In order to convert to cylindrical coordinates, we need to recall the conversion equations.

\(\displaystyle r=\sqrt{x^2+y^2}\)

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

\(\displaystyle z=z\)

Now lets apply this to our problem.

\(\displaystyle r=\sqrt{(\sin(t))^2+(\cos(t))^2}=\sqrt{1}=1\)

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

\(\displaystyle z=\tan(t)\)

To convert a point \(\displaystyle (x,y,z)\) into cylindrical corrdinates, the transformation equations are

\(\displaystyle x = r\cos\theta\)

\(\displaystyle y = r\sin\theta\)

\(\displaystyle z = z\).

Choices for \(\displaystyle r, \theta\) may vary depending on the situation, but the \(\displaystyle z\) coordinate remains the same.

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (5,12,7)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow13\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow67.38^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=7\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (4,-4,8)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow5.66\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow -45^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=8\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (1,\sqrt{3},4)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow 2\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow 60^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=4\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (-9,40,11)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow41\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow 102.68^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=11\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (-3,-4,10)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow5\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow -126.87^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=10\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (-4,8,10)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow8.94\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow 116.57^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=10\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (6,8,10)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow10\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow 53.13^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=10\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). The formula for it is as follows: 

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

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

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

For our coordinates \(\displaystyle (-5,-5,-5)\)

\(\displaystyle r=\sqrt{x^2+y^2}\rightarrow7.07\)

\(\displaystyle \theta=arctan(\frac{y}{x})\rightarrow -135^{\circ}\) (Bearing in mind sign convention)

\(\displaystyle z=-5\)