1. ## Cylindrical Tank Draining

Hi! I'm new here, and studying for a test. I was wondering if someone can help me out.

3.) When the valve at the bottom of a cylindrical tank 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" minutes after the valve is opened, water drains from the tank according to the differential equation dy/dt = -k √y for some positive constant k that depends on the size of the drain.

a) Suppose that y(0) = 9 and y(20) = 4. Find an equation for y(t).

I'm assuming you have to seperate the variables and integrate.

√y dy = -k dt

But I'm not exactly sure how the first bits of information fit in. Some help, please?

b)At what time is the water level dropping at a rate of .1 feet per minute?

Help here, as well.

Thanks a bunch. I appreciate it.

2. dy/dt = -k*sqrt(y)
Clear the fractipon, multiply both sides by dt,
dy = -k*sqrt(y) *dt

Divide both sides by sqrt(y)
dy / sqrt(y) = -k*dt -------------that is what it should be.

y^(-1/2) dy = -k dt
Integrate both sides,
2*y^(1/2) = -k*t +C
2sqrt(y) = -kt +C -------(i)

a) Suppose that y(0) = 9 and y(20) = 4. Find an equation for y(t).

y(0) = 9
meaning, when t=0, y=9
So, into (i),
2sqrt(9) = -k*0 +C
C = 2sqrt(9) = 2*3 = 6

y(20) = 4,
2sqrt(4) = -k(20) +6 -----C=6, remember.
2*2 = -20k +6
4 -6 = -20k
k = -2/(-20) = 0.1 --------------------***

Substitute those into (i),
2sqrt(y) = -kt +C -------(i)
2sqrt(y) = -0.1t +6
Divide both sides by 2,
sqrt(y) = -0.05t +3 ------------------***
Square both sides,
y = 0.0025t^2 -0.3t +9
In function form,
y(t) = 0.0025t^2 -0.3t +9 ---------answer.

b)At what time is the water level dropping at a rate of .1 feet per minute?

dy/dt = -k*sqrt(y)
0.1 = -0.1*(-0.05t +3)
1 = -(-0.05t +3)
1 = 0.05t -3
1 +3 = 0.05t
t = 4 / (0.05)
t = 80 min

Therefore, after 80 minutes, the water level is dropping at the rate of 0.1 ft/min. ----answer.