Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #2131 : Calculus 3

\(\displaystyle \begin{align*}&\text{A point in space is located, in spherical coordinates, at } (19,-94^{\circ},33^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}\)

Possible Answers:

\(\displaystyle (-0.24,-0.06,19)\)

\(\displaystyle (-0.72,-10.32,15.93)\)

\(\displaystyle (18.42,4.66,-0.25)\)

\(\displaystyle (-1.11,-15.9,10.35)\)

Correct answer:

\(\displaystyle (-0.72,-10.32,15.93)\)

Explanation:

\(\displaystyle \begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=19sin(33^{\circ})cos(-94^{\circ})=-0.72\\&y=10.3481sin(33^{\circ})sin(-94^{\circ})=-10.32\\&z=19cos(33^{\circ})15.93\end{align*}\)

Example Question #472 : 3 Dimensional Space

\(\displaystyle \begin{align*}&\text{A point in space is located, in spherical coordinates, at } (63,145^{\circ},8^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}\)

Possible Answers:

\(\displaystyle (55.09,29.15,-9.17)\)

\(\displaystyle (-51.1,35.78,8.77)\)

\(\displaystyle (-8.1,-4.29,62.33)\)

\(\displaystyle (-7.18,5.03,62.39)\)

Correct answer:

\(\displaystyle (-7.18,5.03,62.39)\)

Explanation:

\(\displaystyle \begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=63sin(8^{\circ})cos(145^{\circ})=-7.18\\&y=8.76791sin(8^{\circ})sin(145^{\circ})=5.03\\&z=63cos(8^{\circ})62.39\end{align*}\)

Example Question #473 : 3 Dimensional Space

\(\displaystyle \begin{align*}&\text{A point in space is located, in spherical coordinates, at } (142,-4^{\circ},88^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}\)

Possible Answers:

\(\displaystyle (141.57,-9.9,4.96)\)

\(\displaystyle (-3.29,3.8,141.91)\)

\(\displaystyle (-92.76,107.4,5.03)\)

\(\displaystyle (4.94,-0.35,141.91)\)

Correct answer:

\(\displaystyle (141.57,-9.9,4.96)\)

Explanation:

\(\displaystyle \begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=142sin(88^{\circ})cos(-4^{\circ})=141.57\\&y=141.913sin(88^{\circ})sin(-4^{\circ})=-9.9\\&z=142cos(88^{\circ})4.96\end{align*}\)

Example Question #2132 : Calculus 3

\(\displaystyle \begin{align*}&\text{A point in space is located, in spherical coordinates, at } (51,-47^{\circ},162^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}\)

Possible Answers:

\(\displaystyle (49.52,6.17,10.53)\)

\(\displaystyle (-10.45,-1.3,-49.9)\)

\(\displaystyle (10.75,-11.53,-48.5)\)

\(\displaystyle (-33.08,35.47,15.76)\)

Correct answer:

\(\displaystyle (10.75,-11.53,-48.5)\)

Explanation:

\(\displaystyle \begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=51sin(162^{\circ})cos(-47^{\circ})=10.75\\&y=15.7599sin(162^{\circ})sin(-47^{\circ})=-11.53\\&z=51cos(162^{\circ})-48.5\end{align*}\)

Example Question #131 : Spherical Coordinates

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

Possible Answers:

\(\displaystyle (-4.12,-4.6,-15.84)\)

\(\displaystyle (10.56,11.8,6.17)\)

\(\displaystyle (-10.12,8.49,10.7)\)

\(\displaystyle (8.2,-6.88,-13.21)\)

Correct answer:

\(\displaystyle (8.2,-6.88,-13.21)\)

Explanation:

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

Example Question #476 : 3 Dimensional Space

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

Possible Answers:

\(\displaystyle (-2.29,27.14,-25.04)\)

\(\displaystyle (-29.33,-19.79,10.82)\)

\(\displaystyle (-2.1,24.95,-27.24)\)

\(\displaystyle (-8.97,-6.05,35.38)\)

Correct answer:

\(\displaystyle (-29.33,-19.79,10.82)\)

Explanation:

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

Example Question #477 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

Example Question #478 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

Example Question #479 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

Example Question #480 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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