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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 #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 #1 : Graphing Polar Form

Graph the equation $r=cos(\theta )$ where $0\leq \theta \leq2\pi$ .

Possible Answers:

R_cosx_1

R_cos2x

R_sinx

R_cosx

Faker_cosx

Correct answer:

R_cosx

Explanation:

At angle the graph as a radius of . As it approaches $\frac{\pi }{2}$ , the radius approaches .

As the graph approaches $\pi$ , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between $\frac{3\pi }{2}$ and $2\pi$ .

Between $\pi$ and $\frac{3\pi }{2}$ , the radius approaches from and redraws the curve in the first quadrant.

Between $\frac{3\pi }{2}$ and $2\pi$ , the graph redraws the curve in the fourth quadrant as the radius approaches from .

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Example Question #2 : Graphing Polar Form

Draw the graph of $r=cos(2\theta )$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R_cosx_1

R_cos2x

R_sin2x

R2_cos2x

R_sin2x

Correct answer:

R_cos2x

Explanation:

Because this function has a period of $\pi$ , the x-intercepts of the graph y=cos(2x) happen at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches from .

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches from .

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches from .

Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches from and is draw in the second quadrant.

Finally between $\frac{7\pi }{4}$ and $2\pi$ , the radius approaches from .

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Example Question #1 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=3\theta+e^\theta$

Possible Answers:

$3+e^\theta$

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)+(3\theta+e^\theta)\sin(\theta)}$

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

$\frac{(3+e^\theta)\sin(\theta)-(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)+(3\theta+e^\theta)\sin(\theta)}$

Correct answer:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

Explanation:

The derivative of a polar function is given by the following:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find $\frac{dr}{d\theta}=3+e^\theta$

We found the derivative using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Finally, we plug in the above derivative and the original function into the above formula:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

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Example Question #12 : Polar Form

What is the derivative of $r=4sin\theta+2$ ?

Possible Answers:

$\frac{8cos\theta\sin\theta+2\cos\theta}{4+8\sin^{2}\theta-2sin\theta}$

$\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta+2sin\theta}$

$\frac{8cos\theta\sin\theta-2\cos\theta}{4+8\sin^{2}\theta+2sin\theta}$

$\frac{8cos\theta\sin\theta-2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

$\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

Correct answer:

$\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4sin\theta+2$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=4cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+(4sin\theta+2)cos\theta}{4cos\theta\cos\theta-(4sin\theta+2)sin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+4sin\theta\cos\theta+2\cos\theta}{4cos\theta\cos\theta-4sin\theta\sin\theta-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(cos^{2}\theta-sin^{2}\theta)-2sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4((1-\sin^{2}\theta)-sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(1-2\sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

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Example Question #31 : Parametric, Polar, And Vector Functions

What is the derivative of $r=2+5\sin\theta$ ?

Possible Answers:

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5+10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta+2\sin\theta}$

$\frac{10\cos\theta\sin\theta-2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta-2\cos\theta}{5+10\sin^{2}\theta+2\sin\theta}$

Correct answer:

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=2+5\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=5\cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{(5\cos\theta)\sin\theta+(2+5\sin\theta)\cos\theta}{(5\cos\theta)\cos\theta-(2+5\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{5\cos\theta\sin\theta+2\cos\theta+5\sin\theta\cos\theta}{5\cos^{2}\theta-2\sin\theta-5\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(\cos^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-\sin^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-2\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

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Example Question #1 : Vector Form

Find the vector form of (6,3,1) to (0,1,3) .

Possible Answers:

$\overrightarrow{v}=[6,2,-2]$

$\overrightarrow{v}=[-6,-2,-2]$

$\overrightarrow{v}=[6,-2,-2]$

$\overrightarrow{v}=[6,2,2]$

$\overrightarrow{v}=[6,2,-3]$

Correct answer:

$\overrightarrow{v}=[6,2,-2]$

Explanation:

When we are trying to find the vector form we need to remember the formula which states to take the difference between the ending and starting point.

Thus we would get:

Given (a,b,c) and (d,e,f)

$\overrightarrow{v}=[d-a, e-b, f-c]$

In our case we have ending point at (6,3,1) and our starting point at (0,1,3) .

Therefore we would set up the following and simplify.

$\overrightarrow{v}=[6-0,3-1,1-3]=[6,2,-2]$

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Example Question #63 : Vectors & Spaces

Given points (2,4,5) and (9,1,0) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 7,3,5\right \rangle$

$\left \langle -7,-3,-5\right \rangle$

$\left \langle 7,-3,-5\right \rangle$

$\left \langle -7,3,-5\right \rangle$

$\left \langle 7,3,-5\right \rangle$

Correct answer:

$\left \langle 7,-3,-5\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point

$a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ ,

the distance is the vector

$v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (2,4,5) and (9,1,0) , we get:

$v=\left \langle 9-2,1-4,0-5\right \rangle$

$v=\left \langle 7,-3,-5\right \rangle$

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Example Question #113 : Linear Algebra

Given points (4,2,-2) and (7,1,0) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 3,1,2\right \rangle$

$\left \langle -3,1,-2\right \rangle$

$\left \langle 3,-1,2\right \rangle$

$\left \langle -3,1,2\right \rangle$

$\left \langle 3,1,-2\right \rangle$

Correct answer:

$\left \langle 3,-1,2\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (4,2,-2) and (7,1,0) , we get:

$v=\left \langle 7-4,1-2,0-(-2)\right \rangle$

$v=\left \langle 3,-1,2\right \rangle$

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Example Question #101 : Vector

The graph of the vector function $r(t)=\left \langle \frac{1}{2}t,\sin(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=-\sin(2x)$

$y=\sin(\frac{x}2{})$

$y=-\sin(\frac{x}2{})$

$y=2\sin(x)$

$y=\sin(2x)$

Correct answer:

$y=\sin(2x)$

Explanation:

We can find the graph of $r(t)=\left \langle \frac{1}{2}t,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

$x=\frac{1}{2}t$

t=2x

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$

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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 #11 : Polar Form

Example Question #1 : Graphing Polar Form

Example Question #2 : Graphing Polar Form

Example Question #1 : Derivatives Of Polar Form

Example Question #12 : Polar Form

Example Question #31 : Parametric, Polar, And Vector Functions

Example Question #1 : Vector Form

Example Question #63 : Vectors & Spaces

Example Question #113 : Linear Algebra

Example Question #101 : Vector

All AP Calculus BC Resources

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