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Calculus 2 : Polar

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Example Questions

Example Question #36 : Polar Form

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

Possible Answers:

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

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

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

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

None of the above

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:

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

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

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

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

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

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:

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

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

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

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

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

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 $r\cos\theta$ ,

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

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

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

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

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Example Question #39 : Polar Form

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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Example Question #40 : Polar Form

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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Example Question #11 : Polar Form

What is the polar form of y=2x-7 ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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-7 , then:

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

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

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

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

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Example Question #41 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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-x , then:

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

$r\sin\theta+r\cos \theta=1$

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

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

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Example Question #42 : Polar

What is the polar form of $y=\frac{3}{10}x^{2}$ ?

Possible Answers:

$r=\frac{10}{3}\tan \theta\sec\theta$

$r=\frac{3}{10}\tan \theta\sec\theta$

$r=-\frac{10}{3}\tan \theta\sec\theta$

$r=\tan\frac{3}{10} \theta\sec\theta$

$r=-\frac{3}{10}\tan \theta\sec\theta$

Correct answer:

$r=\frac{10}{3}\tan \theta\sec\theta$

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 $y=\frac{3}{10}x^{2}$ , then:

$r\sin\theta=\frac{3}{10}(r\cos \theta)^{2}$

$r\sin\theta=\frac{3}{10}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{3}{10}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{3}{10}r$

$\tan \theta\sec\theta=\frac{3}{10}r$

$r=\frac{10}{3}\tan \theta\sec\theta$

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Example Question #43 : Polar

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

Possible Answers:

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

$r=\frac{1}{\sin2\theta+\cos \theta}$

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

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

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

Correct answer:

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

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

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

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

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

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

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Example Question #44 : Polar

What is the polar form of y=19+5x ?

Possible Answers:

$r=\frac{19}{\sin\theta-5\cos \theta}$

$r=\frac{19}{5\sin\theta+\cos \theta}$

$r=-\frac{19}{\sin\theta-5\cos \theta}$

$r=\frac{19}{\sin\theta+5\cos \theta}$

$r=-\frac{19}{\sin\theta+5\cos \theta}$

Correct answer:

$r=\frac{19}{\sin\theta-5\cos \theta}$

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=19+5x , then:

$r\sin\theta=19+5(r\cos \theta)$

$r\sin\theta-5(r\cos \theta)=19$

$r(\sin\theta-5\cos \theta)=19$

$r=\frac{19}{\sin\theta-5\cos \theta}$

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #36 : Polar Form

Example Question #37 : Polar Form

Example Question #38 : Polar Form

Example Question #39 : Polar Form

Example Question #40 : Polar Form

Example Question #11 : Polar Form

Example Question #41 : Polar

Example Question #42 : Polar

Example Question #43 : Polar

Example Question #44 : Polar

All Calculus 2 Resources

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