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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #138 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(14,-61,74).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(96.92,-77.07^{\circ},49.78^{\circ})$

$(96.92,102.93^{\circ},130.22^{\circ})$

$(96.92,102.93^{\circ},139.78^{\circ})$

$(96.92,-77.07^{\circ},40.22^{\circ})$

Correct answer:

$(96.92,-77.07^{\circ},40.22^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(14,-61,74)\\&\rho=\sqrt{(14)^2+(-61)^2+(74)^2}=96.92\\&\theta=arctan(\frac{-61}{14})=-77.07^{\circ}\\&\phi=arccos(\frac{74}{96.92})=40.22^{\circ}\end{align*}$

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Example Question #139 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-94,56,-95).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(144.9,149.22^{\circ},130.97^{\circ})$

$(144.9,149.22^{\circ},139.03^{\circ})$

$(144.9,-30.78^{\circ},49.03^{\circ})$

$(144.9,-30.78^{\circ},40.97^{\circ})$

Correct answer:

$(144.9,149.22^{\circ},130.97^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-94,56,-95)\\&\rho=\sqrt{(-94)^2+(56)^2+(-95)^2}=144.9\\&\theta=arctan(\frac{56}{-94})=149.22^{\circ}\\&\phi=arccos(\frac{-95}{144.9})=130.97^{\circ}\end{align*}$

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Example Question #140 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-40,38,84).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(100.5,136.47^{\circ},33.3^{\circ})$

$(100.5,-43.53^{\circ},146.7^{\circ})$

$(100.5,-43.53^{\circ},123.3^{\circ})$

$(100.5,136.47^{\circ},56.7^{\circ})$

Correct answer:

$(100.5,136.47^{\circ},33.3^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-40,38,84)\\&\rho=\sqrt{(-40)^2+(38)^2+(84)^2}=100.5\\&\theta=arctan(\frac{38}{-40})=136.47^{\circ}\\&\phi=arccos(\frac{84}{100.5})=33.3^{\circ}\end{align*}$

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Example Question #141 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-126,129,83).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(198.51,134.33^{\circ},65.28^{\circ})$

$(198.51,134.33^{\circ},24.72^{\circ})$

$(198.51,-45.67^{\circ},114.72^{\circ})$

$(198.51,-45.67^{\circ},155.28^{\circ})$

Correct answer:

$(198.51,134.33^{\circ},65.28^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-126,129,83)\\&\rho=\sqrt{(-126)^2+(129)^2+(83)^2}=198.51\\&\theta=arctan(\frac{129}{-126})=134.33^{\circ}\\&\phi=arccos(\frac{83}{198.51})=65.28^{\circ}\end{align*}$

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Example Question #142 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-4,-19,-16).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(25.16,-101.89^{\circ},129.49^{\circ})$

$(25.16,-101.89^{\circ},140.51^{\circ})$

$(25.16,78.11^{\circ},39.49^{\circ})$

$(25.16,78.11^{\circ},50.51^{\circ})$

Correct answer:

$(25.16,-101.89^{\circ},129.49^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-4,-19,-16)\\&\rho=\sqrt{(-4)^2+(-19)^2+(-16)^2}=25.16\\&\theta=arctan(\frac{-19}{-4})=-101.89^{\circ}\\&\phi=arccos(\frac{-16}{25.16})=129.49^{\circ}\end{align*}$

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Example Question #143 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-58,3,3).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(58.15,-2.96^{\circ},177.04^{\circ})$

$(58.15,177.04^{\circ},87.04^{\circ})$

$(58.15,-2.96^{\circ},92.96^{\circ})$

$(58.15,177.04^{\circ},2.96^{\circ})$

Correct answer:

$(58.15,177.04^{\circ},87.04^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-58,3,3)\\&\rho=\sqrt{(-58)^2+(3)^2+(3)^2}=58.15\\&\theta=arctan(\frac{3}{-58})=177.04^{\circ}\\&\phi=arccos(\frac{3}{58.15})=87.04^{\circ}\end{align*}$

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Example Question #144 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(96,89,43).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(137.79,42.83^{\circ},71.82^{\circ})$

$(137.79,42.83^{\circ},18.18^{\circ})$

$(137.79,-137.17^{\circ},161.82^{\circ})$

$(137.79,-137.17^{\circ},108.18^{\circ})$

Correct answer:

$(137.79,42.83^{\circ},71.82^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(96,89,43)\\&\rho=\sqrt{(96)^2+(89)^2+(43)^2}=137.79\\&\theta=arctan(\frac{89}{96})=42.83^{\circ}\\&\phi=arccos(\frac{43}{137.79})=71.82^{\circ}\end{align*}$

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Example Question #145 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-37,94,10).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(101.51,111.49^{\circ},84.35^{\circ})$

$(101.51,111.49^{\circ},5.65^{\circ})$

$(101.51,-68.51^{\circ},174.35^{\circ})$

$(101.51,-68.51^{\circ},95.65^{\circ})$

Correct answer:

$(101.51,111.49^{\circ},84.35^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-37,94,10)\\&\rho=\sqrt{(-37)^2+(94)^2+(10)^2}=101.51\\&\theta=arctan(\frac{94}{-37})=111.49^{\circ}\\&\phi=arccos(\frac{10}{101.51})=84.35^{\circ}\end{align*}$

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Example Question #146 : Spherical Coordinates

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #147 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-45,-91,-75).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(126.22,63.69^{\circ},53.54^{\circ})$

$(126.22,-116.31^{\circ},143.54^{\circ})$

$(126.22,-116.31^{\circ},126.46^{\circ})$

$(126.22,63.69^{\circ},36.46^{\circ})$

Correct answer:

$(126.22,-116.31^{\circ},126.46^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-45,-91,-75)\\&\rho=\sqrt{(-45)^2+(-91)^2+(-75)^2}=126.22\\&\theta=arctan(\frac{-91}{-45})=-116.31^{\circ}\\&\phi=arccos(\frac{-75}{126.22})=126.46^{\circ}\end{align*}$

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #138 : Spherical Coordinates

Example Question #139 : Spherical Coordinates

Example Question #140 : Spherical Coordinates

Example Question #141 : Spherical Coordinates

Example Question #142 : Spherical Coordinates

Example Question #143 : Spherical Coordinates

Example Question #144 : Spherical Coordinates

Example Question #145 : Spherical Coordinates

Example Question #146 : Spherical Coordinates

Example Question #147 : Spherical Coordinates

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