Rates Of Change

Question

A conical vessel with a base radius 3 cm and a vertical height of 20 cm is arranged with its vertex facing downwards. Water is being added at a constant rate of 1 cm^3/s and leaks away through a small hole in the vertex at a constant rate of 1.5cm^3/s. Find the rate of change of the depth when the depth is 8 cm.

Answer 1

Place the cone with its vertex at the origin; coordinate x is the radius, y is the height; V = (pi x²y)/3 

3/20 = x/y  so x = 3y/20

 

The problem gives dV/dt as -0.5 cu cm per sec (net loss)

 

dV/dt = d/dt {(pi/3)•y[(3y)/20]²} = 9pi/400 y² (dy/dt) 

 

Now plug in y = 8 and dV/dt as -0.5 and solve for dy/dt -- CAREFULLY!  Answer will be negative, of course, and is units are cm per second.

 

Also...check my work, too...all this tricky typing could have been a distraction.