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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 #425 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(129,59,25).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(133.28,155.42^{\circ},100^{\circ})$

$(133.28,24.58^{\circ},80^{\circ})$

$(144.04,24.58^{\circ},80^{\circ})$

$(144.04,155.42^{\circ},80^{\circ})$

Correct answer:

$(144.04,24.58^{\circ},80^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(129,59,25)\\&\rho=\sqrt{(129)^2+(59)^2+(25)^2}=144.04\\&\theta=arctan(\frac{59}{129})=24.58^{\circ}\\&\phi=arccos(\frac{25}{144.04})=80^{\circ}\end{align*}$

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Example Question #426 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(95,114,147).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(208.88,129.81^{\circ},45.27^{\circ})$

$(208.88,50.19^{\circ},45.27^{\circ})$

$(176.68,129.81^{\circ},134.73^{\circ})$

$(176.68,50.19^{\circ},45.27^{\circ})$

Correct answer:

$(208.88,50.19^{\circ},45.27^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(95,114,147)\\&\rho=\sqrt{(95)^2+(114)^2+(147)^2}=208.88\\&\theta=arctan(\frac{114}{95})=50.19^{\circ}\\&\phi=arccos(\frac{147}{208.88})=45.27^{\circ}\end{align*}$

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Example Question #427 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-150,110,34).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(126.09,36.25^{\circ},100.36^{\circ})$

$(126.09,143.75^{\circ},79.64^{\circ})$

$(189.09,36.25^{\circ},79.64^{\circ})$

$(189.09,143.75^{\circ},79.64^{\circ})$

Correct answer:

$(189.09,143.75^{\circ},79.64^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-150,110,34)\\&\rho=\sqrt{(-150)^2+(110)^2+(34)^2}=189.09\\&\theta=arctan(\frac{110}{-150})=143.75^{\circ}\\&\phi=arccos(\frac{34}{189.09})=79.64^{\circ}\end{align*}$

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Example Question #428 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(147,8,-6).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(123,3.12^{\circ},92.33^{\circ})$

$(147.34,176.88^{\circ},92.33^{\circ})$

$(123,176.88^{\circ},87.67^{\circ})$

$(147.34,3.12^{\circ},92.33^{\circ})$

Correct answer:

$(147.34,3.12^{\circ},92.33^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(147,8,-6)\\&\rho=\sqrt{(147)^2+(8)^2+(-6)^2}=147.34\\&\theta=arctan(\frac{8}{147})=3.12^{\circ}\\&\phi=arccos(\frac{-6}{147.34})=92.33^{\circ}\end{align*}$

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Example Question #429 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(91,-82,-1).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(58.69,137.98^{\circ},89.53^{\circ})$

$(58.69,-42.02^{\circ},90.47^{\circ})$

$(122.5,137.98^{\circ},90.47^{\circ})$

$(122.5,-42.02^{\circ},90.47^{\circ})$

Correct answer:

$(122.5,-42.02^{\circ},90.47^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(91,-82,-1)\\&\rho=\sqrt{(91)^2+(-82)^2+(-1)^2}=122.5\\&\theta=arctan(\frac{-82}{91})=-42.02^{\circ}\\&\phi=arccos(\frac{-1}{122.5})=90.47^{\circ}\end{align*}$

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Example Question #430 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(121,22,104).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.72,10.3^{\circ},49.78^{\circ})$

$(161.06,169.7^{\circ},49.78^{\circ})$

$(161.06,10.3^{\circ},49.78^{\circ})$

$(142.72,169.7^{\circ},130.22^{\circ})$

Correct answer:

$(161.06,10.3^{\circ},49.78^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(121,22,104)\\&\rho=\sqrt{(121)^2+(22)^2+(104)^2}=161.06\\&\theta=arctan(\frac{22}{121})=10.3^{\circ}\\&\phi=arccos(\frac{104}{161.06})=49.78^{\circ}\end{align*}$

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Example Question #431 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(72,26,-76).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(107.87,19.86^{\circ},134.79^{\circ})$

$(36.38,160.14^{\circ},45.21^{\circ})$

$(107.87,160.14^{\circ},134.79^{\circ})$

$(36.38,19.86^{\circ},134.79^{\circ})$

Correct answer:

$(107.87,19.86^{\circ},134.79^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(72,26,-76)\\&\rho=\sqrt{(72)^2+(26)^2+(-76)^2}=107.87\\&\theta=arctan(\frac{26}{72})=19.86^{\circ}\\&\phi=arccos(\frac{-76}{107.87})=134.79^{\circ}\end{align*}$

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Example Question #81 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(50,-125,38).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(121.04,111.8^{\circ},105.76^{\circ})$

$(121.04,-68.2^{\circ},74.24^{\circ})$

$(139.89,111.8^{\circ},74.24^{\circ})$

$(139.89,-68.2^{\circ},74.24^{\circ})$

Correct answer:

$(139.89,-68.2^{\circ},74.24^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(50,-125,38)\\&\rho=\sqrt{(50)^2+(-125)^2+(38)^2}=139.89\\&\theta=arctan(\frac{-125}{50})=-68.2^{\circ}\\&\phi=arccos(\frac{38}{139.89})=74.24^{\circ}\end{align*}$

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Example Question #2091 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(48,69,118).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(127.69,55.18^{\circ},35.46^{\circ})$

$(144.88,55.18^{\circ},35.46^{\circ})$

$(127.69,124.82^{\circ},144.54^{\circ})$

$(144.88,124.82^{\circ},35.46^{\circ})$

Correct answer:

$(144.88,55.18^{\circ},35.46^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(48,69,118)\\&\rho=\sqrt{(48)^2+(69)^2+(118)^2}=144.88\\&\theta=arctan(\frac{69}{48})=55.18^{\circ}\\&\phi=arccos(\frac{118}{144.88})=35.46^{\circ}\end{align*}$

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Example Question #91 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(145,81,25).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(153.2,29.19^{\circ},81.44^{\circ})$

$(167.96,29.19^{\circ},81.44^{\circ})$

$(167.96,150.81^{\circ},81.44^{\circ})$

$(153.2,150.81^{\circ},98.56^{\circ})$

Correct answer:

$(167.96,29.19^{\circ},81.44^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(145,81,25)\\&\rho=\sqrt{(145)^2+(81)^2+(25)^2}=167.96\\&\theta=arctan(\frac{81}{145})=29.19^{\circ}\\&\phi=arccos(\frac{25}{167.96})=81.44^{\circ}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #425 : 3 Dimensional Space

Example Question #426 : 3 Dimensional Space

Example Question #427 : 3 Dimensional Space

Example Question #428 : 3 Dimensional Space

Example Question #429 : 3 Dimensional Space

Example Question #430 : 3 Dimensional Space

Example Question #431 : 3 Dimensional Space

Example Question #81 : Spherical Coordinates

Example Question #2091 : Calculus 3

Example Question #91 : Spherical Coordinates

All Calculus 3 Resources

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