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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #58 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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

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

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

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

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Example Question #56 : Polar

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

Possible Answers:

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

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

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

None of the above

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #57 : Polar

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

Possible Answers:

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

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

None of the above

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

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #61 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #62 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #63 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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Example Question #64 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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

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

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

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Example Question #64 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #227 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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

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 #67 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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

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

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

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #58 : Polar

Example Question #56 : Polar

Example Question #57 : Polar

Example Question #61 : Polar

Example Question #62 : Polar

Example Question #63 : Polar

Example Question #64 : Polar

Example Question #64 : Polar

Example Question #227 : Parametric, Polar, And Vector

Example Question #67 : Polar

All Calculus 2 Resources

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