Topic awaiting preservation: math problem (Page 1 of 1)

gah! i cannot seem to solve this problem; it's giving me much trouble. i know conceptually, kind of, what i want to do, though:

On triangle (YSP) y parallels the y-axis. Side S is decreasing at a rate of -.5 ft/sec. Find the rate of vertical + horizontal change at (x, y), i.e. the uppermost end of side P, when Y = 20.



I think I need two seperate equations. I would need to divide the triangle into two seperate triangles from point (x, y) at y = 20. Then I would differentiate the equations to compute (x, y). However it doesn't work. I cannot seem to logically fit it together.

Some help would be nice. I'm not requesting anyone to solve the equation, but it'd be nice to be pointed in the right direction.

Hmm, sounds like a related rates problem... I think we had a thread (started by yannah?) on those a while back; that might be a good starting place.

I'm sure this is doable but it looks like it might take some time to work out.

I can't find that thread...here's what I have so far. I assume it must be an easy problem because it's only like a fifth of the way through my AP calc book.

If the line h from point (x, y) splits the triangle into two parts y1 and y2 at y=20, two triangles remain. One on top described by (y2^2)+(h^2)=(s^2) and one on the bottom described by (y1^2)+(h^2)=(s^2).

The derivative of the upper triangle is 2*(sqrt(1200))*(dh/dt) = -1 ft/sec
the derivative of the lower triangle is 2h*(dh/dt) = (2p)*(dp/dt)

I don't get how these two relate. Since I need to find (x, y) I think I need the actual variable h and the variable y1 in the two equations. This is where I get confused.

Isn't this question about Optimisation?

------------------------------------------------------------------------------------------------------------
Deviations | 5464 | My Poetry Cell | My Own Domain | Xanga| Support and advice needed. Now!
------------------------------------------------------------------------------------------------------------
| "The past will always attack the present with the pain of your memories." - Seiichi Kirima |
---------------------------------------------------------------------------------------------------------

Aha. Well I just took a break from staring at it, spent two days getting a concussion, a bottle of vicodin, and the best of the Vaselines album. Then i returned and saw how incredibly easy it was.

I just renamed side y to b (for building) and p to l (for ladder). Because that's what the problem described and I have a problem with consitency. I don't like it. Instead of using two of the pythagorean theorums, I just used a cosine law because that related all there sides.

(S^2) = (L^2) + (B^2) - (2*B*L*cos x)

(S^2) = - 2BL (Y/L) L^2 and B^2 are constants, so they can go away, and cos x = Y/L. Y is now half of the original y, which is 20. (half of B)

2s(ds/dt) = -2B (dy/dt)

.5 ft/sec = dy/dt (YAY!)

and then substitute into (X^2) + (Y^2) = (L^2) because this relates dy/dx, dx/dy, and dl/dy

2y(dy/dt) + 2x(dx/dt) = 0 because L is a constant

2(20) + 2*20*sqrt(3)*(dx/dt) = 0

which simplifies to dx/dt = sqrt(3)/6 (YAY!)

(Edited by cfb on 11-02-2005 05:26)

(Edited by cfb on 11-02-2005 05:26)