Check out some similar questions!

galactus · Dec 28, 2006, 10:05 AM

"Find the value of k so that the areas of 'area 1' and 'area 2' are equal".

Capuchin · Dec 28, 2006, 10:32 AM

Interesting :)

You've labelled pi in the wrong place I believe, it should be on the x axis.

I'm too busy to work on this :(

galactus · Dec 28, 2006, 10:38 AM

Yes, DUH! thanks for the heads up. I will try to fix

asterisk_man · Dec 28, 2006, 02:44 PM

so we need to find k such that:
$\int_0^{x_1} {k-\sin x \;dx}=\int_{x_1}^{x_2}\sin x -k \;dx$
with x1 and x2 being the first and second intersection points. The RHS is a mirror image around pi/2 and x1 is arcsin(k) so we can have
$\int_0^{\arcsin k} {k-\sin x \;dx}=2\int_{\arcsin k}^{\frac \pi 2}\sin x -k \;dx$
If we integrate we get:
$kx+\cos x \mid_0^{\arcsin k}=2*\left(-kx-\cos x\mid_{\arcsin k}^{\frac \pi 2}\right)$
If we finish the integration we get:
$\left(\cos \left(\arcsin k\right) + k \arcsin k\right) - 1 = <br /> 2*\left(\left(\frac {\pi k} 2\right) - \left(\cos \left(arcsin k\right) + k \arcsin k\right)\right)<br />$

This is all I have time for now and my integration skills are _VERY_ rusty so someone else should go over what I have. The next step seems to be to solve for k... I'm really not sure how that would be done. Maybe someone else has some suggestions or a better route to the answer.
Good luck!

galactus · Dec 28, 2006, 05:42 PM

Thanks for the post, Asterisk Man.

Here's my method:

If we let (a,k), where $\frac{\pi}{2} < a < {\pi}$ , be the coordinates of the intersection point of y=k and y=sin(x).

Then we have k=sin(a) and if 'area 1' and 'area 2' are equal:

$\int_{0}^{a}(k-sin(x))dx=\int_{0}^{a}(sin(a)-sin(x))dx=a*sin(a)+cos(a)-1=0$

I used Newton's method to find that a=2.331122.

Therefore, k=sin(a)=0.724611.......

asterisk_man · Dec 28, 2006, 06:57 PM

galactus:

I've used your value for k in my original equations just to verify that I started at the right place and I do get two equal areas, about 0.276433 square units.

I like the fact that you integrated the whole range, 0 to the second intersection, and let everything cancel and equal 0. and saw that k must equal sin(a). I'm going to look over your work and see if I maybe can replace some of my arcsin(k) with some a to get a reasonable answer.

Also, I'd like to point out that $a*\sin a + \cos a -1 = 0$ has another solution for a which is not a valid answer. Namely, a=0. Obviously if a=0 then k=0 and the two areas are not equal.