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AP Calculus BC : Vector Form

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

Example Questions

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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Example Question #1 : 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:

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

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

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

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

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

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 #31 : Functions, Graphs, And Limits

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

Possible Answers:

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

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

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

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

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 #1 : Vector Form

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

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

Possible Answers:

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

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

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

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

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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AP Calculus BC : Vector Form

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Vector Form

Example Question #63 : Vectors & Spaces

Example Question #113 : Linear Algebra

Example Question #101 : Vector

Example Question #1 : Vector Form

Example Question #31 : Functions, Graphs, And Limits

Example Question #1 : Vector Form

All AP Calculus BC Resources

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