Parametric, Polar, and Vector Functions - AP Calculus BC

Card 0 of 984

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 :

$y=\sin(t)$

$y=\sin(2x)$

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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 :

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

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Question

Write in Cartesian form:

$x = 3t, y = 2t^{2}$

Answer

, so

$x^{2} = (3t)^{2} = 9t^{2} = \frac{9}{2} \cdot 2t^{2} = \frac{9}{2} y$ .

$\frac{9}{2} y= x^{2}$ , so

$\frac{2}{9}\cdot \frac{9}{2} y = \frac{2}{9}x^{2}$

$y = \frac{2}{9}x^{2}$

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Question

Write in Cartesian form:

Answer

,

so the Cartesian equation is

.

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Question

Rewrite as a Cartesian equation:

$x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]$

Answer

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

$x = \left ( t + 1\right )^{2}$

$\pm \sqrt{x} = t + 1$

So

$\sqrt{x} = t + 1$ or $-\sqrt{x} = t + 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so is nonnegative; we choose

$\sqrt{x} = t + 1$ .

Also,

$y = t^{2} - 2t + 1$

$y= \left ( t - 1\right )^{2}$

$\sqrt{y} = \pm \left ( t - 1\right )$

So

$\sqrt{y} = t- 1$ or $-\sqrt{y} = t- 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so is nonpositive; we choose

$-\sqrt{y} = t- 1$

or equivalently,

$\sqrt{y} = -t+ 1$

to make nonpositive.

Then,

$\sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and

$\sqrt{x}+ \sqrt{y} = 2$

$\sqrt{y} = 2 - \sqrt{x}$

$y =\left ( 2 - \sqrt{x} \right )^2$

$y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$y = x + 4 - 4\sqrt{x}$

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Question

Write in Cartesian form:

$x = \ln (t+1)$

$t \in (-1, \infty )$

Answer

$x = \ln (t+1)$

so

$e ^{x} = \left [ \ln (t+1) \right ] ^{x}$

$e ^{x} = t + 1$

$y = t^2 + 2t + 2 = t^2 + 2t + 1 + 1 = (t + 1)^{2}+ 1 = \left (e^{x} \right )^{2} + 1 = e^{2x} + 1$

Therefore the Cartesian equation is $y = e^{2x} + 1$ .

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Question

Write in Cartesian form:

$x = \cos 2t, y = \sin t$

Answer

Rewrite using the double-angle formula:

$x = \cos 2t =1 - 2\sin ^{2} t = 1 - 2y^{2}$

Then

$x = 1 - 2y^{2}$

$x + 2y^{2} = 1$

which is the correct choice.

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Question

Rewrite as a Cartesian equation:

$x =10^ {t}, y = \sinh t, t \in (0,\infty )$

Answer

$y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$ , so

$y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$ .

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Question

$\small x=3t+4$ and $\small y=1/t$ . What is $\small y$ in terms of $\small x$ (rectangular form)?

Answer

In order to solve this, we must isolate $\small t$ in both equations.

$\small \small x=3t+4\rightarrow t=(x-4)/3$ and

$\small y=1/t\rightarrow t=1/y$ .

Now we can set the right side of those two equations equal to each other since they both equal $\small t$ .

$\small (x-4)/3=1/y$ .

By multiplying both sides by $\small 3y/(x-4)$ , we get $\small y=3/(x-4)$ , which is our equation in rectangular form.

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Question

If and , what is in terms of (rectangular form)?

Answer

Given and , we can find in terms of by isolating in both equations:

$x=3+t\rightarrow t=x-3$

$y=4t+7\rightarrow t=\frac{y-7}{4}$

Since both of these transformations equal , we can set them equal to each other:

$\frac{y-7}{4}=x-3$

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Question

If and , what is in terms of (rectangular form)?

Answer

Given and , we can find in terms of by isolating in both equations:

$x=6t+5\rightarrow t=\frac{x-5}{6}$

$y=5t+6\rightarrow t=\frac{y-6}{5}$

Since both of these transformations equal , we can set them equal to each other:

$\frac{y-6}{5}=\frac{x-5}{6}$

${y-6}=5\left(\frac{x-5}{6}\right)$

$y=\frac{5}{6}(x-5)+6$

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Question

If and , what is in terms of (rectangular form)?

Answer

Given and , we can find in terms of by isolating in both equations:

$x=4t-3\rightarrow t=\frac{x+3}{4}$

$y=2t+7\rightarrow t=\frac{y-7}{2}$

Since both of these transformations equal , we can set them equal to each other:

$\frac{y-7}{2}=\frac{x+3}{4}$

$y-7=2\left(\frac{x+3}{4}\right)$

$y-7=\frac{x+3}{2}$

$y=\frac{x+3}{2}+7$

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Question

Given and , what is in terms of (rectangular form)?

Answer

In order to find with respect to , we first isolate in both equations: $x=7t-5\rightarrow t=\frac{x+5}{7}$

$y=2t+9\rightarrow t=\frac{y-9}{2}$

Since both equations equal , we can then set them equal to each other and solve for :

$\frac{y-9}{2}=\frac{x+5}{7}$

${y-9}=2(\frac{x+5}{7})$

$y=\frac{2}{7}({x+5})+9$

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Question

Given and , what is in terms of (rectangular form)?

Answer

In order to find with respect to , we first isolate in both equations: $x=4t+9\rightarrow t=\frac{x-9}{4}$

$y=-t+2\rightarrow t=2-y$

Since both equations equal , we can then set them equal to each other and solve for :

$2-y=\frac{x-9}{4}$

$y=2-\frac{x-9}{4}$

$y=\frac{8}{4}-\frac{x-9}{4}$

$y=\frac{8-(x-9)}{4}$

$y=\frac{8-(x-9)}{4}$

$y=\frac{8-x+9}{4}$

$y=\frac{17-x}{4}$

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Question

Given and , what is in terms of (rectangular form)?

Answer

In order to find with respect to , we first isolate in both equations:

$x=10t-3\rightarrow t=\frac{x+3}{10}$

$y=7t+8\rightarrow t=\frac{y-8}{7}$

Since both equations equal , we can then set them equal to each other and solve for :

$\frac{y-8}{7}=\frac{x+3}{10}$

$y-8=7(\frac{x+3}{10})$

$y=\frac{7}{10}({x+3)+8$

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