<p>There is a rectangular solid. The side walls are moving apart at a rate of .1 m/sec but the end walls are moving together a rate of .3 m/sec. The volume of the liquid inside is 20 cubic meters, a constant.
a) Write an equation expressing the rate of change of the depth of the liquid in terms of the width and length of the region inside the compactor.
b) When the side walls are 5 m long and the end walls are 2m long, Is the depth increasing or decreasing at what rate?</p>
<p>that's calc...</p>
<p>Hmmm...I don't really remember this from calc but here is my guess:</p>
<p>x=side distance y=end distance</p>
<p>A=xy
A'=(dx/dt)y+(dy/dx)x</p>
<p>(dx/dt)=.1m/s
(dy/dt)=-.3m/s
A'=(.1m/s)(2)+(-.3)(5). From this, Area is changing at rate of +.05 m^2. This means the depth is decreasing.</p>
<p>V=Ad 20=10d d=2 A = 10
v'=(dA/dt)d+(dd/dt)A
0=(.05)2+(dd/dt)10</p>
<p>dd/dt=-.01 m/s if I did everything correctly</p>
<p>bump im curious about how to get the answer</p>
<p>sat ii math doesn't give you part a and b</p>
<p>yeah, there isn't any calculus in SAT 2 math 2.</p>
<p>Dsmo:
I think your answer needs to be tweaked a bit.</p>
<p>A'=(.1m/s)(y)+(-.3)(x), which =(.1m/s)(2)+(-.3)(5) when y=2, x=5.</p>
<p>However, the formula for depth = V/A = 20 / (xy)
d' = 20(-1)(1/xy)^2 (A')
= -20 (0.1 y - 0.3x) / (xy)^2</p>
<p>For part (2), evaluate d' when x=5, y=2:
d'(x=5,y=2) = -20( 0.2 - 1.5) / (100) = 26/100 = 0.26 m/sec</p>
<p>SAT II's don't have free response. Are you sure this isn't from an AP test?</p>
<p>This was a practice test from my school for preparation for Math IIc. I guess they wanted me to apply many concepts all at once.</p>
<p>tell your school that their practice problems dont apply to the math2c.</p>
<p>Yeah I messed up some addition there. Optimizerdad has the correct answer.</p>
<p>This definitely looks like an AP test question though.</p>
<p>aren't you forgetting the denominator and also the zero.
For example: the numerator: 0-20(-109xy0^-2 (derivative of A)
Denominator: (xy)^2</p>
<p>should be (-10)(-1) and so on</p>
<p>I'm treating the formula as d= 20 (A^-1), where A=xy</p>
<p>So d' = 20 (-1)(A^-2) A'
= -20 (1/xy)^2 (0.1y - 0.3x)</p>
<p>which does away with having to remember all that nasty denominator stuff...</p>