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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Polar Form

What is the equation $y=-\frac{1}{2}x^{2}$ in polar form?

Possible Answers:

$r=-2\tan \theta\sec\theta$

$r=-\frac{1}{2}\tan \theta\sec\theta$

$r=2\tan \theta\sec\theta$

$r=-\tan 2\theta\sec\theta$

$r=\frac{1}{2}\tan \theta\sec\theta$

Correct answer:

$r=-2\tan \theta\sec\theta$

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 $y=-\frac{1}{2}x^{2}$ , then:

$r\sin\theta=-\frac{1}{2}(r\cos \theta)^{2}$

$r\sin\theta=-\frac{1}{2}r^{2}\cos^{2} \theta$

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

$\tan \theta=-\frac{1}{2}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=-\frac{1}{2}r$

$\tan \theta\sec\theta=-\frac{1}{2}r$

$r=-2\tan \theta\sec\theta$

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Example Question #692 : Calculus Ii

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

$r\sin\theta=-7r^{2}\cos^{2}\theta$

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

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

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

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

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

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Example Question #6 : Polar Form

What is the polar form of y=3-4x ?

Possible Answers:

$r=-\frac{3}{\sin\theta+4\cos\theta}$

$r=-\frac{3}{\sin\theta-4\cos\theta}$

$r=\frac{3}{\sin\theta+4\cos\theta}$

None of the above

$r=\frac{3}{\sin\theta-4\cos\theta}$

Correct answer:

$r=\frac{3}{\sin\theta+4\cos\theta}$

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=3-4x , then:

$r\sin\theta=3-4(r\cos\theta)$

$r\sin\theta+4(r\cos\theta)=3$

$r(\sin\theta+4\cos\theta)=3$

$r=\frac{3}{\sin\theta+4\cos\theta}$

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Example Question #7 : Polar Form

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

Possible Answers:

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

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

None of the above

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

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

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 #2 : Polar Form

What is the polar form of y=-10x+5 ?

Possible Answers:

$r=\frac{5}{\sin\theta-10\cos\theta}$

$r=-\frac{5}{\sin\theta-10\cos\theta}$

$r=-\frac{5}{\sin\theta+10\cos\theta}$

$r=\frac{5}{\sin\theta+10\cos\theta}$

None of the above

Correct answer:

$r=\frac{5}{\sin\theta+10\cos\theta}$

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=-10x+5 , then:

$r\sin\theta=-10(r\cos\theta)+5$

$r\sin\theta+10(r\cos\theta)=5$

$r(\sin\theta+10\cos\theta)=5$

$r=\frac{5}{\sin\theta+10\cos\theta}$

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Example Question #21 : Polar

What is the polar form of $y=\frac{2}{3}x+9$ ?

Possible Answers:

$r=-\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

$r=-\frac{9}{\sin\theta+\frac{2}{3}\cos\theta}$

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

$r=\frac{9}{\sin\theta+\frac{2}{3}\cos\theta}$

None of the above

Correct answer:

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

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 $y=\frac{2}{3}x+9$ , then:

$r\sin\theta= \frac{2}{3}(r\cos\theta)+9$

$r\sin\theta-\frac{2}{3}(r\cos\theta)=9$

$r(\sin\theta-\frac{2}{3}\cos\theta)=9$

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

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Example Question #41 : Polar Form

What is the polar form of y=2x-7 ?

Possible Answers:

$r=-\frac{7}{\sin\theta-2\cos \theta}$

$r=-\frac{7}{\sin\theta+2\cos \theta}$

$r=\frac{7}{2\sin\theta-\cos \theta}$

$r=\frac{7}{\sin\theta+2\cos \theta}$

$r=\frac{7}{\sin\theta-2\cos \theta}$

Correct answer:

$r=-\frac{7}{\sin\theta-2\cos \theta}$

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=2x-7 , then:

$r\sin\theta=2(r\cos \theta)-7$

$r\sin\theta-2(r\cos \theta)=-7$

$r(\sin\theta-2\cos \theta)=-7$

$r=-\frac{7}{\sin\theta-2\cos \theta}$

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Example Question #21 : Parametric, Polar, And Vector Functions

What is the polar form of $y=\frac{3}{9}x^{2}$ ?

Possible Answers:

$r=-\frac{1}{3}\tan \theta\sec\theta$

$r=\frac{1}{3}\tan \theta\sec\theta$

$r=\tan \frac{1}{3}\theta\sec\theta$

$r=-3\tan \theta\sec\theta$

$r=3\tan \theta\sec\theta$

Correct answer:

$r=3\tan \theta\sec\theta$

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 $y=\frac{3}{9}x^{2}$ , then:

$r\sin\theta=\frac{3}9{}(r\cos \theta)^{2}$

$r\sin\theta=\frac{3}{9}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{3}9{}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{3}{9}r$

$\tan \theta\sec\theta=\frac{3}{9}r$

$r=\frac{9}{3}\tan \theta\sec\theta$

$r=3\tan \theta\sec\theta$

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Example Question #11 : Polar Form

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

Possible Answers:

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

(8,-74.74)

$(\sqrt{130},74.74)$

(8,74.74)

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 #71 : Polar Form

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

Possible Answers:

$(\sqrt{230},62.8)$

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

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

$(\sqrt{225},54.8)$

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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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #4 : Polar Form

Example Question #692 : Calculus Ii

Example Question #6 : Polar Form

Example Question #7 : Polar Form

Example Question #2 : Polar Form

Example Question #21 : Polar

Example Question #41 : Polar Form

Example Question #21 : Parametric, Polar, And Vector Functions

Example Question #11 : Polar Form

Example Question #71 : Polar Form

All AP Calculus BC Resources

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