3-Dimensional Space - AP Calculus BC

Card 0 of 2605

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(17)^2+(18)^2}=24.76\&\theta=arctan(\frac{18}{17})=46.64^{\circ}\&z=-20\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(-8)^2+(-5)^2}=9.43\&\theta=arctan(\frac{-5}{-8})=-147.99^{\circ}\&z=-2\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(-14)^2+(-18)^2}=22.8\&\theta=arctan(\frac{-18}{-14})=-127.87^{\circ}\&z=-10\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(10)^2+(14)^2}=17.2\&\theta=arctan(\frac{14}{10})=54.46^{\circ}\&z=-10\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(0)^2+(16)^2}=16\&\theta=arctan(\frac{16}{0})=90^{\circ}\&z=20\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(12)^2+(-5)^2}=13\&\theta=arctan(\frac{-5}{12})=-22.62^{\circ}\&z=31\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(-24)^2+(10)^2}=26\&\theta=arctan(\frac{10}{-24})=157.38^{\circ}\&z=32\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(20)^2+(14)^2}=24.41\&\theta=arctan(\frac{14}{20})=34.99^{\circ}\&z=25\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(-9)^2+(12)^2}=15\&\theta=arctan(\frac{12}{-9})=126.87^{\circ}\&z=15\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(36)^2+(-160)^2}=164\&\theta=arctan(\frac{-160}{36})=-77.32^{\circ}\&z=18\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(-1.5)^2+(-2)^2}=2.5\&\theta=arctan(\frac{-2}{-1.5})=-126.87^{\circ}\&z=10\end{align}$

Compare your answer with the correct one above

Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in cylindrical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

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

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$\begin{align}&r=\sqrt{(-50)^2+(120)^2}=130\&\theta=arctan(\frac{120}{-50})=112.62^{\circ}\&z=100\end{align}$

Compare your answer with the correct one above

Question

Determine the length of the curve $\vec{r}(t)=\left \langle t,4\sin(2t),4\cos(2t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$

Answer

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+64}=\sqrt{65}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left | \vec{v}(t)\right |dt$

$L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}$

Compare your answer with the correct one above

Question

Determine the length of the curve $\vec{r}(t)=\left \langle t,3\sin(4t),3\cos(4t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$

Answer

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 12\cos(4t), -12\sin(4t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(12\cos(4t))^2+(-12\sin(4t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+144\cos^2(4t)+144\sin^2(4t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+144(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+144}=\sqrt{145}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left | \vec{v}(t)\right |dt$

$L=\int_{0}^{2\pi} \sqrt{145}\ dt=t\sqrt{145}\Big|_{0}^{2\pi}=2\pi\sqrt{145}$

Compare your answer with the correct one above

Question

Find the length of the curve $<e^{2t},e^{-2t},2\sqrt{2}t>$ , from , to

Answer

The formula for the length of a parametric curve in 3-dimensional space is $l = \int_{t_0}^{t_1}\sqrt{(\frac{dx_1}{dt})^2+(\frac{dx_1}{dt})^2+(\frac{dx_3}{dt})^2} dt$

Taking dervatives and substituting, we have

$l = \int_{0}^{5}\sqrt{(2e^{2t})^2+(-2e^{-2t})^2+(2\sqrt{2})^2} dt$

$l = \int_{0}^{5}\sqrt{4e^{4t}+4e^{-4t}+8} dt$

$l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2} dt$ . Factor a out of the square root.

$l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2e^{2t}e^{-2t}} dt$ . "Uncancel" an $e^{2t}, e^{-2t}$ next to the . Now there is a perfect square inside the square root.

$l = 2\int_{0}^{5}\sqrt{(e^{2t}+e^{-2t})^2} dt$ . Factor

$l = 2[\frac{1}{2}e^{2t}-\frac{1}{2}e^{-2t}]_{0}^{5}$ . Take the square root, and integrate.

$=2(\frac{e^{10}}{2}-\frac{e^{-10}}{2})-2({0})$

$=e^{10}-e^{-10}$

Compare your answer with the correct one above

Question

Find the length of the arc drawn out by the vector function $F(t) = <a\cos t, a\sin t, at>$ with from to $t =\pi$ .

Answer

To find the arc length of a function, we use the formula

$\int_{t_{0}}^{t_{1}} \sqrt{(\frac{dx}{dt})^2 +(\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$ .

Using $t_0 =0, t_1 =\pi$ we have

$=\int_0^\pi \sqrt{(-a\sin t)^2+(a\cos t)^2 +(a)^2} dt$

$=\int_0^\pi \sqrt{a^2\sin^2 t+a^2\cos^2 t +a^2} dt$

$=\int_0^\pi \sqrt{a^2(\sin^2 t+\cos^2 t) +a^2} dt$

$=\int_0^\pi \sqrt{a^2 +a^2} dt$

$=[\sqrt2at]_0^\pi$

$=\sqrt{2} a \pi$

Compare your answer with the correct one above

Question

Evaluate the curvature of the function at the point .

Answer

The formula for curvature of a Cartesian equation is $\kappa(x)= \frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ . (It's not the easiest to remember, but it's the most convenient form for Cartesian equations.)

We have , hence

$\kappa(x)= \frac{|20|}{[1+(20x)^2]^{3/2}}$

and $\kappa(1)= \frac{|20|}{[1+(20(1))^2]^{3/2}} = \frac{20}{401^{3/2}}$ .

Compare your answer with the correct one above

Question

Find the length of the parametric curve

$\vec{r}(t)= \left< \frac{\sqrt{10}}{2}t^2, \frac{1}{3}t^3,5t+14\right>$

for $0\leq t \leq 2$ .

Answer

To find the solution, we need to evaluate

$L = \int_0^2 \left| \vec{r}(t)' \right| dt$ .

First, we find

$\vec{r}(t)' = \left< \sqrt{10}t, t^2, 5\right>$ , which leads to

$\left| \vec{r}(t)' \right| = \sqrt{(\sqrt{10}t)^2+(t^2)^2+(5)^2}$

$\left| \vec{r}(t)' \right| = \sqrt{t^4+10t^2+25}=\sqrt{(t^2+5)^2}=t^2+5$ .

So we have a final expression to integrate for our answer

$L= \int_0^2 \left| \vec{r}(t)' \right| dt = \int_0^2 \left(t^2+5 \right )dt= \frac{(2)^3}{3}+5(2)=\frac{38}{3}$

Compare your answer with the correct one above

Question

Determine the length of the curve given below on the interval 0<t<2

$\mathbf{r}=[2\sin t,\sqrt{5} t, 2\cos t]$

Answer

The length of a curve r is given by:

$L=\int_{t_1}^{t_2}|\mathbf{r'}(t)|dt$

To solve:

$\mathbf{r}'(t)=(2\cos t, \sqrt{5}, 2 \sin t)$

$|\mathbf{r'}(t)|=\sqrt{4\cos^2t+5+4\sin^2t}=\sqrt{4+5}=\sqrt{9}=3$

$L=\int^2_0 3 dt=3t|^2_0=6$

Compare your answer with the correct one above

Question

Find the arc length of the curve

on the interval

$0\leq t\leq 10$

Answer

To find the arc length of the curve function

on the interval $a\leq t\leq b$

we follow the formula

$\int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2},dt$

For the curve function in this problem we have

$\frac{\mathrm{dx} }{\mathrm{d} t}=3$

$\frac{\mathrm{dy} }{\mathrm{d} t}=4cos(2t)$

$\frac{\mathrm{dz} }{\mathrm{d} t}=-4sin(2t)$

and following the arc length formula we solve for the integral

$\int_0^{10} \sqrt{(3)^2+(4cos(2t))^2+(-4sin(2t))^2},dt$

$=\int_0^{10} \sqrt{9+16sin^2(2t)+16cos^2(2t)},dt$

$=\int_0^{10} \sqrt{9+16(sin^2(2t)+cos^2(2t))},dt$

$=\int_0^{10} \sqrt{9+16},dt$

$=\int_0^{10} \sqrt{25},dt$

$=\int_0^{10} 5,dt$

$=5t|_0^{10}$

Hence the arc length is

Compare your answer with the correct one above

Tap the card to reveal the answer