Functions, Graphs, and Limits - AP Calculus BC

Card 0 of 1344

Question

What is the vector form of ?

Answer

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for , we can derive the vector form $\left \langle -8,8,-1\right \rangle$ .

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Question

What is the vector form of ?

Answer

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for , we can derive the vector form $\left \langle 7,2,-7\right \rangle$ .

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Question

What is the vector form of ?

Answer

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for , we can derive the vector form $\left \langle 1,-10,0\right \rangle$ .

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Question

What is the vector form of ?

Answer

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle -1,9,3\right \rangle$ .

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Question

What is the vector form of ?

Answer

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle 1,0,-3\right \rangle$ .

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Question

What is the vector form of ?

Answer

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle 2,-1,0\right \rangle$ .

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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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Question

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

Answer

Knowing that and , we can isolate in both equations as follows:

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

$y=5t+1\rightarrow t=\frac{y-1}{5}$

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

$\frac{y-1}{5}=\frac{x+2}{9}$

${y-1}=5(\frac{x+2}{9})$

$y=\frac{5}{9}({x+2}})+1$

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