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Calculus 2 : Polar

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Example Questions

Example Question #242 : Parametric, Polar, And Vector

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

Possible Answers:

89.87

81.03

75.22

84.31

Correct answer:

81.03

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=3

y=2x^3+3

y=2(3)^3+3

$y=2\cdot8+3$

y=19

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

$\theta=\arctan(\frac{19}{3})$

$\theta=81.03$

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Example Question #75 : Polar Form

What is the polar form of $y=\frac{x^{2}}{7}$ ?

Possible Answers:

$r=-\frac{1}{}7\tan\theta\sec\theta$

$r=7\tan\theta\sec\theta$

$r=\frac{1}{}7\tan\theta\sec\theta$

$r=-7\tan\theta\sec\theta$

$r=\tan7\theta\sec\theta$

Correct answer:

$r=7\tan\theta\sec\theta$

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 $y=\frac{x^{2}}{7}$ , then:

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

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

$\tan\theta=\frac{r\cos \theta}{7}$

$r\cos\theta=7\tan\theta$

$r=\frac{7\tan\theta}{\cos\theta}$

$r=7\tan\theta\sec\theta$

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Example Question #76 : Polar Form

What is the polar form of y=3x+10 ?

Possible Answers:

$r=\frac{10}{\sin\theta+3\cos \theta}$

$r=-\frac{10}{\sin\theta-3\cos \theta}$

$r=-\frac{10}{\sin\theta+3\cos \theta}$

$r=\frac{10}{\sin\theta-3\cos \theta}$

$r=\frac{10}{3\sin\theta-\cos \theta}$

Correct answer:

$r=\frac{10}{\sin\theta-3\cos \theta}$

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=3x+10 , then:

$r\sin\theta=3(r\cos \theta)+10$

$r\sin\theta-3r\cos \theta=10$

$r(\sin\theta-3\cos \theta)=10$

$r=\frac{10}{\sin\theta-3\cos \theta}$

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Example Question #81 : Polar

What is the polar form of y=12-5x ?

Possible Answers:

$r=-\frac{12}{\sin\theta-5\cos \theta}$

$r=\frac{12}{\sin\theta-5\cos \theta}$

$r=\frac{12}{5\sin\theta+\cos \theta}$

$r=-\frac{12}{\sin\theta+5\cos \theta}$

$r=\frac{12}{\sin\theta+5\cos \theta}$

Correct answer:

$r=\frac{12}{\sin\theta+5\cos \theta}$

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=12-5x , then:

$r\sin\theta=12-5(r\cos \theta)$

$r\sin\theta+5(r\cos \theta)=12$

$r(\sin\theta+5\cos \theta)=12$

$r=\frac{12}{\sin\theta+5\cos \theta}$

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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

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

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

$\theta=89.09$

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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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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #242 : Parametric, Polar, And Vector

Example Question #75 : Polar Form

Example Question #76 : Polar Form

Example Question #81 : Polar

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

All Calculus 2 Resources

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