3-Dimensional Space

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AP Calculus BC › 3-Dimensional Space

Questions 1 - 10

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(6,-46,-105).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(114.79,-82.57^{\circ},156.16^{\circ})$

$(114.79,97.43^{\circ},156.16^{\circ})$

$(46.39,-82.57^{\circ},156.16^{\circ})$

$(46.39,97.43^{\circ},23.84^{\circ})$

$(46.39,97.43^{\circ},-23.84^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(6,-46,-105)\&\rho=\sqrt{(6)^2+(-46)^2+(-105)^2}=114.79\&\theta=arctan(\frac{-46}{6})=-82.57^{\circ}\&\phi=arccos(\frac{-105}{114.79})=156.16^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (19,-94^{\circ},33^{\circ})\text{ of form: } (\rho,\theta,\phi)\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Explanation

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\&x=\rho sin(\phi)cos(\theta)\&y=\rho sin(\phi)sin(\theta)\&z=\rho cos(\phi)\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\& \text{imperative that the correct units (degrees or radians) are specified}\& \text{for the input!}\&\text{For our coordinates}\&x=19sin(33^{\circ})cos(-94^{\circ})=-0.72\&y=19sin(33^{\circ})sin(-94^{\circ})=-10.32\&z=19cos(33^{\circ})15.93\end{align*}$

Find an equation of a sphere that has the line segment joining (0,4,2) and (6,0,2) as a diameter.

None of the Above.

Explanation

To find the diameter we need to find the distance between the two points, where:

$D=\sqrt{(6-0)^2+(0-4)^2+(2-2)^2}$

$D=\sqrt{52}=2\sqrt{13}$

Since the radius is half of the Diameter,

$r=\sqrt{13}$

To find the the Center point we use the mid-point formula. So,

$Center = \left ( \frac{0+6}{2},\frac{4+0}{2},\frac{2+2}{2} \right )=\left ( 3,2,2\right )$

So the equation of the sphere is:

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(95,-77,129).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(177.75,-39.03^{\circ},43.47^{\circ})$

$(177.75,-39.03^{\circ},46.53^{\circ})$

$(177.75,140.97^{\circ},133.47^{\circ})$

$(177.75,140.97^{\circ},136.53^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(95,-77,129)\&\rho=\sqrt{(95)^2+(-77)^2+(129)^2}=177.75\&\theta=arctan(\frac{-77}{95})=-39.03^{\circ}\&\phi=arccos(\frac{129}{177.75})=43.47^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(23,-133,-80).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(156.9,-80.19^{\circ},120.66^{\circ})$

$(156.9,-80.19^{\circ},149.34^{\circ})$

$(156.9,99.81^{\circ},30.66^{\circ})$

$(156.9,99.81^{\circ},59.34^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(23,-133,-80)\&\rho=\sqrt{(23)^2+(-133)^2+(-80)^2}=156.9\&\theta=arctan(\frac{-133}{23})=-80.19^{\circ}\&\phi=arccos(\frac{-80}{156.9})=120.66^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (121,148^{\circ},78^{\circ})\text{ of form: } (\rho,\theta,\phi)\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Explanation

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\&x=\rho sin(\phi)cos(\theta)\&y=\rho sin(\phi)sin(\theta)\&z=\rho cos(\phi)\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\& \text{imperative that the correct units (degrees or radians) are specified}\& \text{for the input!}\&\text{For our coordinates}\&x=121sin(78^{\circ})cos(148^{\circ})=-100.37\&y=118.356sin(78^{\circ})sin(148^{\circ})=62.72\&z=121cos(78^{\circ})25.16\end{align*}$ a

Express the three-dimensional (x,y,z) Cartesian coordinates as cylindrical coordinates (r, θ, z):

$\left ( 2,1, -2\right )$

$\left ( \sqrt{5}, \tan^{-1}\frac{1}{2}, -2 \right )$

$\left ( \sqrt{5}, \tan^{-1}2, -2 \right )$

$\left ( 5, \tan^{-1}\frac{1}{2}, -2 \right )$

$\left ( 5, \tan^{-1}2, -2 \right )$

$\left ( \sqrt{5}, \tan^{-1}\frac{1}{2}, 1 \right )$

Explanation

The coordinates (2, 1, -2) corresponds to: x = 2, y = 1, z = -2, and are to be converted to the cylindrical coordinates in form of (r, θ, z), where:

$\ r = \sqrt{x^{2}+y^{2}} \ \tan(\theta) = \frac{y}{x} \ z = z$

So, filling in for x, y, z:

$\ r = {\sqrt{2^{2}+1^{2}}} = \sqrt{5} \ \tan(\theta) = \frac{1}{2} \Rightarrow \theta = \tan^{-1}\frac{1}{2} \ z = -2$

Then the cylindrical coordinates are represented as:

$\left ( \sqrt{5}, \tan^{-1}\frac{1}{2}, -2 \right )$

$\begin{align*}&\text{A point in space is located, in cylindrical coordinates, at } (105,163^{\circ},-55).\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Explanation

$\begin{align*}&\text{If asked to convert cylindrical coordinates of the form} (r,\theta,z)\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{ and }y\& \text{to the radius, }r\text{, and the angle, }\theta.\text{ The relationships are as follows:}\&x=rcos(\theta)\&y=rsin(\theta)\&\text{Finding }z\text{ is much simpler; it does not change between}\& \text{Cartesian and cylindrical coordinates:}\&z=z\&\text{However, care should be taken when finding }x\text{ and }y\text{; if using a calculator, it is}\& \text{imperative that the correct units (degrees or radians) are specified}\& \text{for the input!}\&\text{For our coordinates}\&x=105cos(163^{\circ})=-100.41\&y=105sin(163^{\circ})=30.7\&z=-55\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at } (95,122,-112).\&\text{What is the position of this point in cylindrical coordinates?}\end{align*}$

$(154.63,52.09^{\circ},-112)$

$(154.63,-127.91^{\circ},-112)$

$(154.63,37.91^{\circ},-112)$

$(154.63,-142.09^{\circ},-112)$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates of the form} (x,y,z)\&\text{to cylindrical coordinates of the form }(r,\theta,z)\text{, the first and third terms}\& \text{are the most straightforward: }\& r=\sqrt{x^2+y^2}\&z=z\&\text{However, it is prudent to be carful when calculating }\theta.\&\text{The formula for it is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Remember, it is important to be mindful of the signs of }x\text{ and }y\text{, bearing in mind}\& \text{which quadrant the point lies in; this determines the value of }\theta.\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\&\text{For our coordinates: }(95,122,-112)\&r=\sqrt{(95)^2+(122)^2}=154.63\&\theta=arctan(\frac{122}{95})=52.09^{\circ}\&z=-112\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at } (95,122,-112).\&\text{What is the position of this point in cylindrical coordinates?}\end{align*}$

$(154.63,52.09^{\circ},-112)$

$(154.63,127.91^{\circ},-112)$

$(217,52.09^{\circ},-112)$

$(217,127.91^{\circ},-112)$

Explanation

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