Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #391 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at \displaystyle (17,11,37). What is the position of this point in spherical coordinates?

Possible Answers:

\displaystyle (20.25,32.91^{\circ},28.69^{\circ})

\displaystyle (20.25,147.09^{\circ},-151.31^{\circ})

\displaystyle (20.25,147.09^{\circ},151.31^{\circ})

\displaystyle (42.18,32.91^{\circ},28.69^{\circ})

\displaystyle (42.18,147.09^{\circ},28.69^{\circ})

Correct answer:

\displaystyle (42.18,32.91^{\circ},28.69^{\circ})

Explanation:

When given Cartesian coordinates of the form \displaystyle (x,y,z) to cylindrical coordinates of the form \displaystyle (\rho,\theta,\phi), it would be useful to calculate the \displaystyle \rho term first, as we'll derive \displaystyle \phi from it.

\displaystyle \rho=\sqrt{x^2+y^2+z^2}

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with \displaystyle \theta, the formula for it is as follows: 

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

 

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

Quadrants

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

To calculate \displaystyle \phi, we can make use of our \displaystyle \rho value. It is found as

\displaystyle \phi=arccos(\frac{z}{\rho}), where \displaystyle 0\leq\phi\leq180^{\circ}

Phirho

For our coordinates 

 \displaystyle \begin{align*}&(17,11,37)\\&\rho=\sqrt{(17)^2+(11)^2+(37)^2}=42.18\\&\theta=arctan(\frac{11}{17})=32.91^{\circ}\\&\phi=arccos(\frac{37}{42.18})=28.69^{\circ}\end{align*}

Example Question #42 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at \displaystyle (-18,-14,35). What is the position of this point in spherical coordinates?

Possible Answers:

\displaystyle (22.8,37.87^{\circ},146.91^{\circ})

\displaystyle (41.77,37.87^{\circ},33.09^{\circ})

\displaystyle (22.8,37.87^{\circ},-146.91^{\circ})

\displaystyle (41.77,-142.13^{\circ},33.09^{\circ})

\displaystyle (22.8,-142.13^{\circ},33.09^{\circ})

Correct answer:

\displaystyle (41.77,-142.13^{\circ},33.09^{\circ})

Explanation:

When given Cartesian coordinates of the form \displaystyle (x,y,z) to cylindrical coordinates of the form \displaystyle (\rho,\theta,\phi), it would be useful to calculate the \displaystyle \rho term first, as we'll derive \displaystyle \phi from it.

\displaystyle \rho=\sqrt{x^2+y^2+z^2}

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with \displaystyle \theta, the formula for it is as follows: 

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

 

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

Quadrants

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

To calculate \displaystyle \phi, we can make use of our \displaystyle \rho value. It is found as

\displaystyle \phi=arccos(\frac{z}{\rho}), where \displaystyle 0\leq\phi\leq180^{\circ}

Phirho

For our coordinates 

 \displaystyle \begin{align*}&(-18,-14,35)\\&\rho=\sqrt{(-18)^2+(-14)^2+(35)^2}=41.77\\&\theta=arctan(\frac{-14}{-18})=-142.13^{\circ}\\&\phi=arccos(\frac{35}{41.77})=33.09^{\circ}\end{align*}

Example Question #41 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(25,12,111).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (27.73,154.36^{\circ},165.97^{\circ})

\displaystyle (27.73,154.36^{\circ},-165.97^{\circ})

\displaystyle (114.41,154.36^{\circ},14.03^{\circ})

\displaystyle (27.73,25.64^{\circ},14.03^{\circ})

\displaystyle (114.41,25.64^{\circ},14.03^{\circ})

Correct answer:

\displaystyle (114.41,25.64^{\circ},14.03^{\circ})

Explanation:

\displaystyle \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 to be careful 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: }(25,12,111)\\&\rho=\sqrt{(25)^2+(12)^2+(111)^2}=114.41\\&\theta=arctan(\frac{12}{25})=25.64^{\circ}\\&\phi=arccos(\frac{111}{114.41})=14.03^{\circ}\end{align*}

Example Question #51 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-71,-55,-115).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (89.81,37.76^{\circ},37.99^{\circ})

\displaystyle (145.91,-142.24^{\circ},142.01^{\circ})

\displaystyle (89.81,-142.24^{\circ},142.01^{\circ})

\displaystyle (145.91,37.76^{\circ},142.01^{\circ})

\displaystyle (89.81,37.76^{\circ},-37.99^{\circ})

Correct answer:

\displaystyle (145.91,-142.24^{\circ},142.01^{\circ})

Explanation:

\displaystyle \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: }(-71,-55,-115)\\&\rho=\sqrt{(-71)^2+(-55)^2+(-115)^2}=145.91\\&\theta=arctan(\frac{-55}{-71})=-142.24^{\circ}\\&\phi=arccos(\frac{-115}{145.91})=142.01^{\circ}\end{align*}

Example Question #52 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(132,44,-6).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (139.14,161.57^{\circ},87.53^{\circ})

\displaystyle (139.14,161.57^{\circ},-87.53^{\circ})

\displaystyle (139.27,18.43^{\circ},92.47^{\circ})

\displaystyle (139.27,161.57^{\circ},92.47^{\circ})

\displaystyle (139.14,18.43^{\circ},92.47^{\circ})

Correct answer:

\displaystyle (139.27,18.43^{\circ},92.47^{\circ})

Explanation:

\displaystyle \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: }(132,44,-6)\\&\rho=\sqrt{(132)^2+(44)^2+(-6)^2}=139.27\\&\theta=arctan(\frac{44}{132})=18.43^{\circ}\\&\phi=arccos(\frac{-6}{139.27})=92.47^{\circ}\end{align*}

Example Question #53 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(13,67,7).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (68.61,100.98^{\circ},84.14^{\circ})

\displaystyle (68.25,79.02^{\circ},84.14^{\circ})

\displaystyle (68.25,100.98^{\circ},-95.86^{\circ})

\displaystyle (68.25,100.98^{\circ},95.86^{\circ})

\displaystyle (68.61,79.02^{\circ},84.14^{\circ})

Correct answer:

\displaystyle (68.61,79.02^{\circ},84.14^{\circ})

Explanation:

\displaystyle \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: }(13,67,7)\\&\rho=\sqrt{(13)^2+(67)^2+(7)^2}=68.61\\&\theta=arctan(\frac{67}{13})=79.02^{\circ}\\&\phi=arccos(\frac{7}{68.61})=84.14^{\circ}\end{align*}

Example Question #54 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(149,-85,-119).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (208.77,-29.7^{\circ},124.75^{\circ})

\displaystyle (171.54,-29.7^{\circ},124.75^{\circ})

\displaystyle (171.54,150.3^{\circ},-55.25^{\circ})

\displaystyle (208.77,150.3^{\circ},124.75^{\circ})

\displaystyle (171.54,150.3^{\circ},55.25^{\circ})

Correct answer:

\displaystyle (208.77,-29.7^{\circ},124.75^{\circ})

Explanation:

\displaystyle \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: }(149,-85,-119)\\&\rho=\sqrt{(149)^2+(-85)^2+(-119)^2}=208.77\\&\theta=arctan(\frac{-85}{149})=-29.7^{\circ}\\&\phi=arccos(\frac{-119}{208.77})=124.75^{\circ}\end{align*}

Example Question #55 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-117,-131,-29).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (178.02,-131.77^{\circ},99.38^{\circ})

\displaystyle (175.64,-131.77^{\circ},99.38^{\circ})

\displaystyle (178.02,48.23^{\circ},99.38^{\circ})

\displaystyle (175.64,48.23^{\circ},80.62^{\circ})

\displaystyle (175.64,48.23^{\circ},-80.62^{\circ})

Correct answer:

\displaystyle (178.02,-131.77^{\circ},99.38^{\circ})

Explanation:

\displaystyle \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: }(-117,-131,-29)\\&\rho=\sqrt{(-117)^2+(-131)^2+(-29)^2}=178.02\\&\theta=arctan(\frac{-131}{-117})=-131.77^{\circ}\\&\phi=arccos(\frac{-29}{178.02})=99.38^{\circ}\end{align*}

Example Question #56 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-16,-40,79).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (43.08,68.2^{\circ},-151.39^{\circ})

\displaystyle (89.98,-111.8^{\circ},28.61^{\circ})

\displaystyle (43.08,-111.8^{\circ},28.61^{\circ})

\displaystyle (89.98,68.2^{\circ},28.61^{\circ})

\displaystyle (43.08,68.2^{\circ},151.39^{\circ})

Correct answer:

\displaystyle (89.98,-111.8^{\circ},28.61^{\circ})

Explanation:

\displaystyle \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: }(-16,-40,79)\\&\rho=\sqrt{(-16)^2+(-40)^2+(79)^2}=89.98\\&\theta=arctan(\frac{-40}{-16})=-111.8^{\circ}\\&\phi=arccos(\frac{79}{89.98})=28.61^{\circ}\end{align*}

Example Question #57 : Spherical Coordinates

\displaystyle \begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(38,82,130).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}

Possible Answers:

\displaystyle (158.33,114.86^{\circ},34.81^{\circ})

\displaystyle (90.38,114.86^{\circ},-145.19^{\circ})

\displaystyle (90.38,65.14^{\circ},34.81^{\circ})

\displaystyle (158.33,65.14^{\circ},34.81^{\circ})

\displaystyle (90.38,114.86^{\circ},145.19^{\circ})

Correct answer:

\displaystyle (158.33,65.14^{\circ},34.81^{\circ})

Explanation:

\displaystyle \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: }(38,82,130)\\&\rho=\sqrt{(38)^2+(82)^2+(130)^2}=158.33\\&\theta=arctan(\frac{82}{38})=65.14^{\circ}\\&\phi=arccos(\frac{130}{158.33})=34.81^{\circ}\end{align*}

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