r\sin\theta=-\frac{1}{4}(r\cos \theta)^{2}$

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

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

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

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

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

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Example Question #34 : Polar Form

What is the polar form of y=7-9x ?

Possible Answers:

$r=\frac{7}{\sin\theta-9\cos \theta}$

None of the above

$r=-\frac{7}{\sin\theta-9\cos \theta}$

$r=\frac{7}{\sin\theta+9\cos \theta}$

$r=-\frac{7}{\sin\theta+9\cos \theta}$

Correct answer:

$r=\frac{7}{\sin\theta+9\cos \theta}$

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=7-9x , then:

$r\sin\theta=7-9(r\cos \theta)$

$r\sin\theta+9(r\cos \theta)=7$

$r(\sin\theta+9\cos \theta)=7$

$r=\frac{7}{\sin\theta+9\cos \theta}$

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Example Question #35 : Polar Form

What is the polar form of y=1+10x ?

Possible Answers:

$r=-\frac{1}{\sin\theta+10\cos \theta}$

$r=\frac{1}{\sin\theta-10\cos \theta}$

$r=\frac{1}{\sin\theta+10\cos \theta}$

$r=\frac{1}{\sin\theta-\cos10 \theta}$

$r=-\frac{1}{\sin\theta-10\cos \theta}$

Correct answer:

$r=\frac{1}{\sin\theta-10\cos \theta}$

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=1+10x , then:

$r\sin\theta=1+10(r\cos \theta)$

$r\sin\theta-10(r\cos \theta)=1$

$r(\sin\theta-10\cos \theta)=1$

$r=\frac{1}{\sin\theta-10\cos \theta}$

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Example Question #36 : Polar Form

What is the polar form of $y=\frac{84}{3}x-3$ ?

Possible Answers:

$r=\frac{3}{\sin\theta+\frac{84}{3}\cos \theta}$

$r=\frac{3}{\sin\theta-\frac{84}{3}\cos \theta}$

None of the above

$r=-\frac{3}{\sin\theta-\frac{84}{3}\cos \theta}$

$r=-\frac{3}{\sin\theta+\frac{84}{3}\cos \theta}$

Correct answer:

$r=-\frac{3}{\sin\theta-\frac{84}{3}\cos \theta}$

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 $y=\frac{84}{3}x-3$ , then:

$r\sin\theta=\frac{84}{3}(r\cos \theta)-3$

$r\sin\theta-\frac{84}{3}(r\cos \theta)=-3$

$r(\sin\theta-\frac{84}{3}\cos \theta)=-3$

$r=-\frac{3}{\sin\theta-\frac{84}{3}\cos \theta}$

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Example Question #37 : Polar Form

What is the polar form of y=2x-5 ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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-5 , then:

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

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

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

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

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Example Question #38 : Polar Form

What is the polar form of $y=\frac{1}{5}x^{2}$ ?

Possible Answers:

$r=-5\tan\theta\sec\theta$

$r=\frac{1}{5}\tan\theta\sec\theta$

$r=5\tan\theta\sec\theta$

$r=-\frac{1}{5}\tan\theta\sec\theta$

$r=\tan5\theta\sec\theta$

Correct answer:

$r=5\tan\theta\sec\theta$

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 $y=\frac{1}{5}x^{2}$ , then:

$r\sin\theta=\frac{1}{5}(r\cos \theta)^{2}$

Dividing both sides by $r\cos\theta$ ,

$\tan\theta=\frac{1}{5}(r\cos \theta)$

$\frac{\tan\theta}{\cos\theta}=\frac{1}{5}r$

$\tan\theta\sec\theta=\frac{1}{5}r$

$r=5\tan\theta\sec\theta$

Report an Error

Example Question #39 : Polar Form

What is the polar form of y=7-3x ?

Possible Answers:

$r=\frac{7}{\sin\theta+3\cos \theta}$

$r=-\frac{7}{\sin\theta-3\cos \theta}$

$r=\frac{7}{\sin3\theta+\cos \theta}$

$r=\frac{7}{\sin\theta-3\cos \theta}$

$r=-\frac{7}{\sin\theta+3\cos \theta}$

Correct answer:

$r=\frac{7}{\sin\theta+3\cos \theta}$

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=7-3x , then:

$r\sin\theta=7-3(r\cos \theta)$

$r\sin\theta+3(r\cos \theta)=7$

$r(\sin\theta+3\cos \theta)=7$

$r=\frac{7}{\sin\theta+3\cos \theta}$

Report an Error

Example Question #31 : Polar Form

What is the polar form of $y=11x^{2}$ ?

Possible Answers:

$r=-\frac{1}{11}\tan \theta\sec\theta$

$r=\tan 11\theta\sec\theta$

$r=-11\tan \theta\sec\theta$

$r=11\tan \theta\sec\theta$

$r=\frac{1}{11}\tan \theta\sec\theta$

Correct answer:

$r=\frac{1}{11}\tan \theta\sec\theta$

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 $y=11x^{2}$ , then:

$r\sin\theta=11(r\cos \theta)^{2}$

$r\sin\theta=11r^{2}\cos^{2} \theta$

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

$\tan \theta=11r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=11r$

$\tan \theta\sec\theta=11r$

$r=\frac{1}{11}\tan \theta\sec\theta$

Report an Error

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Calculus 2 : Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #31 : Polar Form

Example Question #32 : Polar Form

Example Question #33 : Polar Form

Example Question #34 : Polar Form

Example Question #35 : Polar Form

Example Question #36 : Polar Form

Example Question #37 : Polar Form

Example Question #38 : Polar Form

Example Question #39 : Polar Form

Example Question #31 : Polar Form

All Calculus 2 Resources

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