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AP Calculus BC : Functions, Graphs, and Limits

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Example Questions

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

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 #32 : 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 #31 : Functions, Graphs, And Limits

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

Possible Answers:

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

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

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

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

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

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 #31 : Parametric, Polar, And Vector Functions

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 #2 : Vector Form

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 #1 : Vector Form

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:

$y=2\sin(x)$

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

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

$y=-\sin(2x)$

$y=\sin(2x)$

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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Example Question #4 : Vector Form

The graph of the vector function $r(t)=\left \langle 3t^{2}-1,\cos(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=-\cos\sqrt{\frac{x-1}{3}}$

$y=\cos\sqrt{\frac{x+1}{3}}$

$y=\cos\sqrt{\frac{x-1}{3}}$

$y=\cos\sqrt{\frac{3}{x-1}}$

$y=-\cos\sqrt{\frac{x+1}{3}}$

Correct answer:

$y=\cos\sqrt{\frac{x+1}{3}}$

Explanation:

We can find the graph of $r(t)=\left \langle 3t^{2}-1,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

$x=3t^{2}-1$

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

$\frac{x+1}{3}=t^{2}$

$t=\sqrt{\frac{x+1}{3}}$

We can now use this value to solve for :

$y=\cos\sqrt{\frac{x+1}{3}}$

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Example Question #1 : Vector Form

$\newline {\vec{r}(t)} = (5t^3,e^{11t},e^{5t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,e^{11t},5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,11,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15,11,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2.11e^{11t},5e^{5t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2.11e^{11t},5e^{5t})$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (5t^3,e^{11t},e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}5t^3=15t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{11t}=11e^{11t}$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{5t}=5e^{5t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,11e^{11t},5e^{5t})$

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Example Question #601 : Parametric, Polar, And Vector

$\newline\vec{v} =(x^2,-x,5)\newline\vec{u} =(2,3x,11)\newline$

Calculate $\vec{v} + \vec{u}$

Possible Answers:

$\vec{v}+\vec{u}=(x^2 + 2,0,16)$

$\vec{v}+\vec{u}=(x^2,2x,3x)$

$\vec{v}+\vec{u}=(x^2,2x,16)$

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

Correct answer:

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

Explanation:

Calculate the sum of vectors.

In general,

$\vec{v} = (v_1,v_2,v_3)$

$\vec{u} = (u_1,u_2,u_3)$

$\vec{v} + \vec{u} = (v_1 + u_1, v_2 + u_2,v_3 + u_3)$

Solution:

$\newline\vec{v} =(x^2,-x,5)\newline\vec{u} =(2,3x,11)\newline$

$\vec{v}+\vec{u}=(x^2,-x,5)+(2,3x,11)$

$\vec{v}+\vec{u}=(x^2+2,-x+3x,5+11)$

$\vec{v}+\vec{u}=(x^2 + 2,2x,16)$

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Example Question #1 : Limits

Evaluate the following limit:

$\lim_{x\rightarrow 3^-}f(x), f(x)=\left\{\begin{matrix} x^2+1, x< 3\\ x, x\geq 3 \end{matrix}\right.$

Possible Answers:

The limit does not exist

Correct answer:

Explanation:

The limit we are given is one sided, meaning we are approaching our x value from one side; in this case, the negative sign exponent indicates that we are approaching 3 from the left side, or using values slightly less than three on approach.

This corresponds to the part of the piecewise function for values less than 3. When we substitute our x value being approached, we get

(3)^2+1=10

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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 #31 : Parametric, Polar, And Vector Functions

Example Question #32 : Parametric, Polar, And Vector Functions

Example Question #31 : Functions, Graphs, And Limits

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

Example Question #2 : Vector Form

Example Question #1 : Vector Form

Example Question #4 : Vector Form

Example Question #1 : Vector Form

Example Question #601 : Parametric, Polar, And Vector

Example Question #1 : Limits

All AP Calculus BC Resources

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