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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 #62 : Polar Form

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

Possible Answers:

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

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

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

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

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 #231 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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 #23 : Parametric, Polar, And Vector Functions

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

Possible Answers:

(8,-74.74)

(8,74.74)

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

$(\sqrt{130},74.74)$

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 #74 : Polar

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

Possible Answers:

$(\sqrt{475},77.77)$

$(\sqrt{471},65.32)$

$(\sqrt{481},65.77)$

$(\sqrt{480},64.77)$

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 #63 : Polar Form

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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Example Question #751 : Calculus Ii

Convert the following cartesian coordinates into polar form: (6,-14)

Possible Answers:

$(\sqrt{232},-66.8)$

$(\sqrt{230},62.8)$

$(\sqrt{210},-42.8)$

$(\sqrt{225},54.8)$

Correct answer:

$(\sqrt{232},-66.8)$

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=6

y=-14

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

$r=\sqrt{(6)^2 + (-14)^2}$

$r=\sqrt{36+196}$

$r=\sqrt{232}$

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

$\theta=\arctan(\frac{-14}{6})$

$\theta=-66.8$

$(\sqrt{232},-66.8)$

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Example Question #71 : Polar Form

Convert the following cartesian coordinates into polar form: (-7,-5)

Possible Answers:

$(\sqrt{63},74.16)$

$(\sqrt{15},83.0)$

$(\sqrt{39},123.73)$

$(\sqrt{74},35.54)$

Correct answer:

$(\sqrt{74},35.54)$

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=-7

y=-5

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

$r=\sqrt{(-7)^2 + (-5)^2}$

$r=\sqrt{49+25}$

$r=\sqrt{74}$

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

$\theta=\arctan(\frac{-5}{-7})$

$\theta=\arctan(\frac{5}{7})$

$\theta=35.54$

$(\sqrt{74},35.54)$

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Example Question #71 : Polar Form

Given y=11x^4-x^3+6 calculate $\theta$ in polar form if x=12

Possible Answers:

76.88

97.48

112.32

88.62

Correct answer:

88.62

Explanation:

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

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

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

Solution:

x=12

y=11x^4-x^3+6

$y=11\cdot(12)^4-(12)^3+6$

$y=11\cdot48-36+6$

y=498

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

$\theta=\arctan(\frac{498}{12})$

$\theta= 88.62$

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Example Question #71 : Polar

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

Possible Answers:

-82.87

-74.65

-98.12

-32.77

Correct answer:

-82.87

Explanation:

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

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

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

Solution:

x=-5

y=x^2-x+10

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

y=25+5+10

y=40

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

$\theta=\arctan(\frac{40}{-5})$

$\theta=\arctan(-8)$

$\theta=-82.87$

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Example Question #71 : Polar Form

Given $y=2 \sqrt{x} -1$ calculate $\theta$ in polar form if x=16

Possible Answers:

36.97

23.63

26.74

17.53

Correct answer:

23.63

Explanation:

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

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

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

Solution:

x=16

$y=2\sqrt{x}-1$

$y=2\sqrt{16}-1$

$y=2\cdot4-1$

y=7

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

$\theta=\arctan(\frac{7}{16})$

$\theta=23.63$

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #62 : Polar Form

Example Question #231 : Parametric, Polar, And Vector

Example Question #23 : Parametric, Polar, And Vector Functions

Example Question #74 : Polar

Example Question #63 : Polar Form

Example Question #751 : Calculus Ii

Example Question #71 : Polar Form

Example Question #71 : Polar Form

Example Question #71 : Polar

Example Question #71 : Polar Form

All Calculus 2 Resources

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