Math / Geometry

QuestionQUESTION 4
Convert the following rectangular coordinates to cylindrical coordinates. Give angles in terms of Pi. If your answer is two-thirds Pi, you would type 2 pil3. It might look familiar. Keep the square root(s) in your answer- do not use decimals. Rectangular: $(2,-2,2 s q r t 2)=$ Cylindrical: $\square$ $\square$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to transform a point from rectangular coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$ . Watch out! The angle $\theta$ needs to be in radians in terms of $\pi$ , and we must keep the square roots in our answer.

STEP 2

1. Find r
2. Find θ
3. Find z

STEP 3

Let's start by finding $r$ , the radial distance.
Remember, $r$ measures the distance from the origin to the point in the $xy$ -plane.
It's like finding the hypotenuse of a right triangle!

STEP 4

We can use the Pythagorean theorem: $r = \sqrt{x^2 + y^2}$ .
Here, our $x$ is 2 and our $y$ is -2.

STEP 5

Let's plug those values in: $r = \sqrt{(\mathbf{2})^2 + (\mathbf{-2})^2} = \sqrt{4 + 4} = \sqrt{8}$ .

STEP 6

We can simplify $\sqrt{8}$ to $2\sqrt{2}$ .
So, our $r$ is $\mathbf{2\sqrt{2}}$ .
Awesome!

STEP 7

Now, let's find $\theta$ , the angle. $\theta$ is the angle between the positive x-axis and the line connecting the origin to the point in the $xy$ -plane.

STEP 8

We can use the formula $\tan(\theta) = \frac{y}{x}$ .
With $y = \mathbf{-2}$ and $x = \mathbf{2}$ , we get $\tan(\theta) = \frac{\mathbf{-2}}{\mathbf{2}} = \mathbf{-1}$ .

STEP 9

Since $\tan(\theta) = -1$ , we know $\theta$ is in either the second or fourth quadrant.
Since our $x$ is positive and our $y$ is negative, we're in the fourth quadrant.

STEP 10

The angle whose tangent is $-1$ in the fourth quadrant is $-\frac{\pi}{4}$ .
We can also express this as $\frac{7\pi}{4}$ .
Let's stick with $\frac{7\pi}{4}$ for now.

STEP 11

Finally, the $z$ -coordinate in cylindrical coordinates is the same as the $z$ -coordinate in rectangular coordinates.
So, $z = \mathbf{2\sqrt{2}}$ .
Easy peasy!

STEP 12

Our cylindrical coordinates are $(2\sqrt{2}, \frac{7\pi}{4}, 2\sqrt{2})$ .

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Studdy Solution

STEP 1

What is this asking? We need to transform a point from rectangular coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$ . Watch out! The angle $\theta$ needs to be in radians in terms of $\pi$ , and we must keep the square roots in our answer.

STEP 2

1. Find r
2. Find θ
3. Find z

STEP 3

Let's start by finding $r$ , the radial distance.
Remember, $r$ measures the distance from the origin to the point in the $xy$ -plane.
It's like finding the hypotenuse of a right triangle!

STEP 4

We can use the Pythagorean theorem: $r = \sqrt{x^2 + y^2}$ .
Here, our $x$ is 2 and our $y$ is -2.

STEP 5

Let's plug those values in: $r = \sqrt{(\mathbf{2})^2 + (\mathbf{-2})^2} = \sqrt{4 + 4} = \sqrt{8}$ .

STEP 6

We can simplify $\sqrt{8}$ to $2\sqrt{2}$ .
So, our $r$ is $\mathbf{2\sqrt{2}}$ .
Awesome!

STEP 7

Now, let's find $\theta$ , the angle. $\theta$ is the angle between the positive x-axis and the line connecting the origin to the point in the $xy$ -plane.

STEP 8

We can use the formula $\tan(\theta) = \frac{y}{x}$ .
With $y = \mathbf{-2}$ and $x = \mathbf{2}$ , we get $\tan(\theta) = \frac{\mathbf{-2}}{\mathbf{2}} = \mathbf{-1}$ .

STEP 9

Since $\tan(\theta) = -1$ , we know $\theta$ is in either the second or fourth quadrant.
Since our $x$ is positive and our $y$ is negative, we're in the fourth quadrant.

STEP 10

The angle whose tangent is $-1$ in the fourth quadrant is $-\frac{\pi}{4}$ .
We can also express this as $\frac{7\pi}{4}$ .
Let's stick with $\frac{7\pi}{4}$ for now.

STEP 11

Finally, the $z$ -coordinate in cylindrical coordinates is the same as the $z$ -coordinate in rectangular coordinates.
So, $z = \mathbf{2\sqrt{2}}$ .
Easy peasy!

STEP 12

Our cylindrical coordinates are $(2\sqrt{2}, \frac{7\pi}{4}, 2\sqrt{2})$ .

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Studdy Solution

STEP 1

What is this asking? We need to transform a point from rectangular coordinates (x,y,z)(x, y, z)(x,y,z) to cylindrical coordinates (r,θ,z)(r, \theta, z)(r,θ,z). Watch out! The angle θ\thetaθ needs to be in radians in terms of π\piπ, and we must keep the square roots in our answer.

STEP 2

1. Find *r*2. Find *θ*3. Find *z*

STEP 3

Let's **start** by finding rrr, the radial distance.Remember, rrr measures the distance from the origin to the point in the xyxyxy-plane.It's like finding the hypotenuse of a right triangle!

STEP 4

We can use the Pythagorean theorem: r=x2+y2r = \sqrt{x^2 + y^2}r=x2+y2​.Here, our xxx is **2** and our yyy is **-2**.

STEP 5

Let's **plug** those values in: r=(2)2+(−2)2=4+4=8r = \sqrt{(\mathbf{2})^2 + (\mathbf{-2})^2} = \sqrt{4 + 4} = \sqrt{8}r=(2)2+(−2)2​=4+4​=8​.

STEP 6

We can **simplify** 8\sqrt{8}8​ to 222\sqrt{2}22​.So, our rrr is 22\mathbf{2\sqrt{2}}22​.Awesome!

STEP 7

Now, let's **find** θ\thetaθ, the angle. θ\thetaθ is the angle between the positive x-axis and the line connecting the origin to the point in the xyxyxy-plane.

STEP 8

We can use the **formula** tan⁡(θ)=yx\tan(\theta) = \frac{y}{x}tan(θ)=xy​.With y=−2y = \mathbf{-2}y=−2 and x=2x = \mathbf{2}x=2, we get tan⁡(θ)=−22=−1\tan(\theta) = \frac{\mathbf{-2}}{\mathbf{2}} = \mathbf{-1}tan(θ)=2−2​=−1.

STEP 9

Since tan⁡(θ)=−1\tan(\theta) = -1tan(θ)=−1, we know θ\thetaθ is in either the second or fourth quadrant.Since our xxx is positive and our yyy is negative, we're in the **fourth quadrant**.

STEP 10

The angle whose tangent is −1-1−1 in the fourth quadrant is −π4-\frac{\pi}{4}−4π​.We can also express this as 7π4\frac{7\pi}{4}47π​.Let's **stick with** 7π4\frac{7\pi}{4}47π​ for now.

STEP 11

Finally, the zzz-coordinate in cylindrical coordinates is the **same** as the zzz-coordinate in rectangular coordinates.So, z=22z = \mathbf{2\sqrt{2}}z=22​.Easy peasy!

STEP 12

Our cylindrical coordinates are (22,7π4,22)(2\sqrt{2}, \frac{7\pi}{4}, 2\sqrt{2})(22​,47π​,22​).

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What is this asking? We need to transform a point from rectangular coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$ . Watch out! The angle $\theta$ needs to be in radians in terms of $\pi$ , and we must keep the square roots in our answer.

1. Find r
2. Find θ
3. Find z

Let's start by finding $r$ , the radial distance.
Remember, $r$ measures the distance from the origin to the point in the $xy$ -plane.
It's like finding the hypotenuse of a right triangle!

We can use the Pythagorean theorem: $r = \sqrt{x^2 + y^2}$ .
Here, our $x$ is 2 and our $y$ is -2.

Let's plug those values in: $r = \sqrt{(\mathbf{2})^2 + (\mathbf{-2})^2} = \sqrt{4 + 4} = \sqrt{8}$ .

We can simplify $\sqrt{8}$ to $2\sqrt{2}$ .
So, our $r$ is $\mathbf{2\sqrt{2}}$ .
Awesome!

Now, let's find $\theta$ , the angle. $\theta$ is the angle between the positive x-axis and the line connecting the origin to the point in the $xy$ -plane.

We can use the formula $\tan(\theta) = \frac{y}{x}$ .
With $y = \mathbf{-2}$ and $x = \mathbf{2}$ , we get $\tan(\theta) = \frac{\mathbf{-2}}{\mathbf{2}} = \mathbf{-1}$ .

Since $\tan(\theta) = -1$ , we know $\theta$ is in either the second or fourth quadrant.
Since our $x$ is positive and our $y$ is negative, we're in the fourth quadrant.

The angle whose tangent is $-1$ in the fourth quadrant is $-\frac{\pi}{4}$ .
We can also express this as $\frac{7\pi}{4}$ .
Let's stick with $\frac{7\pi}{4}$ for now.

Finally, the $z$ -coordinate in cylindrical coordinates is the same as the $z$ -coordinate in rectangular coordinates.
So, $z = \mathbf{2\sqrt{2}}$ .
Easy peasy!

Our cylindrical coordinates are $(2\sqrt{2}, \frac{7\pi}{4}, 2\sqrt{2})$ .