Vector Form - AP Calculus BC

Card 0 of 70

Find the vector form of to .

Tap to see back →

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 and

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

In our case we have ending point at and our starting point at .

Therefore we would set up the following and simplify.

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

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

Tap to see back →

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}$)$

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

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

Tap to see back →

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$

Given points and , what is the vector form of the distance between the points?

Tap to see back →

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 and , we get:

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

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

Given points and , what is the vector form of the distance between the points?

Tap to see back →

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 and , we get:

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

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

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?

Tap to see back →

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 :

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

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$

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?

Tap to see back →

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 :

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

We can now use this value to solve for :

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

Find the vector form of to .

Tap to see back →

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 and

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

In our case we have ending point at and our starting point at .

Therefore we would set up the following and simplify.

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

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

Tap to see back →

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}$)$

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

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

Tap to see back →

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$

Given points and , what is the vector form of the distance between the points?

Tap to see back →

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 and , we get:

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

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

Given points and , what is the vector form of the distance between the points?

Tap to see back →

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 and , we get:

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

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

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?

Tap to see back →

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 :

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

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$

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?

Tap to see back →

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 :

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

We can now use this value to solve for :

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

Find the vector form of to .

Tap to see back →

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 and

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

In our case we have ending point at and our starting point at .

Therefore we would set up the following and simplify.

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

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

Tap to see back →

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}$)$

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

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

Tap to see back →

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$

Given points and , what is the vector form of the distance between the points?

Tap to see back →

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 and , we get:

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

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

Given points and , what is the vector form of the distance between the points?

Tap to see back →

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 and , we get:

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

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

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?

Tap to see back →

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 :

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

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$