Rate of change: soda?

Question

A soda glass has the shape of the surface generated by revolving the graph of y=6x^2 for 0<=x<=1 about the y-axis. Soda is extracted from the glass through a straw at the rate of 1/2 cubic inch per second. How fast is the soda level in the glass dropping when the level is 3 inches? (Answer should be implicitly in units of inches per second. Do not put units in your answer. Also your answer should be positive, since we are asking for the rate at which the level DROPS rather than rises.)

Any suggestions?


Thanks

Answer 1

y=6x^2, When x = 0, y = 0 When x = 1, y = 6

Since rotating about the x-axis
V = π∫(from y = 0 to h) x² dy
= π∫(from y = 0 to h) y/6 dy
= πh²/12

When h = 6 V = π*36/12
So volume in glass (at time t and height h) is given by

V = π/12 * (36 - h²); 0 ≤ h ≤ 6

So dV/dh = π/12 * (-2h)
= - πh/6

Now dV/dt = dV/dh * dh/dt
- ½ = - π*3/6 *dh/dt
dh/dt = 1/π

So the level is dropping at 1/π (in/s)

Answer 2

First generate the equation for the volume of the glass as a function of x.

I think that you solve dV/dx = dV/dy*dy/dx for dV/dy :

dV/dy = dy/dx*dV/dx

dy/dx comes from your curve, dV/dx comes by differentiating the V(x)

Btw, you want the reciprocal to get dy/dV, then dy/dt = dy/dV*dV/dt where dV/dt is given( 0.5 ft^3/sec).