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Calculus 2 : Polar Form

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Example Questions

Example Question #11 : 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 #97 : Polar

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

Possible Answers:

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

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

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

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

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

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 #91 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 #92 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 #93 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Polar Form

Example Question #97 : Polar

Example Question #91 : Polar Form

Example Question #92 : Polar Form

Example Question #93 : Polar Form

All Calculus 2 Resources

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