#25--The volume of water in the pool is 6*cross sectional area of the water in the pool.  Since this cross section is a triangle similar to a right triangle with legs 12 and 2 we can get the volume of water as a function of the depth.  (This will of course only hold until the depth reaches 2 meters, but that is not of concern as we are only interested in the rate of change of the depth when the depth is 1 meter).

#29--Draw a perpendicular from the point (x,y) to the building.  Then we will have x² + (12 - y)² = s².  We can eliminate x from this relation by noting that we also have x² + y² = 12 ²

#37--This is a nice mechanical movement.  x moves to the left and right as given by x(t) = sin(t/6).  As x moves y will move up or down depending on the position of x and its direction of movement.  Since x² + y² = 1², and we have a formula for x we can differentiate to get a relation involving the derivative of y.

#39--v(t) = (4/3)r(t)³ which we can differentiate with respect to t.  But this derivative will be equal to a constant times the surface area 4r² because we are given that the rate of change of volume is directly proportional to the surface area.

#43--Let h(t) be the height of the balloon at time t.  Then we have               h(t)/30=tan(t) where (t) is the angle of elevation from the observer to the balloon.  Now differentiate with respect to t.

I hope that these hints help.

                  --CFW