Spherical Coordinates - AP Calculus BC

Card 0 of 890

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Answer

In order to convert to spherical coordinates , we need to remember the conversion equations.

$\rho=\sqrt{x^2+y^2+z^2}$

$\theta=\tan^{-1}\Big(\frac{y}{x}\Big)$

$\phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$

Now lets apply this to our problem.

$\rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}$

$\theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(9,3,2)\&\rho=\sqrt{(9)^2+(3)^2+(2)^2}=9.7\&\theta=arctan(\frac{3}{9})=18.43^{\circ}\&\phi=arccos(\frac{2}{9.7})=78.1^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-12,4,-3)\&\rho=\sqrt{(-12)^2+(4)^2+(-3)^2}=13\&\theta=arctan(\frac{4}{-12})=161.57^{\circ}\&\phi=arccos(\frac{-3}{13})=103.34^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(5,1,-6)\&\rho=\sqrt{(5)^2+(1)^2+(-6)^2}=7.87\&\theta=arctan(\frac{1}{5})=11.31^{\circ}\&\phi=arccos(\frac{-6}{7.87})=139.64^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(5,-8,3)\&\rho=\sqrt{(5)^2+(-8)^2+(3)^2}=9.9\&\theta=arctan(\frac{-8}{5})=-57.99^{\circ}\&\phi=arccos(\frac{3}{9.9})=72.36^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-9,-9,-9)\&\rho=\sqrt{(-9)^2+(-9)^2+(-9)^2}=15.59\&\theta=arctan(\frac{-9}{-9})=-135^{\circ}\&\phi=arccos(\frac{-9}{15.59})=125.26^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(11,0,-5)\&\rho=\sqrt{(11)^2+(0)^2+(-5)^2}=12.08\&\theta=arctan(\frac{0}{11})=0^{\circ}\&\phi=arccos(\frac{-5}{12.08})=114.44^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(85,10,13)\&\rho=\sqrt{(85)^2+(10)^2+(13)^2}=86.57\&\theta=arctan(\frac{10}{85})=6.71^{\circ}\&\phi=arccos(\frac{13}{86.57})=81.36^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(3,-23,50)\&\rho=\sqrt{(3)^2+(-23)^2+(50)^2}=55.12\&\theta=arctan(\frac{-23}{3})=-82.57^{\circ}\&\phi=arccos(\frac{50}{55.12})=24.89^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(11,12,13)\&\rho=\sqrt{(11)^2+(12)^2+(13)^2}=20.83\&\theta=arctan(\frac{12}{11})=47.49^{\circ}\&\phi=arccos(\frac{13}{20.83})=51.39^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(14,-12,-20)\&\rho=\sqrt{(14)^2+(-12)^2+(-20)^2}=27.2\&\theta=arctan(\frac{-12}{14})=-40.6^{\circ}\&\phi=arccos(\frac{-20}{27.2})=137.33^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(11,24,15)\&\rho=\sqrt{(11)^2+(24)^2+(15)^2}=30.36\&\theta=arctan(\frac{24}{11})=65.38^{\circ}\&\phi=arccos(\frac{15}{30.36})=60.4^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-8,-11,15)\&\rho=\sqrt{(-8)^2+(-11)^2+(15)^2}=20.25\&\theta=arctan(\frac{-11}{-8})=-126.03^{\circ}\&\phi=arccos(\frac{15}{20.25})=42.2^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-26,-8,2)\&\rho=\sqrt{(-26)^2+(-8)^2+(2)^2}=27.28\&\theta=arctan(\frac{-8}{-26})=-162.9^{\circ}\&\phi=arccos(\frac{2}{27.28})=85.8^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(10,-48,-42)\&\rho=\sqrt{(10)^2+(-48)^2+(-42)^2}=64.56\&\theta=arctan(\frac{-48}{10})=-78.23^{\circ}\&\phi=arccos(\frac{-42}{64.56})=130.58^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(8,41,-38)\&\rho=\sqrt{(8)^2+(41)^2+(-38)^2}=56.47\&\theta=arctan(\frac{41}{8})=78.96^{\circ}\&\phi=arccos(\frac{-38}{56.47})=132.29^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-3,2,-28)\&\rho=\sqrt{(-3)^2+(2)^2+(-28)^2}=28.23\&\theta=arctan(\frac{2}{-3})=146.31^{\circ}\&\phi=arccos(\frac{-28}{28.23})=172.66^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-44,-37,-7)\&\rho=\sqrt{(-44)^2+(-37)^2+(-7)^2}=57.91\&\theta=arctan(\frac{-37}{-44})=-139.94^{\circ}\&\phi=arccos(\frac{-7}{57.91})=96.94^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(-24,-34,24)\&\rho=\sqrt{(-24)^2+(-34)^2+(24)^2}=48.04\&\theta=arctan(\frac{-34}{-24})=-125.22^{\circ}\&\phi=arccos(\frac{24}{48.04})=60.03^{\circ}\end{align}$

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Question

A point in space is located, in Cartesian coordinates, at . What is the position of this point in spherical coordinates?

Answer

When given Cartesian coordinates of the form to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

For our coordinates

$\begin{align}&(34,43,-21)\&\rho=\sqrt{(34)^2+(43)^2+(-21)^2}=58.7\&\theta=arctan(\frac{43}{34})=51.67^{\circ}\&\phi=arccos(\frac{-21}{58.7})=110.96^{\circ}\end{align}$

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