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Calculus 2 : Polar Form

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

Example Questions

Example Question #66 : Polar

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:

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

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

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

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

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

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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Example Question #68 : Polar

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

Possible Answers:

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

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

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

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

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

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 #69 : Polar

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 #70 : Polar

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

Possible Answers:

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

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

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

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

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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Example Question #71 : Polar

Calculate the polar form hypotenuse of the following cartesian equation: $y=9x^{20}$

Possible Answers:

$\sqrt[19]{\frac{\tan(\theta)}{9\cos(\theta)^{19}}}$

$\sqrt[20]{\frac{\tan(\theta)}{9\cos^{20}(\theta)}}$

$\sqrt[20]{\frac{\tan(\theta)}{9\cos^{19}(\theta)}}$

$\sqrt[19]{\frac{\tan(\theta)}{9\cos^{20}(\theta)}}$

Correct answer:

$\sqrt[19]{\frac{\tan(\theta)}{9\cos(\theta)^{19}}}$

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:

$y=9x^{20} = r\sin(\theta)$

$y=9(r\cos(\theta))^{20} = 9r^{20}\cos^{20}(\theta)$

$r\sin(\theta) = 9r^{20}\cos^{20}(\theta)$

$r = \sqrt[19]{\frac{\tan(\theta)}{9\cos^{19}(\theta)}}$

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Example Question #72 : Polar

Calculate the polar form hypotenuse of the following cartesian equation: $y=\frac{x^3}{81}$

Possible Answers:

$\sqrt{\frac{9\tan(\theta)}{\cos(\theta)}}$

$\sqrt{\frac{9\tan(\theta)}{\cos^2(\theta)}}$

$\sqrt{\frac{81\tan(\theta)}{\cos^2(\theta)}}$

$\sqrt{\frac{81\tan(\theta)}{\cos(\theta)}}$

Correct answer:

$\sqrt{\frac{81\tan(\theta)}{\cos^2(\theta)}}$

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:

$y=\frac{x^3}{81} = r\sin(\theta)$

$x = r\cos(\theta)$

$y=\frac{(r\cos(\theta))^3}{81} = \frac{r^3\cos^3(\theta)}{81}$

$r\sin(\theta) = \frac{r^3\cos^3(\theta)}{81}$

$r = \sqrt{\frac{81\tan(\theta)}{\cos^2(\theta)}}$

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Example Question #13 : Polar Form

Convert the following cartesian coordinates into polar form: (-3,11)

Possible Answers:

$(\sqrt{130},-74.74)$

(8,74.74)

(8,-74.74)

$(\sqrt{130},74.74)$

Correct answer:

$(\sqrt{130},-74.74)$

Explanation:

Cartesian coordinates have x and y, represented as (x,y). Polar coordinates have $(r,\theta)$

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

$r = \sqrt{x^2+y^2}$

$\theta = \arctan(\frac{y}{x})$

Solution:

x=-3

y=11

$r = \sqrt{x^2+y^2}$

$r = \sqrt{(-3)^2+11^2 }= \sqrt{(9+121)}$

$r = \sqrt{130}$

$\theta = \arctan(\frac{y}{x})$

$\theta = \arctan(-\frac{11}{3}) = -74.74$

$(\sqrt{130},-74.74)$

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Example Question #73 : Polar

Convert the following cartesian coordinates into polar form: (9,20)

Possible Answers:

$(\sqrt{471},65.32)$

$(\sqrt{481},65.77)$

$(\sqrt{480},64.77)$

$(\sqrt{475},77.77)$

Correct answer:

$(\sqrt{481},65.77)$

Explanation:

Cartesian coordinates have x and y, represented as (x,y). Polar coordinates have $(r,\theta)$

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

$r = \sqrt{x^2+y^2}$

$\theta = \arctan(\frac{y}{x})$

Solution:

x=9

y=20

$r = \sqrt{x^2+y^2}$

$r=\sqrt{(9)^2 + (20)^2}$

$r=\sqrt{81+400}$

$r=\sqrt{481}$

$\theta=\arctan(\frac{y}{x})$

$\theta=\arctan(\frac{20}{9})$

$\theta=65.77$

$(\sqrt{481},65.77)$

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Example Question #74 : Polar

Calculate the polar form hypotenuse of the following cartesian equation: $y=\frac{x^2}{144}$

Possible Answers:

$\frac{144\tan(12\theta)}{\cos(\theta)}$

$\frac{12\tan(\theta)}{\cos(\theta)}$

$\frac{144\tan(\theta)}{\cos(\theta)}$

$\frac{12\tan(12\theta)}{\cos(\theta)}$

Correct answer:

$\frac{144\tan(\theta)}{\cos(\theta)}$

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:

$y=\frac{x^2}{144} = r\sin(\theta)$

$x = r\cos(\theta)$

$y=\frac{(r\cos(\theta))^2}{144}$

$r\sin(\theta) = \frac{r^2\cos^2(\theta)}{144}$

$r = \frac{144\tan(\theta)}{\cos(\theta)}$

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Calculus 2 : Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #66 : Polar

Example Question #67 : Polar

Example Question #68 : Polar

Example Question #69 : Polar

Example Question #70 : Polar

Example Question #71 : Polar

Example Question #72 : Polar

Example Question #13 : Polar Form

Example Question #73 : Polar

Example Question #74 : Polar

All Calculus 2 Resources

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