Differentiation / Volume Problem (coffee pot)

[paulxzt]

A coffee pot has a shape of a cylinder with radius 5. Let h be the depth of the coffee measured in inches, where h is a function of time t, measured in seconds. The volume V of coffee in the pot is changing at the rate of -5(pi)sqrt[h] cube inches per second. Vcylinder = pir^2h

1. show that dh/dt = ( -sqrt[h] )/ 5

This seems so easy but for some reason I keep getting -sqrt[h] over 2.
Am I supposed to use product rule? (Either way I get the wrong answer)
or
dV/dt = -5(pi)sqrt[h] = 2pir(dh/dt)
and when I solve for dh/dt I keep getting -sqrt[h] divided by 2.

Any help, thanks.
I'm probably making a stupid mistake.

2. At what time t is the coffee pot empty? do I set h = 0 ?

thank you

 

[galactus]

You're correct, paulxt, you need the product rule:

\(\displaystyle \L\\\frac{dV}{dt}={\pi}[r^{2}\frac{dh}{dt}+2rh\frac{dr}{dt}]\)

\(\displaystyle \L\\\frac{dV}{dt}={\pi}[(5)^{2}\frac{dh}{dt}+0]\)

dr/dt is not changing, so we have 0.

Now you can see:

\(\displaystyle \L\\\frac{-5{\pi}\sqrt{h}}{25{\pi}}=\frac{-\sqrt{h}}{5}=\frac{dh}{dt}\)


For part b, are you sure you don't have an initial condition, say, at t=0, the height h is 12 inches or something like that.

 

[paulxzt]

You're correct, paulxt, you need the product rule:

\(\displaystyle \L\\\frac{dV}{dt}={\pi}[r^{2}\frac{dh}{dt}+2rh\frac{dr}{dt}]\)

\(\displaystyle \L\\\frac{dV}{dt}={\pi}[(5)^{2}\frac{dh}{dt}+0]\)

dr/dt is not changing, so we have 0.

Now you can see:

\(\displaystyle \L\\\frac{-5{\pi}\sqrt{h}}{25{\pi}}=\frac{-\sqrt{h}}{5}=\frac{dh}{dt}\)


For part b, are you sure you don't have an initial condition, say, at t=0, the height h is 12 inches or something like that.

Oh on a previous question it said given that h = 17 at t= 0. Given this, would solve for h and then set V = 0 ?

 

[skeeter]

\(\displaystyle \L \frac{dh}{dt} = -\frac{\sqrt{h}}{5}\)

seperate variables ...

\(\displaystyle \L \frac{dh}{\sqrt{h}} = - \frac{dt}{5}\)

integrate ...

\(\displaystyle \L 2\sqrt{h} = -\frac{t}{5} + C\)

you have an initial condition ... h(0) = 17, now determine h as a function of t ... then solve for t when h = 0.