Differential Equations and Rates

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When the valve at the bottom of cylindrical tanik is opened, the rate at which the level of liquid in the tank drops is proportional to the square root of the depth of the liquid. Thus if y(t) is the liquid's depth at time t, in minutes, after the valve is opened, water drains from the tank according to the differential equation dy/dt = -k(Square Root of Y) for some positive constant k that depends on the size of the drain.

A. Find a general solution for the differential equation.
B. Suppose that y(0) = 0 feet and y (20) = 4 feet. Find an equation for y(t)
C. At what time is the water level dropping at a rate of .1 feet per minute?

Here's what I have so far, am I on the right track?

A. Square Root of Y = -dy/dt + 1/k y = (dy/dt * 1/k) /\2
B. 9 = (-k SquareRootof 9) /\2 * 1/k/\2
C. I have NO idea

Help please?
 
\(\displaystyle \L\\\frac{dy}{dt}=-k\sqrt{y}\)

\(\displaystyle \L\\\frac{dy}{\sqrt{y}}=-kdt\)

\(\displaystyle \L\\\int{\frac{1}{\sqrt{y}}}dy=-k\int{dt}\)

\(\displaystyle \L\\2\sqrt{y}={-}kt+C\)

\(\displaystyle \L\\y=\frac{(kt-C)^{2}}{4}\)


This looks like Torricelli's law \(\displaystyle A(h)\frac{dh}{dt}=-k\sqrt{h}\)
 
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