# Related rates help :/

1. Mar 27, 2007

### joe32725

1. The problem statement, all variables and given/known data

2. Coffee is draining from a conical filter into a cylindrical coffee pot at the rate of 10 in3/min.

a) How fast is the level in the pot rising? ____________

b) How fast is the level in the cone falling when the level in the cone is 5 in.? _________

2. Relevant equations

i know i need to use the volume formulas for a cylinder and a cone, but i dont know how to differentiate them. heres the wrong answers i got:

V= pi(r^2)h --> d/dt[V]= pi(2r)(dh/dt)--> dV/dt= 2(pi)(r)*dh/dt
dh/dt= (1/(2(pi)(r)))*dV/dt

dV/dt= 2/3(pi)(r)(dh/dt)
3/(2(pi)(r))*(dV/dt)

3. The attempt at a solution

when i differentiate both sides with respect to time i get:

d/dt[V]= 2/3(Pi)(r)*(dh/dt) ---> dV/dt= 2/3(pi)(r)(dh/dt)

dh/dt= 3/(2(pi)(r))*(dV/dt)

dh/dt= 3/(2(pi)(r))*(10in^3/min)

30/2(pi)r = 15/(pi)(r) in^3/min

p.s. this is really getting frustrating. if calc 1 is throwing me curveballs like this that i cant solve, my quest to be a mechanical engineer seams very bleak.

2. Mar 27, 2007

### HallsofIvy

Staff Emeritus
Okay, that's how fast the height of the coffee in the pot is increasing.

?? $V= (1/3)\pi r^2 h$. You seem to have differentiated r2 with respect to r but then multiplied by dh/dt rather than dr/dt. Using the product rule
[tex]\frac{dV}{dt}= (2/3)\pi r h \frac{dr}{dt}+ (1/3)\pi r^2 \frac{dh}{dt}[/itex]

No, as I said above, you you have to use the product rule. Remember that the cone formed by the water always has the ratio of r to h that the filter has. Are you given the height and radius of the filter cone in the problem?

You need to know the ratio of radius to height of the original filter cone and you haven't given that here.

Last edited: Mar 27, 2007