Cylindrical Coordinates - AP Calculus BC

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Question

Convert the following into Cylindrical coordinates.

$x=\sin(t)$

$y=\cos(t)$

$z=\tan(t)$

Answer

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

$r=\sqrt{x^2+y^2}$

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

Now lets apply this to our problem.

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

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

$z=\tan(t)$

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Question

When converting rectangular coordinates to cylindrical coordinates, which variable remains fixed?

Answer

To convert a point into cylindrical corrdinates, the transformation equations are

$x = r\cos\theta$

$y = r\sin\theta$

.

Choices for $r, \theta$ may vary depending on the situation, but the coordinate remains the same.

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow13$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow5.66$

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

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Question

A point in space is located, in Cartesian coordinates, at $(1,\sqrt{3},4)$ . What is the position of this point in cylindrical coordinates?

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

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

$r=\sqrt{x^2+y^2}\rightarrow 2$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow41$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow5$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow8.94$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow10$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow7.07$

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

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Question

A point in space is located, in Cartesian coordinates, at $(4,-4\sqrt{3},14)$ . What is the position of this point in cylindrical coordinates?

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates $(4,-4\sqrt{3},14)$

$r=\sqrt{x^2+y^2}\rightarrow 8$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow 15$

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

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow16.4$

$\theta=arctan(\frac{y}{x})\rightarrow 52.43^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow 51.9$

$\theta=arctan(\frac{y}{x})\rightarrow 70.87^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow9.43$

$\theta=arctan(\frac{y}{x})\rightarrow 147.99^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow6.32$

$\theta=arctan(\frac{y}{x})\rightarrow -161.57^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow8.06$

$\theta=arctan(\frac{y}{x})\rightarrow 60.26^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow8.94$

$\theta=arctan(\frac{y}{x})\rightarrow 153.43^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow6.08$

$\theta=arctan(\frac{y}{x})\rightarrow 9.46^{\circ}$ (Bearing in mind sign convention)

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Question

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

Answer

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

$r=\sqrt{x^2+y^2}$

Care should be taken, however, when calculating $\theta$ . 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 $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates

$r=\sqrt{x^2+y^2}\rightarrow9.43$

$\theta=arctan(\frac{y}{x})\rightarrow 147.99^{\circ}$ (Bearing in mind sign convention)

Compare your answer with the correct one above

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