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Calculus 2 : Polar

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Example Questions

Example Question #81 : Polar Form

Given y=5x^3-8x^2+x-9 calculate $\theta$ in polar form if x=7

Possible Answers:

81.2

89.7

92.6

84.5

Correct answer:

89.7

Explanation:

You need to calculate $\theta$ . Before you do so, first find and . You are given and a function , so plug in into .

After you have and , use the trig function $\tan$ .

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

Solution:

y=5x^3-8x^2+x-9

x=7

y=5(7)^3-8(7)^2+7-9

$y=5\cdot343-8\cdot49-2$

y=1715-392-2

y=1321

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

$\theta=\arctan(\frac{1321}{7})$

$\theta=89.7$

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Example Question #87 : Polar Form

Given y=3x^4-4x^3+5x^2+10 calculate $\theta$ in polar form if x=3

Possible Answers:

63.4

74.1

89.1

64.7

Correct answer:

89.1

Explanation:

You need to calculate $\theta$ . Before you do so, first find and . You are given and a function , so plug in into .

After you have and , use the trig function $\tan$ .

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

Solution:

y=3x^4-4x^3+5x^2+10

x=3

y=3(3)^4-4(3)^3+5(3)^2+10

$y=3\cdot81-4\cdot27+5\cdot9+10$

y=243-108+45+10

y=190

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

$\theta=\arctan(\frac{190}{3})$

$\theta=89.1$

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Example Question #82 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation: $y=\frac{x^2}{36}$

Possible Answers:

$\frac{36\tan(\theta)}{\cos^2(\theta)}$

$\frac{6\tan(\theta)}{\cos(\theta)}$

$\frac{6\tan(\theta)}{\cos^2(\theta)}$

$\frac{36\tan(\theta)}{\cos(\theta)}$

Correct answer:

$\frac{36\tan(\theta)}{\cos(\theta)}$

Explanation:

In a cartesian form, the primary parameters are and . 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:

$y=\frac{x^2}{36} = r\sin(\theta)$

$x= r\cos(\theta)$

$y=\frac{r^2\cos^2(\theta)}{36} = r\sin(\theta)$

$r=\frac{36\tan(\theta)}{\cos(\theta)}$

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Example Question #89 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation: $y=\frac{x^3}{100}$

Possible Answers:

$\sqrt\frac{100\tan(\theta)}{\cos(\theta)}$

$\sqrt\frac{10\tan(\theta)}{\cos(\theta)}$

$\sqrt\frac{10\tan(\theta)}{\cos^2(\theta)}$

$\sqrt\frac{100\tan(\theta)}{\cos^2(\theta)}$

Correct answer:

$\sqrt\frac{100\tan(\theta)}{\cos^2(\theta)}$

Explanation:

In a cartesian form, the primary parameters are and . 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:

$y=\frac{x^3}{100} = r\sin(\theta)$

$y=\frac{(r\cos(\theta))^3}{100}$

$y=\frac{r^3\cos^3(\theta)}{100} = r\sin(\theta)$

$r = \sqrt\frac{100\tan(\theta)}{\cos^2(\theta)}$

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Example Question #90 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation: $y=\frac{2x^2}{7}$

Possible Answers:

$\frac{7\tan(\theta)}{2\sin^2(\theta)}$

$\frac{7\tan(\theta)}{2\sin(\theta)}$

$\frac{7\tan(\theta)}{2\cos^2(\theta)}$

$\frac{7\tan(\theta)}{2\cos(\theta)}$

Correct answer:

$\frac{7\tan(\theta)}{2\cos(\theta)}$

Explanation:

In a cartesian form, the primary parameters are and . 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:

$y=\frac{2x^2}{7} = r\sin(\theta)$

$x= r\cos(\theta)$

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

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

$r = \frac{7\tan(\theta)}{2\cos(\theta)}$

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Example Question #15 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation: y^2=100x^8

Possible Answers:

$\sqrt[3]{\frac{\tan(\theta)}{\cos(\theta)}}$

$\frac{\tan(\theta)}{10\cos^3(\theta)}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos(\theta)}}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Correct answer:

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Explanation:

In a cartesian form, the primary parameters are and . 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^2=100x^8

$y=10x^4=r\sin(\theta)$

$y=10(r\cos(\theta))^4$

$y=10r^4\cos^4(\theta) = r\sin(\theta)$

$r^3 = \frac{\tan(\theta)}{10\cos^3(\theta)}$

$r = \sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

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Example Question #771 : Calculus Ii

What is the polar form of $y=\frac{4}{3}x^{2}$

Possible Answers:

$r=\frac{4}{3}\tan \theta\sec\theta$

$r=-\frac{4}{3}\tan \theta\sec\theta$

$r=-\frac{3}{4}\tan \theta\sec\theta$

$r=\frac{3}{4}\tan \theta\sec\theta$

$r=\tan\frac{3}{4} \theta\sec\theta$

Correct answer:

$r=\frac{3}{4}\tan \theta\sec\theta$

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 $y=\frac{4}{3}x^{2}$ , then:

$r\sin\theta=\frac{4}{3}(r\cos \theta)^{2}$

$r\sin\theta=\frac{4}{3}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{4}{3}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{4}{3}r$

$\tan \theta\sec\theta=\frac{4}{3}r$

$r=\frac{3}{4}\tan \theta\sec\theta$

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Example Question #772 : Calculus Ii

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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

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

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

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

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Example Question #773 : Calculus Ii

What is the polar form of y=6+14x ?

Possible Answers:

$r=\frac{6}{\sin\theta-14\cos \theta}$

$r=\frac{6}{\sin\theta+14\cos \theta}$

$r=-\frac{6}{\sin\theta-14\cos \theta}$

$r=\frac{\sin\theta-14\cos \theta}{6}$

$r=-\frac{6}{\sin\theta+14\cos \theta}$

Correct answer:

$r=\frac{6}{\sin\theta-14\cos \theta}$

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=6+14x , then:

$r\sin\theta=6+14(r\cos \theta)$

$r\sin\theta-14(r\cos \theta)=6$

$r(\sin\theta-14\cos \theta)=6$

$r=\frac{6}{\sin\theta-14\cos \theta}$

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Example Question #774 : Calculus Ii

What is the polar form of y=2-9x ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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=2-9x , then:

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

$r\sin\theta+9(r\cos \theta)=2$

$r(\sin\theta+9\cos \theta)=2$

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

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Polar Form

Example Question #87 : Polar Form

Example Question #82 : Polar Form

Example Question #89 : Polar Form

Example Question #90 : Polar Form

Example Question #15 : Polar Form

Example Question #771 : Calculus Ii

Example Question #772 : Calculus Ii

Example Question #773 : Calculus Ii

Example Question #774 : Calculus Ii

All Calculus 2 Resources

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