Question

can anyone please help me with this? Water tank rate problem? How to represent represent h as a function of t if dV/dt = 50 = pi(h)^2(k)^2dh/dt? (Water is flowing at a rate of 50m^3/min, the cone shaped tank is 10m in height, and the diameter is 40m.)

Answer 1

The dV/dt = 50 = pi(h)^2(k)^2dh/dt is something that someone else helped me get. The question for this review problem are: 1. A. Write an expression for the rate of change of the water level with respect to time, in terms of h (the water’s height in the tank). 2. B. Assume that, at t = 0, the tank of water is empty. Find the water level, h, as a function of the time t. 3. C. What is the rate of change of the radius of the cone with respect to time when the water is 8 feet deep?

Answer 2

I have some work on the problem. but I'm pretty much sure that it's going no where and I don't know what to do.

Answer 3

what does the "k" refer to?

Answer 4

The was from some the work that someone else on this site tried to help me with. The idea was: I can derive a formula from the V = 1/3pi(r)^2h, and also on the idea that the radius, r =kh, where k is my constant for the angle of the cone. Meaning that it would plugged into the formula as V = 1/3pi(h)^3(k)^2. If I write it to represent the rate of change of water level, it becomes dV/dt = 50 = pi20^2(10)(h)^2dh/dt.

Answer 5

Or something like that... I mean I tried to work it further, but I think I'm jsut getting further from the solution :(

Answer 6

lets start with volume for cone:
\[V = \frac{1}{3}h*\pi r^{2}\]
since we want function for height...we need to put r in terms of h

so the relationship between radius and height remains proportional all throughout the cone
\[\frac{r}{h} = \frac{20}{10} = 2\]
--> r = 2h
new volume equation is
\[V = \frac{4}{3} \pi h^{3}\]
set up equation for dh/dt
\[\frac{dh}{dt} = \frac{dV}{dt}*\frac{dh}{dV}\]
differentiate volume equation
\[\frac{dV}{dh} = 4 \pi h^{2} \rightarrow \frac{dh}{dV} = \frac{1}{4 \pi h^{2}}\]
\[\rightarrow \frac{dh}{dt} = \frac{50}{4 \pi h^{2}}\]
Now to solve for h(t) you must solve differential equation
separate variables...get h terms on side with dh, dt on other side
\[h^{2} dh = \frac{50}{4\pi} dt\]
integrate both sides
\[\frac{1}{3}h^{3} = \frac{50}{4\pi} t + C\]
\[\rightarrow h = \sqrt[3]{\frac{75}{2\pi}t + C}\]
i think it said initially h=0...so C = 0