# Thread: Hard Dif EQ Word Prob

1. ## Hard Dif EQ Word Prob

A water tank has the form of a right-circular cylinder. Through a whole in the bottom of the water tank, if water is able to pass through this whole with gravity acting upon it, the height, h, of water at time t, is given by the following Dif EQ:

dh/dt = -c*(A_(h)/A_(w))*sqrt(2*g*h) where A_(w) is the cross-sectional area of the water and A_(h) is the cross-sectional area of the hole. Further, c is the friction/contraction factor of the hole.

1.) Solve the Dif EQ if the initial height of the water is 20 ft and A_(w) = 50 ft^2 and A_(h) = 1/4 ft^2

2.) Suppose c = 1. Determine the time at which the tank is empty.

3.) If the friction/contraction factor, that is, if c = 0.6, how long would the tank take to empty?

2. Originally Posted by Ideasman
A water tank has the form of a right-circular cylinder. Through a whole in the bottom of the water tank, if water is able to pass through this whole with gravity acting upon it, the height, h, of water at time t, is given by the following Dif EQ:

dh/dt = -c*(A_(h)/A_(w))*sqrt(2*g*h) where A_(w) is the cross-sectional area of the water and A_(h) is the cross-sectional area of the hole. Further, c is the friction/contraction factor of the hole.

1.) Solve the Dif EQ if the initial height of the water is 20 ft and A_(w) = 50 ft^2 and A_(h) = 1/4 ft^2

2.) Suppose c = 1. Determine the time at which the tank is empty.

3.) If the friction/contraction factor, that is, if c = 0.6, how long would the tank take to empty?
It is difficult only because it is in a word problem form.

Many are afraid of word problems.

dh/dt = -c*(A_(h)/A_(w))*sqrt(2*g*h)

1.) Solve the Dif EQ if the initial height of the water is 20 ft and A_(w) = 50 ft^2 and A_(h) = 1/4 ft^2

I assume g = 32 ft/sec/sec. (It's supposed to be 32.2 ft/sec/sec.)

dh/dt = -c(1/4 /50)sqrt(2*32*h)
dh/dt = -c(1/200)(8sqrt(h)
dh/dt = -c(0.04)sqrt(h)
dh/sqrt(h) = -0.04c*dt
h^(-1/2) *dh = -0.04c *dt
Integrate both sides, the c is a constant,
2sqrt(h) = (-0.04c)t +C ------------------(i)

Given: when t=0, h=20
So,
2sqrt(20) = 0 +C
C = 4sqrt(5) ------------***

Substitute that into (i),
2sqrt(h) = (-0.04c)t +4sqrt(5)
sqrt(h) = (-0.02c)t +2sqrt5
Square both sides,
h = (0.0004c^2)t^2 -0.08sqrt(5)*c*t +20 -----the equation.

2.) Suppose c = 1. Determine the time at which the tank is empty.

When tank is empty, then h=0.

h = (0.0004c^2)t^2 -0.08sqrt(5)*c*t +20
0 = (0.0004*1^2)t^2 -0.08sqrt(5)*1*t +20
0 = 0.0004t^2 -0.08sqrt(5)*t +20
Using the Quadratic formula,
(the discriminant is zero. Umm...)
t = 0.08sqrt(5) / 0.0008
t = 223.6 seconds ------------the tank is empty.

3.) If the friction/contraction factor, that is, if c = 0.6, how long would the tank take to empty?

Again, when tank is empty, then h=0.

h = (0.0004c^2)t^2 -0.08sqrt(5)*c*t +20
0 = (0.0004*0.6^2)t^2 -0.08sqrt(5)*0.6*t +20
0 = 0.000144t^2 -0.048sqrt(5)*t +20
Using the Quadratic formula,
(the discriminant is zero again. )
t = 0.048sqrt(5) / 0.000288
t = 372.7 seconds ------------the tank is empty.