# calculus

posted by
**Freddy**
.

Please check this for me:

the lenght l, width w, & height h of a box change with time. At a certain instant the dimensions are l=1m & w=h=2m. L and W are increasing at a rate of 2m/sec while H is decreasing at a rate of 2m/sec. At that instant, find the rate at which the surface area is changing.

Heres what I got:

We want the derivative of surface area with respect to t (ds/dt)...

ds/dt= 2y*dx/dt + 2x*dy/dt + 2z*dy/dt+ 2y*dz/dt + 2x*dz/dt + 2z*dx/dt

=2(2m)(-3) + 2(1m)(2) + 2(2m)(2)+ 2(2m)(2)+ 2(1m)(2)+2(2m)(-3)

=-12m+4m+8m+8m+4m+12m

=-24m+24m

=O m^2/sec

Is this right??