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Calculus 3 : Calculus 3

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Example Questions

Example Question #2141 : Calculus 3

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (17,-40^{\circ},141^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(10.56,11.8,6.17)

(8.2,-6.88,-13.21)

(-4.12,-4.6,-15.84)

(-10.12,8.49,10.7)

Correct answer:

(8.2,-6.88,-13.21)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=17sin(141^{\circ})cos(-40^{\circ})=8.2\\&y=10.6984sin(141^{\circ})sin(-40^{\circ})=-6.88\\&z=17cos(141^{\circ})-13.21\end{align*}$

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Example Question #133 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (37,-146^{\circ},73^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-2.29,27.14,-25.04)

(-8.97,-6.05,35.38)

(-2.1,24.95,-27.24)

(-29.33,-19.79,10.82)

Correct answer:

(-29.33,-19.79,10.82)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=37sin(73^{\circ})cos(-146^{\circ})=-29.33\\&y=35.3833sin(73^{\circ})sin(-146^{\circ})=-19.79\\&z=37cos(73^{\circ})10.82\end{align*}$

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Example Question #2142 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(23,-133,-80).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(156.9,99.81^{\circ},30.66^{\circ})$

$(156.9,-80.19^{\circ},149.34^{\circ})$

$(156.9,99.81^{\circ},59.34^{\circ})$

$(156.9,-80.19^{\circ},120.66^{\circ})$

Correct answer:

$(156.9,-80.19^{\circ},120.66^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(23,-133,-80)\\&\rho=\sqrt{(23)^2+(-133)^2+(-80)^2}=156.9\\&\theta=arctan(\frac{-133}{23})=-80.19^{\circ}\\&\phi=arccos(\frac{-80}{156.9})=120.66^{\circ}\end{align*}$

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Example Question #2143 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-44,97,-146).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(180.72,114.4^{\circ},143.89^{\circ})$

$(180.72,114.4^{\circ},126.11^{\circ})$

$(180.72,-65.6^{\circ},53.89^{\circ})$

$(180.72,-65.6^{\circ},36.11^{\circ})$

Correct answer:

$(180.72,114.4^{\circ},143.89^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-44,97,-146)\\&\rho=\sqrt{(-44)^2+(97)^2+(-146)^2}=180.72\\&\theta=arctan(\frac{97}{-44})=114.4^{\circ}\\&\phi=arccos(\frac{-146}{180.72})=143.89^{\circ}\end{align*}$

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Example Question #136 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-138,-100,45).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(176.26,-144.07^{\circ},75.21^{\circ})$

$(176.26,35.93^{\circ},104.79^{\circ})$

$(176.26,-144.07^{\circ},14.79^{\circ})$

$(176.26,35.93^{\circ},165.21^{\circ})$

Correct answer:

$(176.26,-144.07^{\circ},75.21^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-138,-100,45)\\&\rho=\sqrt{(-138)^2+(-100)^2+(45)^2}=176.26\\&\theta=arctan(\frac{-100}{-138})=-144.07^{\circ}\\&\phi=arccos(\frac{45}{176.26})=75.21^{\circ}\end{align*}$

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Example Question #137 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(70,44,-15).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(84.03,-147.85^{\circ},10.28^{\circ})$

$(84.03,32.15^{\circ},100.28^{\circ})$

$(84.03,32.15^{\circ},169.72^{\circ})$

$(84.03,-147.85^{\circ},79.72^{\circ})$

Correct answer:

$(84.03,32.15^{\circ},100.28^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(70,44,-15)\\&\rho=\sqrt{(70)^2+(44)^2+(-15)^2}=84.03\\&\theta=arctan(\frac{44}{70})=32.15^{\circ}\\&\phi=arccos(\frac{-15}{84.03})=100.28^{\circ}\end{align*}$

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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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #2141 : Calculus 3

Example Question #133 : Spherical Coordinates

Example Question #2142 : Calculus 3

Example Question #2143 : Calculus 3

Example Question #136 : Spherical Coordinates

Example Question #137 : Spherical Coordinates

Example Question #138 : Spherical Coordinates

Example Question #139 : Spherical Coordinates

Example Question #140 : Spherical Coordinates

Example Question #141 : Spherical Coordinates

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