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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 #44 : Polar

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

Possible Answers:

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

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

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

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

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

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 #722 : Calculus Ii

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

Possible Answers:

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

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

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

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

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

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 #41 : Polar

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

Possible Answers:

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

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

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

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

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

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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Example Question #51 : Polar

What is the polar form of $y=\frac{5}{7}x^{2}$ ?

Possible Answers:

$r=\frac{7}{5}\tan \theta\sec\theta$

$r=\tan \frac{5}{7} \theta\sec\theta$

$r=-\frac{5}{7}\tan \theta\sec\theta$

$r=\frac{5}{7}\tan \theta\sec\theta$

$r=-\frac{7}{5}\tan \theta\sec\theta$

Correct answer:

$r=\frac{7}{5}\tan \theta\sec\theta$

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 $y=\frac{5}{7}x^{2}$ , then:

$r\sin\theta=\frac{5}{7}(r\cos \theta)^{2}$

$r\sin\theta=\frac{5}{7}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{5}{7}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{5}{7}r$

$\tan \theta\sec\theta=\frac{5}{7}r$

$r=\frac{7}{5}\tan \theta\sec\theta$

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Example Question #724 : Calculus Ii

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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

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

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

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Example Question #725 : Calculus Ii

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

Possible Answers:

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

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

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

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

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

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=8-3x , then:

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

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

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

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

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Example Question #726 : Calculus Ii

What is the polar form of $y=\frac{9}{7}x^{2}$ ?

Possible Answers:

$r=-\frac{9}{7}\tan \theta\sec\theta$

$r=\frac{9}{7}\tan \theta\sec\theta$

$r=\tan\frac{7}{9} \theta\sec\theta$

$r=\frac{7}{9}\tan \theta\sec\theta$

$r=-\frac{7}{9}\tan \theta\sec\theta$

Correct answer:

$r=\frac{7}{9}\tan \theta\sec\theta$

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 $y=\frac{9}{7}x^{2}$ , then:

$r\sin\theta=\frac{9}{7}(r\cos \theta)^{2}$

$r\sin\theta=\frac{9}{7}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{9}{7}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{9}{7}r$

$\tan \theta\sec\theta=\frac{9}{7}r$

$r=\frac{7}{9}\tan \theta\sec\theta$

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Example Question #41 : Polar Form

What is the polar form of y=10+4x ?

Possible Answers:

$r=-\frac{10}{\sin\theta-4\cos \theta}$

$r=\frac{10}{\sin\theta-4\cos \theta}$

$r=\frac{10}{\sin\theta+4\cos \theta}$

$r=\frac{10}{4\sin\theta-\cos \theta}$

$r=-\frac{10}{\sin\theta+4\cos \theta}$

Correct answer:

$r=\frac{10}{\sin\theta-4\cos \theta}$

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=10+4x , then:

$r\sin\theta=10+4r\cos \theta$

$r\sin\theta-4r\cos \theta=10$

$r(\sin\theta-4\cos \theta)=10$

$r=\frac{10}{\sin\theta-4\cos \theta}$

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Example Question #51 : Polar Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 , then:

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

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

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

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

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Example Question #52 : Polar Form

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

Possible Answers:

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

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

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

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

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

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

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

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

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

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

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

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #44 : Polar

Example Question #722 : Calculus Ii

Example Question #41 : Polar

Example Question #51 : Polar

Example Question #724 : Calculus Ii

Example Question #725 : Calculus Ii

Example Question #726 : Calculus Ii

Example Question #41 : Polar Form

Example Question #51 : Polar Form

Example Question #52 : Polar Form

All Calculus 2 Resources

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