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Calculus 3 : Calculus 3

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Example Questions

Example Question #484 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #485 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #486 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #487 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #488 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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 #489 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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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Example Question #491 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(35,-8,-45).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(57.57,167.12^{\circ},51.42^{\circ})$

$(57.57,167.12^{\circ},38.58^{\circ})$

$(57.57,-12.88^{\circ},128.58^{\circ})$

$(57.57,-12.88^{\circ},141.42^{\circ})$

Correct answer:

$(57.57,-12.88^{\circ},141.42^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(35,-8,-45)\\&\rho=\sqrt{(35)^2+(-8)^2+(-45)^2}=57.57\\&\theta=arctan(\frac{-8}{35})=-12.88^{\circ}\\&\phi=arccos(\frac{-45}{57.57})=141.42^{\circ}\end{align*}$

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Example Question #492 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(100,26,15).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(104.41,14.57^{\circ},81.74^{\circ})$

$(104.41,14.57^{\circ},8.26^{\circ})$

$(104.41,-165.43^{\circ},98.26^{\circ})$

$(104.41,-165.43^{\circ},171.74^{\circ})$

Correct answer:

$(104.41,14.57^{\circ},81.74^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(100,26,15)\\&\rho=\sqrt{(100)^2+(26)^2+(15)^2}=104.41\\&\theta=arctan(\frac{26}{100})=14.57^{\circ}\\&\phi=arccos(\frac{15}{104.41})=81.74^{\circ}\end{align*}$

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Example Question #493 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(126,-64,77).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(160.94,-26.93^{\circ},28.58^{\circ})$

$(160.94,-26.93^{\circ},61.42^{\circ})$

$(160.94,153.07^{\circ},151.42^{\circ})$

$(160.94,153.07^{\circ},118.58^{\circ})$

Correct answer:

$(160.94,-26.93^{\circ},61.42^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(126,-64,77)\\&\rho=\sqrt{(126)^2+(-64)^2+(77)^2}=160.94\\&\theta=arctan(\frac{-64}{126})=-26.93^{\circ}\\&\phi=arccos(\frac{77}{160.94})=61.42^{\circ}\end{align*}$

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Example Question #151 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(76,-36,20).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(86.44,154.65^{\circ},103.38^{\circ})$

$(86.44,154.65^{\circ},166.62^{\circ})$

$(86.44,-25.35^{\circ},76.62^{\circ})$

$(86.44,-25.35^{\circ},13.38^{\circ})$

Correct answer:

$(86.44,-25.35^{\circ},76.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(76,-36,20)\\&\rho=\sqrt{(76)^2+(-36)^2+(20)^2}=86.44\\&\theta=arctan(\frac{-36}{76})=-25.35^{\circ}\\&\phi=arccos(\frac{20}{86.44})=76.62^{\circ}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #484 : 3 Dimensional Space

Example Question #485 : 3 Dimensional Space

Example Question #486 : 3 Dimensional Space

Example Question #487 : 3 Dimensional Space

Example Question #488 : 3 Dimensional Space

Example Question #489 : 3 Dimensional Space

Example Question #491 : 3 Dimensional Space

Example Question #492 : 3 Dimensional Space

Example Question #493 : 3 Dimensional Space

Example Question #151 : Spherical Coordinates

All Calculus 3 Resources

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