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AP Calculus BC : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

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Example Question #1 : Graphing Polar Form

Draw the curve of $r^2=sin(2\theta )$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R2_cos2x

R_sinx_1

R_sin2x

R2_sin2x

R_sinx

Correct answer:

R2_sin2x

Explanation:

Taking the graph of y=sin(2x) , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to $\frac{\pi }{2}$ and $\pi$ to $\frac{3\pi }{2}$ .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\pm 1$ .

To draw the graph, the radius is 0 at and traces to 1 at $\frac{\pi }{4}$ . As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From $\frac{\pi }4{}$ to $\frac{\pi }{2}$ , the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in $\pi$ to $2\pi$ .

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Example Question #1 : Polar

Rewrite in polar form:

$x^{2} = 3xy + 1$

Possible Answers:

$r = \sqrt{\frac{1}{\sin ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }$

$r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta + 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Correct answer:

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Explanation:

$x^{2} = 3xy + 1$

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

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Example Question #2 : Polar Form

What is the following coordinate in polar form?

(-2,-5)

Provide the angle in degrees.

Possible Answers:

(10,248)

$(\sqrt{29},68)$

(16,112)

$(\sqrt{29},202)$

$(\sqrt{29},248)$

Correct answer:

$(\sqrt{29},248)$

Explanation:

To calculate the polar coordinate, use

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

$r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ .

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Example Question #201 : Ap Calculus Bc

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=2x^{2}$ , then $r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ .

$r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ . Dividing both sides by $r\cos \theta$ ,

$\tan \theta=2r\cos \theta$

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

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

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

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Example Question #171 : Parametric, Polar, And Vector

What is the equation $y=\frac{6}{x}$ in polar form?

Possible Answers:

$r=\sqrt\frac{6}{\cos \theta}$

$r=\sqrt\frac{6}{\tan \theta}$

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

$r=\sqrt\frac{6}{\sin \theta}$

None of the above

Correct answer:

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=\frac{6}{x}$ , then $r\sin \theta=\frac{6}{r\cos \theta}$ . Multiplying both sides by $r\cos \theta$ ,

$r^{2}\sin \theta cos \theta=6$

$r^{2}=\frac{6}{\sin \theta cos \theta}$

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

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Example Question #5 : Polar Form

Convert the following function into polar form:

f(x)=2x^2+5x+2+2y^2

Possible Answers:

$2r^2+5r\cos(\theta )+2$

$r^2+1+5r\cos(\theta)$

$r^2+\sin(\theta)$

$2r^2+5r\sin(\theta)+2$

Correct answer:

$2r^2+5r\cos(\theta )+2$

Explanation:

The following formulas were used to convert the function from polar to Cartestian coordinates:

$x=r\cos(\theta), y=r\sin(\theta), x^2+y^2=r^2$

Note that the last formula is a manipulation of a trignometric identity.

Simply replace these with x and y in the original function.

f(x)=2x^2+5x+2+2y^2

f(x)=(x^2+y^2)2+5x+2

$f(x)=2r^2+5rcos(\theta)+2$

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Example Question #1 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 #7 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 #8 : Polar Form

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

Possible Answers:

None of the above

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

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

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

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

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 #1 : Polar Form

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

Possible Answers:

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

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

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

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

None of the above

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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AP Calculus BC : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Graphing Polar Form

Example Question #1 : Polar

Example Question #2 : Polar Form

Example Question #201 : Ap Calculus Bc

Example Question #171 : Parametric, Polar, And Vector

Example Question #5 : Polar Form

Example Question #1 : Polar Form

Example Question #7 : Polar Form

Example Question #8 : Polar Form

Example Question #1 : Polar Form

All AP Calculus BC Resources

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