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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #431 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(129,24,-145).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(195.56,169.46^{\circ},137.86^{\circ})$

$(195.56,10.54^{\circ},137.86^{\circ})$

$(96.12,169.46^{\circ},42.14^{\circ})$

$(96.12,10.54^{\circ},137.86^{\circ})$

Correct answer:

$(195.56,10.54^{\circ},137.86^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(129,24,-145)\\&\rho=\sqrt{(129)^2+(24)^2+(-145)^2}=195.56\\&\theta=arctan(\frac{24}{129})=10.54^{\circ}\\&\phi=arccos(\frac{-145}{195.56})=137.86^{\circ}\end{align*}$

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Example Question #432 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-114,109,-5).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(157.8,136.28^{\circ},91.82^{\circ})$

$(57.15,136.28^{\circ},91.82^{\circ})$

$(57.15,43.72^{\circ},88.18^{\circ})$

$(157.8,43.72^{\circ},91.82^{\circ})$

Correct answer:

$(157.8,136.28^{\circ},91.82^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-114,109,-5)\\&\rho=\sqrt{(-114)^2+(109)^2+(-5)^2}=157.8\\&\theta=arctan(\frac{109}{-114})=136.28^{\circ}\\&\phi=arccos(\frac{-5}{157.8})=91.82^{\circ}\end{align*}$

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Example Question #433 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(104,-87,16).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(77.77,140.09^{\circ},96.73^{\circ})$

$(136.53,-39.91^{\circ},83.27^{\circ})$

$(77.77,-39.91^{\circ},83.27^{\circ})$

$(136.53,140.09^{\circ},83.27^{\circ})$

Correct answer:

$(136.53,-39.91^{\circ},83.27^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(104,-87,16)\\&\rho=\sqrt{(104)^2+(-87)^2+(16)^2}=136.53\\&\theta=arctan(\frac{-87}{104})=-39.91^{\circ}\\&\phi=arccos(\frac{16}{136.53})=83.27^{\circ}\end{align*}$

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Example Question #434 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(39,-141,35).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(150.42,105.46^{\circ},76.55^{\circ})$

$(150.42,-74.54^{\circ},76.55^{\circ})$

$(139.27,-74.54^{\circ},76.55^{\circ})$

$(139.27,105.46^{\circ},103.45^{\circ})$

Correct answer:

$(150.42,-74.54^{\circ},76.55^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(39,-141,35)\\&\rho=\sqrt{(39)^2+(-141)^2+(35)^2}=150.42\\&\theta=arctan(\frac{-141}{39})=-74.54^{\circ}\\&\phi=arccos(\frac{35}{150.42})=76.55^{\circ}\end{align*}$

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Example Question #2101 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-41,-136,-3).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.08,73.22^{\circ},91.21^{\circ})$

$(137.23,73.22^{\circ},88.79^{\circ})$

$(142.08,-106.78^{\circ},91.21^{\circ})$

$(137.23,-106.78^{\circ},91.21^{\circ})$

Correct answer:

$(142.08,-106.78^{\circ},91.21^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-41,-136,-3)\\&\rho=\sqrt{(-41)^2+(-136)^2+(-3)^2}=142.08\\&\theta=arctan(\frac{-136}{-41})=-106.78^{\circ}\\&\phi=arccos(\frac{-3}{142.08})=91.21^{\circ}\end{align*}$

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Example Question #2102 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-93,-113,-89).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(143.46,-129.45^{\circ},121.31^{\circ})$

$(171.29,-129.45^{\circ},121.31^{\circ})$

$(143.46,50.55^{\circ},58.69^{\circ})$

$(171.29,50.55^{\circ},121.31^{\circ})$

Correct answer:

$(171.29,-129.45^{\circ},121.31^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-93,-113,-89)\\&\rho=\sqrt{(-93)^2+(-113)^2+(-89)^2}=171.29\\&\theta=arctan(\frac{-113}{-93})=-129.45^{\circ}\\&\phi=arccos(\frac{-89}{171.29})=121.31^{\circ}\end{align*}$

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Example Question #441 : 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:

$(57.57,167.12^{\circ},141.42^{\circ})$

$(36.53,-12.88^{\circ},141.42^{\circ})$

$(36.53,167.12^{\circ},38.58^{\circ})$

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

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 #441 : 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:

$(100.69,165.43^{\circ},98.26^{\circ})$

$(104.41,165.43^{\circ},81.74^{\circ})$

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

$(100.69,14.57^{\circ},81.74^{\circ})$

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 #442 : 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:

$(160.94,153.07^{\circ},61.42^{\circ})$

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

$(129.96,153.07^{\circ},118.58^{\circ})$

$(129.96,-26.93^{\circ},61.42^{\circ})$

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 #443 : 3 Dimensional Space

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

Possible Answers:

$(86.44,154.65^{\circ},76.62^{\circ})$

$(73.7,154.65^{\circ},103.38^{\circ})$

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

$(73.7,-25.35^{\circ},76.62^{\circ})$

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 #431 : 3 Dimensional Space

Example Question #432 : 3 Dimensional Space

Example Question #433 : 3 Dimensional Space

Example Question #434 : 3 Dimensional Space

Example Question #2101 : Calculus 3

Example Question #2102 : Calculus 3

Example Question #441 : 3 Dimensional Space

Example Question #441 : 3 Dimensional Space

Example Question #442 : 3 Dimensional Space

Example Question #443 : 3 Dimensional Space

All Calculus 3 Resources

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