Math / Calculus

QuestionFind the area under $y=4 \cos (x)$ and above $y=4 \sin (x)$ for $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$ . (Note that this area may not be defined over the entire interval.) area $=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the area trapped between two wiggly curves, $4 \cos(x)$ and $4 \sin(x)$ , between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ . Watch out! One function might not always be above the other.
We need to find where they switch!

STEP 2

1. Find Intersection
2. Set up Integral
3. Evaluate Integral

STEP 3

Alright, let's find where these two curves meet!
We set $4 \cos(x) = 4 \sin(x)$ .

STEP 4

Dividing both sides by $4 \cos(x)$ gives us $\tan(x) = 1$ .
Since we're looking in the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$ , our intersection happens at $x = \frac{5\pi}{4}$ .

STEP 5

Now, we need to figure out which function is on top in each part of our interval.
Let's test some points!

STEP 6

For $\frac{\pi}{2} \leq x < \frac{5\pi}{4}$ , let's pick $x = \pi$ .
We have $4 \cos(\pi) = -4$ and $4 \sin(\pi) = 0$ .
So $4 \sin(x)$ is above $4 \cos(x)$ in this region.

STEP 7

For $\frac{5\pi}{4} < x \leq \frac{3\pi}{2}$ , let's pick $x = \frac{7\pi}{6}$ .
We have $4 \cos(\frac{7\pi}{6}) = 4(-\frac{\sqrt{3}}{2}) = -2\sqrt{3}$ and $4 \sin(\frac{7\pi}{6}) = 4(-\frac{1}{2}) = -2$ .
Since $-2 > -2\sqrt{3}$ , $4 \sin(x)$ is still greater than $4 \cos(x)$ .

STEP 8

So, $4 \sin(x)$ is always greater than or equal to $4 \cos(x)$ in our interval!
Our integral for the area is $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (4 \sin(x) - 4 \cos(x)) \, dx$ .

STEP 9

Let's do this integral! $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (4 \sin(x) - 4 \cos(x)) \, dx = [-4 \cos(x) - 4 \sin(x)]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$

STEP 10

Plugging in our bounds, we get: $(-4 \cos(\frac{3\pi}{2}) - 4 \sin(\frac{3\pi}{2})) - (-4 \cos(\frac{\pi}{2}) - 4 \sin(\frac{\pi}{2}))$

STEP 11

This simplifies to: $(-4 \cdot 0 - 4 \cdot (-1)) - (-4 \cdot 0 - 4 \cdot 1) = (4) - (-4) = 8$

STEP 12

The area between the curves is 8.

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QuestionFind the area under $y=4 \cos (x)$ and above $y=4 \sin (x)$ for $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$ . (Note that this area may not be defined over the entire interval.) area $=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the area trapped between two wiggly curves, $4 \cos(x)$ and $4 \sin(x)$ , between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ . Watch out! One function might not always be above the other.
We need to find where they switch!

STEP 2

1. Find Intersection
2. Set up Integral
3. Evaluate Integral

STEP 3

Alright, let's find where these two curves meet!
We set $4 \cos(x) = 4 \sin(x)$ .

STEP 4

Dividing both sides by $4 \cos(x)$ gives us $\tan(x) = 1$ .
Since we're looking in the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$ , our intersection happens at $x = \frac{5\pi}{4}$ .

STEP 5

Now, we need to figure out which function is on top in each part of our interval.
Let's test some points!

STEP 6

For $\frac{\pi}{2} \leq x < \frac{5\pi}{4}$ , let's pick $x = \pi$ .
We have $4 \cos(\pi) = -4$ and $4 \sin(\pi) = 0$ .
So $4 \sin(x)$ is above $4 \cos(x)$ in this region.

STEP 7

For $\frac{5\pi}{4} < x \leq \frac{3\pi}{2}$ , let's pick $x = \frac{7\pi}{6}$ .
We have $4 \cos(\frac{7\pi}{6}) = 4(-\frac{\sqrt{3}}{2}) = -2\sqrt{3}$ and $4 \sin(\frac{7\pi}{6}) = 4(-\frac{1}{2}) = -2$ .
Since $-2 > -2\sqrt{3}$ , $4 \sin(x)$ is still greater than $4 \cos(x)$ .

STEP 8

So, $4 \sin(x)$ is always greater than or equal to $4 \cos(x)$ in our interval!
Our integral for the area is $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (4 \sin(x) - 4 \cos(x)) \, dx$ .

STEP 9

Let's do this integral! $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (4 \sin(x) - 4 \cos(x)) \, dx = [-4 \cos(x) - 4 \sin(x)]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}$

STEP 10

Plugging in our bounds, we get: $(-4 \cos(\frac{3\pi}{2}) - 4 \sin(\frac{3\pi}{2})) - (-4 \cos(\frac{\pi}{2}) - 4 \sin(\frac{\pi}{2}))$

STEP 11

This simplifies to: $(-4 \cdot 0 - 4 \cdot (-1)) - (-4 \cdot 0 - 4 \cdot 1) = (4) - (-4) = 8$

STEP 12

The area between the curves is 8.

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QuestionFind the area under y=4cos⁡(x)y=4 \cos (x)y=4cos(x) and above y=4sin⁡(x)y=4 \sin (x)y=4sin(x) for π2≤x≤3π2\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}2π​≤x≤23π​. (Note that this area may not be defined over the entire interval.) area === □\square□

Studdy Solution

STEP 1

STEP 2

1. Find Intersection2. Set up Integral3. Evaluate Integral

STEP 3

Alright, let's **find** where these two curves meet!We set 4cos⁡(x)=4sin⁡(x)4 \cos(x) = 4 \sin(x)4cos(x)=4sin(x).

STEP 4

Dividing both sides by 4cos⁡(x)4 \cos(x)4cos(x) gives us tan⁡(x)=1\tan(x) = 1tan(x)=1.Since we're looking in the interval [π2,3π2][\frac{\pi}{2}, \frac{3\pi}{2}][2π​,23π​], our intersection happens at x=5π4x = \frac{5\pi}{4}x=45π​.

STEP 5

Now, we need to figure out which function is on top in each part of our interval.Let's **test** some points!

STEP 6

For π2≤x<5π4\frac{\pi}{2} \leq x < \frac{5\pi}{4}2π​≤x<45π​, let's pick x=πx = \pix=π.We have 4cos⁡(π)=−44 \cos(\pi) = -44cos(π)=−4 and 4sin⁡(π)=04 \sin(\pi) = 04sin(π)=0.So 4sin⁡(x)4 \sin(x)4sin(x) is above 4cos⁡(x)4 \cos(x)4cos(x) in this region.

STEP 7

STEP 8

So, 4sin⁡(x)4 \sin(x)4sin(x) is always greater than or equal to 4cos⁡(x)4 \cos(x)4cos(x) in our interval!Our **integral** for the area is ∫π23π2(4sin⁡(x)−4cos⁡(x)) dx \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (4 \sin(x) - 4 \cos(x)) \, dx ∫2π​23π​​(4sin(x)−4cos(x))dx.

STEP 9

STEP 10

**Plugging** in our bounds, we get: (−4cos⁡(3π2)−4sin⁡(3π2))−(−4cos⁡(π2)−4sin⁡(π2)) (-4 \cos(\frac{3\pi}{2}) - 4 \sin(\frac{3\pi}{2})) - (-4 \cos(\frac{\pi}{2}) - 4 \sin(\frac{\pi}{2})) (−4cos(23π​)−4sin(23π​))−(−4cos(2π​)−4sin(2π​))

STEP 11

This **simplifies** to: (−4⋅0−4⋅(−1))−(−4⋅0−4⋅1)=(4)−(−4)=8 (-4 \cdot 0 - 4 \cdot (-1)) - (-4 \cdot 0 - 4 \cdot 1) = (4) - (-4) = 8 (−4⋅0−4⋅(−1))−(−4⋅0−4⋅1)=(4)−(−4)=8

STEP 12

The area between the curves is **8**.

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QuestionFind the area under $y=4 \cos (x)$ and above $y=4 \sin (x)$ for $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{2}$ . (Note that this area may not be defined over the entire interval.) area $=$ $\square$

1. Find Intersection
2. Set up Integral
3. Evaluate Integral

Alright, let's find where these two curves meet!
We set $4 \cos(x) = 4 \sin(x)$ .

Dividing both sides by $4 \cos(x)$ gives us $\tan(x) = 1$ .
Since we're looking in the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$ , our intersection happens at $x = \frac{5\pi}{4}$ .

Now, we need to figure out which function is on top in each part of our interval.
Let's test some points!

For $\frac{\pi}{2} \leq x < \frac{5\pi}{4}$ , let's pick $x = \pi$ .
We have $4 \cos(\pi) = -4$ and $4 \sin(\pi) = 0$ .
So $4 \sin(x)$ is above $4 \cos(x)$ in this region.

So, $4 \sin(x)$ is always greater than or equal to $4 \cos(x)$ in our interval!
Our integral for the area is $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (4 \sin(x) - 4 \cos(x)) \, dx$ .

Plugging in our bounds, we get: $(-4 \cos(\frac{3\pi}{2}) - 4 \sin(\frac{3\pi}{2})) - (-4 \cos(\frac{\pi}{2}) - 4 \sin(\frac{\pi}{2}))$

This simplifies to: $(-4 \cdot 0 - 4 \cdot (-1)) - (-4 \cdot 0 - 4 \cdot 1) = (4) - (-4) = 8$

The area between the curves is 8.