Vector Form - AP Calculus BC

Card 0 of 266

Question

Find the vector form of to .

Answer

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.

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

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Question

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

Answer

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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Question

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

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

Answer

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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Question

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

Answer

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:

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

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

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Question

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

Answer

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:

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

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

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Question

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?

Answer

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 :

$y=\sin(t)$

$y=\sin(2x)$

Compare your answer with the correct one above

Question

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?

Answer

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 :

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

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

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

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

We can now use this value to solve for :

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

Compare your answer with the correct one above

Question

Find the vector form of to .

Answer

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.

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

Compare your answer with the correct one above

Question

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

Answer

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 $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

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

Compare your answer with the correct one above

Question

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

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

Answer

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:

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

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

Compare your answer with the correct one above

Question

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?

Answer

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 :

$y=\sin(t)$

$y=\sin(2x)$

Compare your answer with the correct one above

Question

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?

Answer

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 :

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

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

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

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

We can now use this value to solve for :

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

Compare your answer with the correct one above

Question

Find the vector form of to .

Answer

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.

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

Compare your answer with the correct one above

Question

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

Answer

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 $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

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

Compare your answer with the correct one above

Question

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

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

Answer

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:

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

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

Compare your answer with the correct one above

Question

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?

Answer

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 :

$y=\sin(t)$

$y=\sin(2x)$

Compare your answer with the correct one above

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