## Wednesday, May 8, 2013

### Applications of Differential Equations (3): Ordinary Word Problems

By far the lion's share of solving DEs for 1st year students has to do with solving more or less prosaic word problems. In this blog we cover a few of them, then turn blog readers loose to try to solve a few on their own.

Here's one type that's very common to do with rates and concentrations:

A 100 gallon tank is full of pure water. Let pure water run into the tank at the rate of 2 gals/ min. and a brine solution containing 1/2 lb. of salt run in at the rate of 2 gals/min. The mixture flows out of the tank through an outlet tube at the rate of 4 gals/min. Assuming perfect mixing, what is the amount of salt in the tank after t minutes?

Solution:

Let s be the amount of salt in the tank in pounds at time t. Then:

s/ 100 = concentration of salt (i.e. as a proportion of total gallons of pure water in tank initially)

Then: ds/ dt = net rate of change =  (rate of gain in lbs/min -   rate of loss in lbs/min)

We can further write:

ds/dt = 1 -  4s/ 100 = 1 - s/25

Writing the basic differential equation to solve: ds/ (25 - s) =

dt/ 25

This requires integrating both sides:

ò s o  ds/ (25 – s) =   ò t o  dt/25

Note the integration is taken from 0 to s on the left side and from 0 to t on the right. This leads to:

ln(25 – s)  - ln (25) = - t/25

And finally:

s =  25 (1 – e –t/25)

Let's take a time t = 25 minutes, what do we get?

s =  25 (1 – e –25 /25)=  25 (1 – e -1) = 25 (1 – 0.3678) = 25(0.6322) = 15.8 lbs.

Another problem, this one to do with a spring.

A block of mass m = 2.0 kg rests on a smooth horizontal surface attached to a spring. The spring has the property that it is stretched 0.05 m by a force of 10 N. If the block is displaced 0.05 m from the equilibrium position and released, find: the frequency and period of the motion.

Solution: We will apply Newton’s 2nd law of motion, so:

F = ma = -kx where k is the spring constant.

Since F = 10N when x = 0.05 m, then:

k = F/x   =  (10 N)/ (0.05m) = 200 Nm-1

Now rewrite the force balance equation ma = -kx:

m(d2x/dt 2) = - kx  or:     m(d2x/dt 2)  +  kx  = 0

Divide through by the mass, m:

(d2x/dt 2)  +  (k/ m)  x  = 0

Which is the basic equation for a simple harmonic oscillator:

(d2x/dt 2)  +  w 2  x  = 0

Where: w =  Ö(k/ m)  =  Ö(200 Nm-1 / 2 kg) =

Ö (100 Nm-1 /kg)  = 10 s-1   or 10 radians/ sec

The frequency, f is related to  angular frequency w :

w = 2 pf  so:

f = w / 2 p  =  (10 s-1  ) / 2 p   = 5/p  s-1

The period T = 2 p/w = 2 p/(10 s-1 ) = p/5 s = 0.63 s

Another problem, somewhat harder:

A sled and its occupants weigh 1,000 lbs. It is coasting down a  5o 45’ incline. The force of friction is 40.2 lbs. and the air resistance (drag) is at any given time 2 times the velocity in feet per second. Find an expression for the velocity after t seconds from rest and the specific velocity after 10 secs.

Recall: Weight = mass x g

And take g (acceleration of gravity) to be 32.17 f/s/s

The force exerted by the pull of gravity is vertically downward (see if you can sketch a free body diagram with forces acting)  and we may write:

F = mg sin q  = 1000 sin 5o 45’

= 100.2

The total force acting is therefore:

F = 100.2 - 40.2 - 2v = 60 - 2v

WHY is this? We have two negative contributions (40.2 and 2v) on the LHS because the force of friction and drag both act opposite to the direction of motion. '2v' because the drag is stated as 'twice the velocity'.   The starting DE becomes:

(1000)/ g (dv/dt) =       60- 2v

Re-arranging to separate variables:

dv / (30 - v)  =  (32.17) dt/ 500  = 0.06434 dt

For which the solution is obtained by integration, i.e.

ò v o  dv/ (30 – v) = ò t o  (0.06434) dt

Or:

ln (30 - v)  - ln 30 =  -0.06434 t

Taking natural logs of each side:

(30 - v)/ 30 =  e –0.06434t

v   =  30 (1 -  e –0.06434t  )

After 10 seconds:  v = 30 (1 -   e –0.6434 ) = 30 (1 -0.5255) = 30 (0.4745)

v(10) = 14.24 feet/sec

Problems for the Math Maven:

1) A ship weighing 64,000 tons starts from rest under the impetus of a constant propeller thrust of 200,000 lbs.  The resistance of the water is 10,000v lbs.

Find the velocity v as a function of the time and the terminal velocity in miles per hour (let g = 32 f/s/s)

2) A 5 lb. weight hangs vertically on a spring whose spring constant k = 10. The weight is pulled 6 inches farther down and released. Find the equation of motion, its period and the frequency. What would the units be for k?

3) A tank contains 200 gallons of brine with 100 pounds of dissolved salt in solution. Salt water containing 1 pound per salt per of gallon water enters the tank at the rate of 5 gallons per minute and the brine flows out at the same rate.

Find the amount of salt left in the tank at the end of one hour.