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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 #6 : Polar Form

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

Possible Answers:

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

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

None of the above

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

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

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 : Functions, Graphs, And Limits

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

Possible Answers:

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

None of the above

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

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

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

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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Example Question #23 : Polar Form

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

Possible Answers:

$r=-\frac{1}{9}tan\theta\sec\theta$

$r=tan9\theta\sec\theta$

$r=-9tan\theta\sec\theta$

$r=\frac{1}{9}tan\theta\sec\theta$

$r=9tan\theta\sec\theta$

Correct answer:

$r=-\frac{1}{9}tan\theta\sec\theta$

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 $y=-9x^{2}$ , then:

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

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

$\tan\theta=-9r\cos\theta$

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

$r=-\frac{1}{9}tan\theta\sec\theta$

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Example Question #24 : Polar Form

What is the polar form of y=14-8x ?

Possible Answers:

$r=\frac{14}{\sin\theta+8\cos\theta}$

$r=-\frac{14}{\sin\theta-8\cos\theta}$

$r=\frac{14}{\sin\theta-8\cos\theta}$

$r=-\frac{14}{\sin\theta+8\cos\theta}$

None of the above

Correct answer:

$r=\frac{14}{\sin\theta+8\cos\theta}$

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=14-8x , then:

$r\sin\theta=14-8(r\cos\theta)$

$r\sin\theta+8(r\cos\theta)=14$

$r(\sin\theta+8\cos\theta)=14$

$r=\frac{14}{\sin\theta+8\cos\theta}$

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Example Question #25 : Polar Form

What is the polar form of $y=\frac{5}{4}x+2$ ?

Possible Answers:

None of the above

$r=\frac{2}{\sin\theta+\frac{5}{4}\cos \theta}$

$r=-\frac{2}{\sin\theta-\frac{5}{4}\cos \theta}$

$r=-\frac{2}{\sin\theta+\frac{5}{4}\cos \theta}$

$r=\frac{2}{\sin\theta-\frac{5}{4}\cos \theta}$

Correct answer:

$r=\frac{2}{\sin\theta-\frac{5}{4}\cos \theta}$

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 $y=\frac{5}{4}x+2$ , then:

$r\sin\theta=\frac{5}{4}(r\cos \theta)+2$

$r\sin\theta-\frac{5}{4}(r\cos \theta)=2$

$r(\sin\theta-\frac{5}{4}\cos \theta)=2$

$r=\frac{2}{\sin\theta-\frac{5}{4}\cos \theta}$

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Example Question #26 : Polar Form

What is the polar form of $y=-15x^{2}$ ?

Possible Answers:

$r=15\tan\theta\sec\theta$

$r=\frac{1}{15}\tan\theta\sec\theta$

$r=-\frac{1}{15}\tan\theta\sec\theta$

$r=\tan15\theta\sec\theta$

$r=-15\tan\theta\sec\theta$

Correct answer:

$r=-\frac{1}{15}\tan\theta\sec\theta$

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 $y=-15x^{2}$ , then:

$r\sin\theta=-15(r\cos \theta)^{2}$

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

$\tan\theta=-15(r\cos \theta)$

$r=-\frac{\tan\theta}{15\cos \theta}$

$r=-\frac{1}{15}\tan\theta\sec\theta$

Report an Error

Example Question #27 : Polar Form

What is the polar form of y=3x-8 ?

Possible Answers:

$r=\frac{8}{\sin\theta-3\cos\theta}$

$r=\frac{8}{\sin\theta+3\cos\theta}$

None of the above

$r=-\frac{8}{\sin\theta+3\cos\theta}$

$r=-\frac{8}{\sin\theta-3\cos\theta}$

Correct answer:

$r=-\frac{8}{\sin\theta-3\cos\theta}$

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=3x-8 , then:

$r\sin\theta=3(r\cos \theta)-8$

$r\sin\theta-3(r\cos \theta)=-8$

$r(\sin\theta-3\cos \theta)=-8$

$r=-\frac{8}{\sin\theta-3\cos\theta}$

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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:

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

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

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

None of the above

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

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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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #6 : Polar Form

Example Question #21 : Functions, Graphs, And Limits

Example Question #22 : Polar Form

Example Question #23 : Polar Form

Example Question #24 : Polar Form

Example Question #25 : Polar Form

Example Question #26 : Polar Form

Example Question #27 : Polar Form

Example Question #31 : Polar Form

Example Question #32 : Polar Form

All Calculus 2 Resources

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