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Calculus 2 : Polar Form

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Example Questions

Example Question #14 : Polar Form

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 #75 : Polar

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

Possible Answers:

$(\sqrt{74},35.54)$

$(\sqrt{39},123.73)$

$(\sqrt{15},83.0)$

$(\sqrt{63},74.16)$

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 #241 : Parametric, Polar, And Vector

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

Possible Answers:

76.88

88.62

112.32

97.48

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 #242 : Parametric, Polar, And Vector

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

Possible Answers:

-82.87

-98.12

-32.77

-74.65

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 #243 : Parametric, Polar, And Vector

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

Possible Answers:

23.63

36.97

17.53

26.74

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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Example Question #244 : Parametric, Polar, And Vector

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

Possible Answers:

81.03

84.31

75.22

89.87

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 #245 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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

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 #246 : Parametric, Polar, And Vector

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 #247 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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

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 #248 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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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Calculus 2 : Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #14 : Polar Form

Example Question #75 : Polar

Example Question #241 : Parametric, Polar, And Vector

Example Question #242 : Parametric, Polar, And Vector

Example Question #243 : Parametric, Polar, And Vector

Example Question #244 : Parametric, Polar, And Vector

Example Question #245 : Parametric, Polar, And Vector

Example Question #246 : Parametric, Polar, And Vector

Example Question #247 : Parametric, Polar, And Vector

Example Question #248 : Parametric, Polar, And Vector

All Calculus 2 Resources

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