Question 3
The use of a graphing calculator is allowed.
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, in feet, at time t minutes after the valve is opened, water drains from the tank according to the differential equation 
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) = 9 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 0.1 feet per minute?
Solution
Part a)
Multiply both sides of equation by 1/√y
(1/√y) dy/dt = -k
Multiply both sides of equation by dt
(1/√y) dy = (-k) dt
Antidifferentiate both sides
2√(y) = (-k)t + C
Multiply by (1/2)
√(y) = (-k/2)t + C
Square both sides of equation
√(y)² = [ (-k/2)t + C ]²
Part b)
Use y(0) = 9 to solve for C by substituting it into the equation
9 = [ (-k/2) (0) + C ]²
Square root both sides
(+/-) √9 = √[ (-k/2) (0) + C ]²
Must take the positive square root of 9 because their cannot be a negative value in terms of feet
3 = C
y = [ (-k/2)t + 3 ]²
Substitute the value of y(20) = 4 into the equation to solve for k
4 = [ (-k/2) (20) + 3 ]²
Take the square root of both sides
√4 = √ (-10)k + 3 ²
2 = (-10)k + 3
Solve for k
-1 = (-10)k
Multiply by -1/10
1/10 = k
k = 0.1
y = [ (-0.1/2) t + 3 ]²
y = [ (-0.05)t + 3 ]²
Part c)
dy/dt = -(1/10) √ (y)
Let dy/dt equal -0.1 and solve for y
(-0.1) = -(1/10) √y
Multiply by -10
(-0.1) (-10) = √y
1 = √y
Square both sides
1² = (√y)²
1 = y
Use the value of y = 1 to solve for t, while using the equation y(t)
y = [ (-0.05)t + 3 ]²
1 = [ (-0.05)t + 3 ]²
Take the square root of both sides
√1 = √ [ (-0.05)t + 3 ]²
1 = (-0.05)t + 3
-2 = (-0.05)t
Multiply by (-1/0.05)
(-2)(-1/0.05) = t
t = 40 min.