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Calculus 2 : Polar Form

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #51 : Polar Form

What is the polar form of y=6x-11 $\displaystyle y=6x-11$?

Possible Answers:

$r=-\frac{11}{6\sin\theta-\cos \theta}$ $\displaystyle r=-\frac{11}{6\sin\theta-\cos \theta}$

$r=-\frac{11}{\sin\theta-6\cos \theta}$ $\displaystyle r=-\frac{11}{\sin\theta-6\cos \theta}$

$r=-\frac{11}{\sin\theta+6\cos \theta}$ $\displaystyle r=-\frac{11}{\sin\theta+6\cos \theta}$

$r=\frac{11}{\sin\theta+6\cos \theta}$ $\displaystyle r=\frac{11}{\sin\theta+6\cos \theta}$

$r=\frac{11}{\sin\theta-6\cos \theta}$ $\displaystyle r=\frac{11}{\sin\theta-6\cos \theta}$

Correct answer:

$r=-\frac{11}{\sin\theta-6\cos \theta}$ $\displaystyle r=-\frac{11}{\sin\theta-6\cos \theta}$

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=6x-11 $\displaystyle y=6x-11$, then:

$r\sin\theta=6r\cos \theta-11$ $\displaystyle r\sin\theta=6r\cos \theta-11$

$r\sin\theta-6r\cos \theta=-11$ $\displaystyle r\sin\theta-6r\cos \theta=-11$

$r(\sin\theta-6\cos \theta)=-11$ $\displaystyle r(\sin\theta-6\cos \theta)=-11$

$r=-\frac{11}{\sin\theta-6\cos \theta}$ $\displaystyle r=-\frac{11}{\sin\theta-6\cos \theta}$

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Example Question #52 : Polar Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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

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Example Question #221 : Parametric, Polar, And Vector

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

Possible Answers:

$r=\frac{1}{3}\tan \theta\sec\theta$ $\displaystyle r=\frac{1}{3}\tan \theta\sec\theta$

$r=\tan \frac{1}{3}\theta\sec\theta$ $\displaystyle r=\tan \frac{1}{3}\theta\sec\theta$

$r=3\tan \theta\sec\theta$ $\displaystyle r=3\tan \theta\sec\theta$

$r=-3\tan \theta\sec\theta$ $\displaystyle r=-3\tan \theta\sec\theta$

$r=-\frac{1}{3}\tan \theta\sec\theta$ $\displaystyle r=-\frac{1}{3}\tan \theta\sec\theta$

Correct answer:

$r=3\tan \theta\sec\theta$ $\displaystyle r=3\tan \theta\sec\theta$

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

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

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

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

$\tan \theta=\frac{3}9{}r\cos\theta$ $\displaystyle \tan \theta=\frac{3}9{}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{3}{9}r$ $\displaystyle \frac{\tan \theta}{\cos\theta}=\frac{3}{9}r$

$\tan \theta\sec\theta=\frac{3}{9}r$ $\displaystyle \tan \theta\sec\theta=\frac{3}{9}r$

$r=\frac{9}{3}\tan \theta\sec\theta$ $\displaystyle r=\frac{9}{3}\tan \theta\sec\theta$

$r=3\tan \theta\sec\theta$ $\displaystyle r=3\tan \theta\sec\theta$

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Example Question #53 : Polar Form

What is the polar form of y=5x+14 $\displaystyle y=5x+14$?

Possible Answers:

None of the above

$r=-\frac{14}{\sin\theta+5\cos \theta}$ $\displaystyle r=-\frac{14}{\sin\theta+5\cos \theta}$

$r=-\frac{14}{\sin\theta-5\cos \theta}$ $\displaystyle r=-\frac{14}{\sin\theta-5\cos \theta}$

$r=\frac{14}{\sin\theta-5\cos \theta}$ $\displaystyle r=\frac{14}{\sin\theta-5\cos \theta}$

$r=\frac{14}{\sin\theta+5\cos \theta}$ $\displaystyle r=\frac{14}{\sin\theta+5\cos \theta}$

Correct answer:

$r=\frac{14}{\sin\theta-5\cos \theta}$ $\displaystyle r=\frac{14}{\sin\theta-5\cos \theta}$

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=5x+14 $\displaystyle y=5x+14$, then:

$r\sin\theta=5r\cos \theta+14$ $\displaystyle r\sin\theta=5r\cos \theta+14$

$r\sin\theta-5r\cos \theta=14$ $\displaystyle r\sin\theta-5r\cos \theta=14$

$r(\sin\theta-5\cos \theta)=14$ $\displaystyle r(\sin\theta-5\cos \theta)=14$

$r=\frac{14}{\sin\theta-5\cos \theta}$ $\displaystyle r=\frac{14}{\sin\theta-5\cos \theta}$

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Example Question #51 : Polar

What is the polar form of y=x-12 $\displaystyle y=x-12$?

Possible Answers:

None of the above

$r=-\frac{12}{\sin\theta-\cos \theta}$ $\displaystyle r=-\frac{12}{\sin\theta-\cos \theta}$

$r=\frac{12}{\sin\theta-\cos \theta}$ $\displaystyle r=\frac{12}{\sin\theta-\cos \theta}$

$r=\frac{12}{\sin\theta+\cos \theta}$ $\displaystyle r=\frac{12}{\sin\theta+\cos \theta}$

$r=-\frac{12}{\sin\theta+\cos \theta}$ $\displaystyle r=-\frac{12}{\sin\theta+\cos \theta}$

Correct answer:

$r=-\frac{12}{\sin\theta-\cos \theta}$ $\displaystyle r=-\frac{12}{\sin\theta-\cos \theta}$

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-12 $\displaystyle y=x-12$, then:

$r\sin\theta=r\cos \theta-12$ $\displaystyle r\sin\theta=r\cos \theta-12$

$r\sin\theta-r\cos \theta=-12$ $\displaystyle r\sin\theta-r\cos \theta=-12$

$r(\sin\theta-\cos \theta)=-12$ $\displaystyle r(\sin\theta-\cos \theta)=-12$

$r=-\frac{12}{\sin\theta-\cos \theta}$ $\displaystyle r=-\frac{12}{\sin\theta-\cos \theta}$

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Example Question #61 : Polar

What is the polar form of y=9x-8 $\displaystyle y=9x-8$?

Possible Answers:

$r=\frac{8}{\sin\theta-9\cos \theta}$ $\displaystyle r=\frac{8}{\sin\theta-9\cos \theta}$

$r=-\frac{8}{\sin\theta-9\cos \theta}$ $\displaystyle r=-\frac{8}{\sin\theta-9\cos \theta}$

$r=-\frac{8}{9\sin\theta-\cos \theta}$ $\displaystyle r=-\frac{8}{9\sin\theta-\cos \theta}$

$r=-\frac{8}{\sin\theta+9\cos \theta}$ $\displaystyle r=-\frac{8}{\sin\theta+9\cos \theta}$

$r=\frac{8}{\sin\theta+9\cos \theta}$ $\displaystyle r=\frac{8}{\sin\theta+9\cos \theta}$

Correct answer:

$r=-\frac{8}{\sin\theta-9\cos \theta}$ $\displaystyle r=-\frac{8}{\sin\theta-9\cos \theta}$

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=9x-8 $\displaystyle y=9x-8$, then:

$r\sin\theta=9(r\cos \theta)-8$ $\displaystyle r\sin\theta=9(r\cos \theta)-8$

$r\sin\theta-9(r\cos \theta)=-8$ $\displaystyle r\sin\theta-9(r\cos \theta)=-8$

$r(\sin\theta-9\cos \theta)=-8$ $\displaystyle r(\sin\theta-9\cos \theta)=-8$

$r=-\frac{8}{\sin\theta-9\cos \theta}$ $\displaystyle r=-\frac{8}{\sin\theta-9\cos \theta}$

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Example Question #62 : Polar

What is the polar form of y=7+x $\displaystyle y=7+x$?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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=7+x $\displaystyle y=7+x$, then:

$r\sin\theta=7+r\cos \theta$ $\displaystyle r\sin\theta=7+r\cos \theta$

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

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

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

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Example Question #63 : Polar

What is the polar form of $y=\frac{5}{11}x^{2}$ $\displaystyle y=\frac{5}{11}x^{2}$?

Possible Answers:

$r=\tan\frac{11}{5}\theta\sec\theta$ $\displaystyle r=\tan\frac{11}{5}\theta\sec\theta$

$r=\frac{11}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{11}{5}\tan\theta\sec\theta$

$r=-\frac{5}{11}\tan\theta\sec\theta$ $\displaystyle r=-\frac{5}{11}\tan\theta\sec\theta$

$r=\frac{5}{11}\tan\theta\sec\theta$ $\displaystyle r=\frac{5}{11}\tan\theta\sec\theta$

$r=-\frac{11}{5}\tan\theta\sec\theta$ $\displaystyle r=-\frac{11}{5}\tan\theta\sec\theta$

Correct answer:

$r=\frac{11}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{11}{5}\tan\theta\sec\theta$

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 $y=\frac{5}{11}x^{2}$ $\displaystyle y=\frac{5}{11}x^{2}$, then:

$r\sin\theta=\frac{5}{11}(r\cos \theta)^{2}$ $\displaystyle r\sin\theta=\frac{5}{11}(r\cos \theta)^{2}$

$\tan\theta=\frac{5}{11}(r\cos \theta)$ $\displaystyle \tan\theta=\frac{5}{11}(r\cos \theta)$

$r=\frac{11}{5}(\frac{\tan\theta}{\cos\theta})$ $\displaystyle r=\frac{11}{5}(\frac{\tan\theta}{\cos\theta})$

$r=\frac{11}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{11}{5}\tan\theta\sec\theta$

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Example Question #64 : Polar

What is the polar form of $y=\frac{6}{5}x^{2}$ $\displaystyle y=\frac{6}{5}x^{2}$?

Possible Answers:

$r=\tan\frac{5}{6} \theta\sec\theta$ $\displaystyle r=\tan\frac{5}{6} \theta\sec\theta$

$r=-\frac{5}{6}\tan \theta\sec\theta$ $\displaystyle r=-\frac{5}{6}\tan \theta\sec\theta$

$r=\frac{6}{5}\tan \theta\sec\theta$ $\displaystyle r=\frac{6}{5}\tan \theta\sec\theta$

$r=-\frac{6}{5}\tan \theta\sec\theta$ $\displaystyle r=-\frac{6}{5}\tan \theta\sec\theta$

$r=\frac{5}{6}\tan \theta\sec\theta$ $\displaystyle r=\frac{5}{6}\tan \theta\sec\theta$

Correct answer:

$r=\frac{5}{6}\tan \theta\sec\theta$ $\displaystyle r=\frac{5}{6}\tan \theta\sec\theta$

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 $y=\frac{6}{5}x^{2}$ $\displaystyle y=\frac{6}{5}x^{2}$, then:

$r\sin\theta=\frac{6}{5}(r\cos \theta)^{2}$ $\displaystyle r\sin\theta=\frac{6}{5}(r\cos \theta)^{2}$

$r\sin\theta=\frac{6}{5}r^{2}\cos^{2} \theta$ $\displaystyle r\sin\theta=\frac{6}{5}r^{2}\cos^{2} \theta$

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

$\tan \theta=3r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{6}{5}r$ $\displaystyle \frac{\tan \theta}{\cos\theta}=\frac{6}{5}r$

$\tan \theta\sec\theta=\frac{6}{5}r$ $\displaystyle \tan \theta\sec\theta=\frac{6}{5}r$

$r=\frac{5}{6}\tan \theta\sec\theta$ $\displaystyle r=\frac{5}{6}\tan \theta\sec\theta$

Report an Error

Example Question #64 : Polar

What is the polar form of y=6x-5 $\displaystyle y=6x-5$?

Possible Answers:

$r=-\frac{5}{6\sin\theta-\cos \theta}$ $\displaystyle r=-\frac{5}{6\sin\theta-\cos \theta}$

$r=\frac{5}{\sin\theta-6\cos \theta}$ $\displaystyle r=\frac{5}{\sin\theta-6\cos \theta}$

$r=\frac{5}{\sin\theta+6\cos \theta}$ $\displaystyle r=\frac{5}{\sin\theta+6\cos \theta}$

$r=-\frac{5}{\sin\theta-6\cos \theta}$ $\displaystyle r=-\frac{5}{\sin\theta-6\cos \theta}$

$r=-\frac{5}{\sin\theta+6\cos \theta}$ $\displaystyle r=-\frac{5}{\sin\theta+6\cos \theta}$

Correct answer:

$r=-\frac{5}{\sin\theta-6\cos \theta}$ $\displaystyle r=-\frac{5}{\sin\theta-6\cos \theta}$

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=6x-5 $\displaystyle y=6x-5$, then:

$r\sin\theta=6(r\cos \theta)-5$ $\displaystyle r\sin\theta=6(r\cos \theta)-5$

$r\sin\theta-6(r\cos \theta)=-5$ $\displaystyle r\sin\theta-6(r\cos \theta)=-5$

$r(\sin\theta-6\cos \theta)=-5$ $\displaystyle r(\sin\theta-6\cos \theta)=-5$

$r=-\frac{5}{\sin\theta-6\cos \theta}$ $\displaystyle r=-\frac{5}{\sin\theta-6\cos \theta}$

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Calculus 2 : Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Polar Form

Example Question #52 : Polar Form

Example Question #221 : Parametric, Polar, And Vector

Example Question #53 : Polar Form

Example Question #51 : Polar

Example Question #61 : Polar

Example Question #62 : Polar

Example Question #63 : Polar

Example Question #64 : Polar

Example Question #64 : Polar

All Calculus 2 Resources

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