Area of a rectangle under a cosine curve?

Printable View

June 25th 2008, 04:39 PM
Myung

Area of a rectangle under a cosine curve?

http://www.cubeupload.com/files/d01fa3capture.jpg

What is the equation of the rectangle under the curve? The curve is 3cos(x).
June 25th 2008, 04:47 PM
Mathstud28

Quote:

Originally Posted by Myung

http://www.cubeupload.com/files/d01fa3capture.jpg

What is the equation of the rectangle under the curve? The curve is 3cos(x).

Equation of the rectangle? If you are asking what

$3\int\cos(x)dx=3\sin(x)+C$
June 25th 2008, 04:51 PM
Myung

Is that all? My guess is 6xcos(x) because x+x times 3cos(x) with something like -pi/2 < x < pi/2 as the restrictions, is that incorrect?
June 25th 2008, 04:56 PM
Mathstud28

Quote:

Originally Posted by Myung

Is that all? My guess is 6xcos(x) because x+x times 3cos(x) with something like -pi/2 < x < pi/2 as the restrictions, is that incorrect?

I have no idea what you area asking, are you asking fo the area between $\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ ?

Given by $3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos(x)$ ?
June 25th 2008, 05:00 PM
Myung

I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start.
June 25th 2008, 05:10 PM
Chris L T521

Quote:

Originally Posted by Myung

http://www.cubeupload.com/files/d01fa3capture.jpg

What is the equation of the rectangle under the curve? The curve is 3cos(x).

Quote:

Originally Posted by Myung

I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start.

The Area of the rectangle is $A(x)=Base*Height=6x\cos(x)$

Are we supposed to find the value of x that maximizes the area?

Thus, $\frac{dA}{dx}=-6x\sin(x)+6\cos(x)$

Setting this equal to zero, we get that $\cot(x)=x$ . Using your calculator, we see that this is the case for $x=\pm.86 \ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

We take the positive value.

Thus, the area of the rectangle is $A(.86)=6(.86)\cos(.86)\approx \color{red}\boxed{3.37}$ .

I believe this is the answer.

Hope that this makes sense to you! :D

--Chris
June 25th 2008, 05:12 PM
Mathstud28

Quote:

Originally Posted by Myung

I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start.

This is a maximum problem, you know the width is $2x$ and the heigth is $3\cos(x)$ since cosine is symetric

So you need to maximize

$A=6x\cos(x)$
June 25th 2008, 05:18 PM
Myung

Quote:

Originally Posted by Chris L T521

The Area of the rectangle is $A(x)=Base*Height=6x\cos(x)$

Are we supposed to find the value of x that maximizes the area?

Thus, $\frac{dA}{dx}=-6x\sin(x)+6\cos(x)$

Setting this equal to zero, we get that $\cot(x)=x$ . Using your calculator, we see that this is the case for $x=\pm.86 \ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

We take the positive value.

Thus, the area of the rectangle is $A(.86)=6(.86)\cos(.86)\approx \color{red}\boxed{3.37}$ .

I believe this is the answer.

Hope that this makes sense to you! :D

--Chris

I think this might be it, not completely sure but i'll check in class. Thanks