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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #79 : Polar Form

Convert the following cartesian coordinates into polar form: (16,-21)

Possible Answers:

$(\sqrt{672},-51.50)$

$(\sqrt{637},-41.54)$

$(\sqrt{697},-52.70)$ s

$(\sqrt{614},-72.76)$

Correct answer:

$(\sqrt{697},-52.70)$ s

Explanation:

Cartesian coordinates have and , 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=16

y=-21

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

$r=\sqrt{(16)^2 + (-21)^2}$

$r=\sqrt{256+441}$

$r=\sqrt{697}$

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

$\theta=\arctan(\frac{-21}{16})$

$\theta=-52.70$

$(\sqrt{697},-52.70)$

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Example Question #85 : Polar

Convert the following cartesian coordinates into polar form:

(15,13)

Possible Answers:

$(\sqrt{363},75.21)$

$(\sqrt{394},40.91)$

$(\sqrt{374},81.70)$

$(\sqrt{390},40.45)$

Correct answer:

$(\sqrt{394},40.91)$

Explanation:

Cartesian coordinates have and , 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=15

y=13

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

$r=\sqrt{(15)^2 + (13)^2}$

$r=\sqrt{225+169}$

$r=\sqrt{394}$

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

$\theta=\arctan(\frac{13}{15})$

$\theta=40.91$

$(\sqrt{394},40.91)$

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Example Question #86 : Polar

Convert the following cartesian coordinates into polar form:

(-12,14)

Possible Answers:

$(\sqrt{365},-74.80)$

$(\sqrt{340},-49.40)$

$(\sqrt{290},-61.00)$

$(\sqrt{320},-57.61)$

Correct answer:

$(\sqrt{340},-49.40)$

Explanation:

Cartesian coordinates have and , 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=-12

y=14

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

$r=\sqrt{(-12)^2 + (14)^2}$

$r=\sqrt{144+196}$

$r=\sqrt{340}$

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

$\theta=\arctan(\frac{14}{-12})$

$\theta=-49.40$

$(\sqrt{340},-49.40)$

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Example Question #87 : Polar

Convert the following cartesian coordinates into polar form: (-11,-13)

Possible Answers:

$(\sqrt{290},49.76)$

$(\sqrt{223},13.27)$

$(\sqrt{251},73.13)$

$(\sqrt{280},53.56)$

Correct answer:

$(\sqrt{290},49.76)$

Explanation:

Cartesian coordinates have and , 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=-11

y=-13

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

$r=\sqrt{(-11)^2 + (-13)^2}$

$r=\sqrt{121+169}$

$r=\sqrt{290}$

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

$\theta=\arctan(\frac{-13}{-11})$

$\theta=49.76$

$(\sqrt{290},49.76)$

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Example Question #88 : Polar

Given y=9x^2-20x+16 calculate $\theta$ in polar form if x=9

Possible Answers:

91.46

85.61

81.23

89.09

Correct answer:

89.09

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=9x^2-20x+16

x=9

$y=9(9)^2-20\cdot9+16$

$y=9\cdot81-180+16$

y = 729-180+16

y=565

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

$\theta=\arctan(\frac{565}{9})$

$\theta=89.09$

Report an Error

Example Question #251 : Parametric, Polar, And Vector

Given y=7x^3+6x^2+2

calculate $\theta$ in polar form if x=6

Possible Answers:

76.9

67.9

89.8

98.9

Correct answer:

89.8

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=7x^3+6x^2+2

x=6

y=7(6)^3+6(6)^2+2

$y=7\cdot216+6\cdot36+2$

y=1512+216+2

y=1730

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

$\theta=\arctan(\frac{1730}{6})$

$\theta=89.8$

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Example Question #91 : Polar

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

Possible Answers:

81.2

89.7

84.5

92.6

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

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

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

$\theta=89.7$

Report an Error

Example Question #92 : Polar

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

Possible Answers:

63.4

74.1

64.7

89.1

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 #251 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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:

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

$x= r\cos(\theta)$

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

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

Report an Error

Example Question #94 : Polar

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

Possible Answers:

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

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

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

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #79 : Polar Form

Example Question #85 : Polar

Example Question #86 : Polar

Example Question #87 : Polar

Example Question #88 : Polar

Example Question #251 : Parametric, Polar, And Vector

Example Question #91 : Polar

Example Question #92 : Polar

Example Question #251 : Parametric, Polar, And Vector

Example Question #94 : Polar

All Calculus 2 Resources

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