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.