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Calculus 2 : Polar

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

Example Questions

Example Question #61 : Polar

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

Possible Answers:

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

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

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

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

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

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 #221 : Parametric, Polar, And Vector

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 #61 : Polar

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

Possible Answers:

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

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

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

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

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

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 #221 : Parametric, Polar, And Vector

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 #61 : Polar Form

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

Report an Error

Example Question #68 : Polar

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

Possible Answers:

$r=-\frac{2}{9}\tan \theta\sec\theta$

$r=\frac{9}{2}\tan \theta\sec\theta$

$r=\tan \frac{2}{9}\theta\sec\theta$

$r=-\frac{9}{2}\tan \theta\sec\theta$

$r=\frac{2}{9}\tan \theta\sec\theta$

Correct answer:

$r=\frac{9}{2}\tan \theta\sec\theta$

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

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

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

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

$\tan \theta=\frac{2}{9}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{2}{9}r$

$\tan \theta\sec\theta=\frac{2}{9}r$

$r=\frac{9}{2}\tan \theta\sec\theta$

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Example Question #61 : Polar Form

What is the polar form of y=7x+2

Possible Answers:

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

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

$r=-\frac{2}{\sin\theta+7\cos \theta}$

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

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

Correct answer:

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

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=7x+2 , then:

$r\sin\theta=7(r\cos \theta)+2$

$r\sin\theta-7r\cos \theta=2$

$r(\sin\theta-7\cos \theta)=2$

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

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Example Question #741 : Calculus Ii

Calculate the polar form hypotenuse of the following cartesian equation: y=8x^3

Possible Answers:

$\sqrt{\frac{\tan(\theta)}{8\cos^2(\theta)}}$

$\frac{\tan(\theta)}{8\cos(\theta)}$

$\sqrt{}\frac{\tan(\theta)}{8\cos(\theta)}}$

$\sqrt{\frac{\tan(\theta)}{\cos(\theta)}}$

Correct answer:

$\sqrt{\frac{\tan(\theta)}{8\cos^2(\theta)}}$

Explanation:

In a cartesian form, the primary parameters are x and y. In polar form, they are and $\theta$

is the hypotenuse, and $\theta$ is the angle created by .

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

$y = 8x^3 = r\sin(\theta)$

$x= r\cos(\theta)$

$y = 8\cdot(r\cos(\theta))^3$

$y = 8\cdot r^3\cdot\cos^3(\theta) = r\sin(\theta)$

$r = \sqrt{\frac{\tan(\theta)}{8\cos^2(\theta)}}$

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Polar

Example Question #221 : Parametric, Polar, And Vector

Example Question #61 : Polar

Example Question #64 : Polar

Example Question #221 : Parametric, Polar, And Vector

Example Question #61 : Polar Form

Example Question #67 : Polar

Example Question #68 : Polar

Example Question #61 : Polar Form

Example Question #741 : Calculus Ii

All Calculus 2 Resources

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