YAMT: Related Rates

[speg]

Eh, I know this is getting to be a daily occurence. *shrug*

A cylindrical tank with height 3m and diameter 1 m is being filled with gasoline at a rate of 5 L/min. At what rate is the fluid level in the tank rising?

So, we know:

dv/dt = 5L/min.
V=Pi * r^2 * h

Now here's where I get a little confused. I want the rate of heights change, so I have to find r in terms of h to make this work right?

~~~ pi * h^3
V= ___________ <----- This could be where I'm going wrong.
~~~~~ 36




So we take the derivative of that....


d/v = ( pi * h^2 /12 ) * dh/dt
5 = (9pi/12) * dh/dt

dh/dt = 2.12

But apparently the answer is supposed to be 0.64cm/min
😕:frown:

 

[bolomite]

so we have dV/dt = 5 L/min, and V = pi * r^2 * h

want dh/dt: using chain rule, dh/dt = (dh/dV)(dV/dt)

rearrange volume formula for h: h = V/(pi * r^2), then differentiate w/ respect to V

----> dh/dV = 1/(pi * r^2) , and dV/dt is given

Careful with units -- when you do conversion, 1 Liter = 1 cubic decimeter (dm^3), and then you should be able to convert to cm