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Calculus 2 : Parametric, Polar, and Vector

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

Example Questions

Example Question #12 : Polar

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

Possible Answers:

$r=\frac{3}{2}\tan \theta\sec\theta$ $\displaystyle r=\frac{3}{2}\tan \theta\sec\theta$

None of the above

$r=\frac{2}{3}\tan \theta\sec\theta$ $\displaystyle r=\frac{2}{3}\tan \theta\sec\theta$

$r=-\frac{3}{2}\tan \theta\sec\theta$ $\displaystyle r=-\frac{3}{2}\tan \theta\sec\theta$

$r=-\frac{2}{3}\tan \theta\sec\theta$ $\displaystyle r=-\frac{2}{3}\tan \theta\sec\theta$

Correct answer:

$r=\frac{3}{2}\tan \theta\sec\theta$ $\displaystyle r=\frac{3}{2}\tan \theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ $\displaystyle y=r\sin\theta$ and $x=r\cos\theta$ $\displaystyle x=r\cos\theta$. Given $y=\frac{2}{3}x^{2}$ $\displaystyle y=\frac{2}{3}x^{2}$, then:

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

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

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

$\tan \theta=\frac{2}{3}r\cos\theta$ $\displaystyle \tan \theta=\frac{2}{3}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{2}{3}r$ $\displaystyle \frac{\tan \theta}{\cos\theta}=\frac{2}{3}r$

$3\tan \theta\sec\theta=2r$ $\displaystyle 3\tan \theta\sec\theta=2r$

$r=\frac{3}{2}\tan \theta\sec\theta$ $\displaystyle r=\frac{3}{2}\tan \theta\sec\theta$

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Example Question #181 : Parametric, Polar, And Vector

What is the polar form of $y=\frac{3}{5}x^{2}$ $\displaystyle y=\frac{3}{5}x^{2}$?

Possible Answers:

None of the above

$r=-\frac{3}{5}\tan\theta\sec\theta$ $\displaystyle r=-\frac{3}{5}\tan\theta\sec\theta$

$r=\frac{5}{3}\tan\theta\sec\theta$ $\displaystyle r=\frac{5}{3}\tan\theta\sec\theta$

$r=\frac{3}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{3}{5}\tan\theta\sec\theta$

$r=-\frac{5}{3}\tan\theta\sec\theta$ $\displaystyle r=-\frac{5}{3}\tan\theta\sec\theta$

Correct answer:

$r=\frac{5}{3}\tan\theta\sec\theta$ $\displaystyle r=\frac{5}{3}\tan\theta\sec\theta$

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 $y=\frac{3}{5}x^{2}$ $\displaystyle y=\frac{3}{5}x^{2}$, then:

$r\sin\theta=\frac{3}{5}(r\cos\theta)^{2}$ $\displaystyle r\sin\theta=\frac{3}{5}(r\cos\theta)^{2}$

$r\sin\theta=\frac{3}{5}r^{2}\cos^{2}\theta$ $\displaystyle r\sin\theta=\frac{3}{5}r^{2}\cos^{2}\theta$

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

$\tan\theta=\frac{3}{5}r\cos\theta$ $\displaystyle \tan\theta=\frac{3}{5}r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=\frac{3}{5}r$ $\displaystyle \frac{\tan\theta}{\cos\theta}=\frac{3}{5}r$

$\tan\theta\sec\theta=\frac{3}{5}r$ $\displaystyle \tan\theta\sec\theta=\frac{3}{5}r$

$r=\frac{5}{3}\tan\theta\sec\theta$ $\displaystyle r=\frac{5}{3}\tan\theta\sec\theta$

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Example Question #12 : Polar

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

Possible Answers:

$r=-7\tan\theta\sec\theta$ $\displaystyle r=-7\tan\theta\sec\theta$

$r=-\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta$

$r=\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{7}\tan\theta\sec\theta$

$r=\tan7\theta\sec\theta$ $\displaystyle r=\tan7\theta\sec\theta$

$r=7\tan\theta\sec\theta$ $\displaystyle r=7\tan\theta\sec\theta$

Correct answer:

$r=-\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta$

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 $y=-7x^{2}$ $\displaystyle y=-7x^{2}$, then:

$r\sin\theta=-7(r\cos\theta)^{2}$ $\displaystyle r\sin\theta=-7(r\cos\theta)^{2}$

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

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

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

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

$\tan\theta\sec\theta=-7r$ $\displaystyle \tan\theta\sec\theta=-7r$

$r=-\frac{1}{7}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta$

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Example Question #12 : Polar Form

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

Possible Answers:

$r=-2\tan\theta\sec\theta$ $\displaystyle r=-2\tan\theta\sec\theta$

$r=\frac{1}{2}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{2}\tan\theta\sec\theta$

None of the above

$r=-\frac{1}{2}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{2}\tan\theta\sec\theta$

$r=2\tan\theta\sec\theta$ $\displaystyle r=2\tan\theta\sec\theta$

Correct answer:

$r=2\tan\theta\sec\theta$ $\displaystyle r=2\tan\theta\sec\theta$

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

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

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

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

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

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

$\tan\theta\sec\theta=\frac{1}{2}r$ $\displaystyle \tan\theta\sec\theta=\frac{1}{2}r$

$r=2\tan\theta\sec\theta$ $\displaystyle r=2\tan\theta\sec\theta$

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Example Question #21 : Polar

What is the polar form of $y=4x^{2}$ $\displaystyle y=4x^{2}$?

Possible Answers:

$r=-4\tan\theta\sec\theta$ $\displaystyle r=-4\tan\theta\sec\theta$

$r=\frac{1}{4}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{4}\tan\theta\sec\theta$

$r=-\frac{1}{4}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{4}\tan\theta\sec\theta$

$r=\tan4\theta\sec\theta$ $\displaystyle r=\tan4\theta\sec\theta$

Correct answer:

$r=\frac{1}{4}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{4}\tan\theta\sec\theta$

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 $y=4x^{2}$ $\displaystyle y=4x^{2}$, then:

$r\sin\theta=4(r\cos\theta)^{2}$ $\displaystyle r\sin\theta=4(r\cos\theta)^{2}$

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

$\tan\theta=4r\cos\theta$ $\displaystyle \tan\theta=4r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=4r$ $\displaystyle \frac{\tan\theta}{\cos\theta}=4r$

$\tan\theta\sec\theta=4r$ $\displaystyle \tan\theta\sec\theta=4r$

$r=\frac{1}{4}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{4}\tan\theta\sec\theta$

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Example Question #7 : Polar Form

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

Possible Answers:

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

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

None of the above

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

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

Correct answer:

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

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

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

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

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

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

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

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

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Example Question #13 : Polar Form

What is the polar form of $y=\frac{1}{6}x^{2}$ $\displaystyle y=\frac{1}{6}x^{2}$?

Possible Answers:

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

$r=\frac{1}{6}\tan\theta\sec\theta$ $\displaystyle r=\frac{1}{6}\tan\theta\sec\theta$

$r=-6\tan\theta\sec\theta$ $\displaystyle r=-6\tan\theta\sec\theta$

$r=-\frac{1}{6}\tan\theta\sec\theta$ $\displaystyle r=-\frac{1}{6}\tan\theta\sec\theta$

$r=6\tan\theta\sec\theta$ $\displaystyle r=6\tan\theta\sec\theta$

Correct answer:

$r=6\tan\theta\sec\theta$ $\displaystyle r=6\tan\theta\sec\theta$

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 $y=\frac{1}{6}x^{2}$ $\displaystyle y=\frac{1}{6}x^{2}$, then:

$r\sin\theta=\frac{1}{6}(r\cos\theta)^{2}$ $\displaystyle r\sin\theta=\frac{1}{6}(r\cos\theta)^{2}$

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

$\tan\theta=\frac{1}{6}r\cos\theta$ $\displaystyle \tan\theta=\frac{1}{6}r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=\frac{1}{6}r$ $\displaystyle \frac{\tan\theta}{\cos\theta}=\frac{1}{6}r$

$\tan\theta\sec\theta=\frac{1}{6}r$ $\displaystyle \tan\theta\sec\theta=\frac{1}{6}r$

$r=6\tan\theta\sec\theta$ $\displaystyle r=6\tan\theta\sec\theta$

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Example Question #21 : Polar

What is the polar form of $y=\frac{5}{3}x^{2}$ $\displaystyle y=\frac{5}{3}x^{2}$?

Possible Answers:

$r=-\frac{3}{5}\tan\theta\sec\theta$ $\displaystyle r=-\frac{3}{5}\tan\theta\sec\theta$

$r=\frac{5}{3}\tan\theta\sec\theta$ $\displaystyle r=\frac{5}{3}\tan\theta\sec\theta$

$r=\frac{3}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{3}{5}\tan\theta\sec\theta$

$r=-\frac{5}{3}\tan\theta\sec\theta$ $\displaystyle r=-\frac{5}{3}\tan\theta\sec\theta$

None of the above

Correct answer:

$r=\frac{3}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{3}{5}\tan\theta\sec\theta$

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 $y=\frac{5}{3}x^{2}$ $\displaystyle y=\frac{5}{3}x^{2}$, then:

$r\sin\theta=\frac{5}{3}(r\cos\theta)^{2}$ $\displaystyle r\sin\theta=\frac{5}{3}(r\cos\theta)^{2}$

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

$\tan\theta=\frac{5}{3}(r\cos\theta)$ $\displaystyle \tan\theta=\frac{5}{3}(r\cos\theta)$

$\frac{\tan\theta}{\cos\theta}=\frac{5}{3}r$ $\displaystyle \frac{\tan\theta}{\cos\theta}=\frac{5}{3}r$

$\tan\theta\sec\theta=\frac{5}{3}r$ $\displaystyle \tan\theta\sec\theta=\frac{5}{3}r$

$r=\frac{3}{5}\tan\theta\sec\theta$ $\displaystyle r=\frac{3}{5}\tan\theta\sec\theta$

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Example Question #21 : Polar Form

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

Possible Answers:

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

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

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

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

None of the above

Correct answer:

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

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

$r\sin\theta=2(r\cos\theta)-1$ $\displaystyle r\sin\theta=2(r\cos\theta)-1$

$r\sin\theta-2(r\cos\theta)=-1$ $\displaystyle r\sin\theta-2(r\cos\theta)=-1$

$r(\sin\theta-2\cos\theta)=-1$ $\displaystyle r(\sin\theta-2\cos\theta)=-1$

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

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Example Question #6 : Polar Form

What is the polar form of y=3-4x $\displaystyle y=3-4x$?

Possible Answers:

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

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

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

None of the above

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

Correct answer:

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

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=3-4x $\displaystyle y=3-4x$, then:

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

$r\sin\theta+4(r\cos\theta)=3$ $\displaystyle r\sin\theta+4(r\cos\theta)=3$

$r(\sin\theta+4\cos\theta)=3$ $\displaystyle r(\sin\theta+4\cos\theta)=3$

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

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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 #12 : Polar

Example Question #181 : Parametric, Polar, And Vector

Example Question #12 : Polar

Example Question #12 : Polar Form

Example Question #21 : Polar

Example Question #7 : Polar Form

Example Question #13 : Polar Form

Example Question #21 : Polar

Example Question #21 : Polar Form

Example Question #6 : Polar Form

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