# 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

Multiply both sides of equation by dt

Antidifferentiate both sides

Multiply by (1/2)

Square both sides of equation

**Answer to Part a) y = [ (-k/2)t + C ]²**

Part b)

Use y(0) = 9 to solve for C by substituting it into the equation

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

Take the square root of both sides

Solve for k

Multiply by -1/10

1/10 = k

k = 0.1

y = [ (-0.1/2) t + 3 ]²

y = [ (-0.05)t + 3 ]²

**Answer to Part b) y = [ (-0.05)t + 3 ]²**

Part c)

**dy/dt = -(1/10) √ (y)**

Let dy/dt equal -0.1 and solve for y

Multiply by -10

Square both sides

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.**

**Answer to Part c) When t = 40 minutes**

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