When an object falls under the influence of gravity and air resistance, the gravitational force is downward in the resistance is upward.

• The air resistance depends in the velocity v with which the object is falling.
• We might assume an air resistance in some sort of proportion to the velocity.

• We denote the force of air resistance by F, and we note the possible proportionalities F = k v, k v^2, k v^3.

Assuming that F = - k v, which as it turns out is reasonably accurate for a smooth object which isn't falling very fast, we can represent the net force as Fnet = m g - k v.

• In this expression we are implicitly assuming that the force of gravity and therefore the velocity is in the positive direction; since the force of air resistance is in the opposite direction this force must therefore be negative.
• As usual we combine the expression for net force with Newtons Second Law in the form Fnet = m dv/dt.

The resulting differential equation is indicated in the first line below.

The proceed to separate the variables and solve the equation in the usual manner.

Our result is v = m g / k - C e^(-kt/m), where C is an arbitrary nonzero constant.

We see immediately then as t increases, v will eventually approach v = mg / k as an asymptote.

• This asymptote is called the terminal velocity.

The value of C can be determine from the initial velocity of the object.

• For example if the initial velocity is zero, then we have v(0) = 0, which gives us C = m g / k.
• This solution yields velocity function v(t) = mg / k ( 1 - e^(-kt/m) ).

`00

A graph of this solution shows the asymptote approach of the velocity to the terminal velocity mg / k.

• The 'half-life' of this approach, i.e., the time required for velocity to become twice as close to terminal velocity, is the same as the helf-life of the exponential function e^(-kt/m).
• Recall that this half-life is obtained by setting e^(-kt/m) = 1/2 and solving for t.

If we have an initial velocity v(0) = 1.1 mg / k , then setting v(0) = 1.1 we would obtain C = .1 mg / k .

• The resulting function would be v(t) = m g / k ( 1 - .1 e^(-kt/m) ).
• The graph of this function shows how we start closer to the asymptote than for initial velocity 0 (i.e., 1.1 m g / k is closer to m g / k than is 0) but still approach the asymptote with the same half-life.

The graph below shows the entire family v(t) = m g / k - C e^( - k t / m) = m g / k (1 - C e^(-k t / m ) ).

• The value of C depends on the initial velocity, with ( 1 - C ) m g / k equal to the initial velocity.
• The v intercept of the graph will therefore represent the initial velocity.

Note that the proportionalities F = k v^2 and F = k v^3 are more nearly appropriate for most following objects, and that they yield differential equations which are also easily separated.

• The resulting integrals are a little more work to evaluate, but are well within the range difficulty of integrals encountered in this course.
• The solutions we get for these proportionalities are not exponential in nature.

`01

In the figure below we consider a lake with volume .8 km^3, with water flowing in from two sources.

• One source carries 30 kg of a pollutant per km^3 and flows at .00001 km^3 / day, while the second source carries unpolluted water at .0002 km^3/day, 20 times the flow rate of the first.

Water flows out of the lake at .00021 km^3/day, which is the same as the rate of inflow, so that the amount of water in the lake remains constant.

We wish to set up an equation involving the amount P of pollutant as a function of time t.

• We know the rate at which pollutant flows into the lake, and we hope we can determine some relationship involving P, t and the rate at which pollutant flows out.

• In this case we will be able to obtain an equation for the rate dP / dt at which the amount P of pollutant in the lake changes.

• We easily see that the inflow rate is equal to the product of the rate at which polluted water flows into the lake and the concentration of pollution in that water.

• We therefore see that the inflow rate for pollutant is .0003 kg / day.

The outflow rate will depend on the .0021 km^3/day rate at which water flows from the lake and on the concentration, or density, of pollutant in the lake.

• The concentration of the pollution is equal to the amount of pollutant divided by the volume of the lake, which is P / .8 km^3.
• We therefore obtain the differential equation in the figure below.

`02

The differential equation is solved by the usual means.

• We obtain P = 1.2 - C e^(-.00036 t).
• This equation can also be expressed as P = 1.2 (1 - C e^(-.00036 t). Note the similarity with the solution to the terminal velocity problem.
• Our solution indicates a limiting concentration of 1.2 kg / km^3.
• We note that the half-life of e^(-.00036 t) is a couple of thousand days (solve e^(-.00036 t) = 1/2 to find an accurate figure)--not quite a decade but on that order.  

`03

Using DERIVE to obtain a power-series solution for certain integral involved with a differential equation, we obtain the terms indicated below.

We see that the ratio of coefficients changes in an irregular manner, which is fairly typical for the first several coefficients of a power series.

• For the given terms the ratios of coefficients appear to have a tendency to increase faster and faster.
• This would lead to an infinite limit of the ratios, which would apply radius the convergence R = 0. (Recall that the radius of convergence is the reciprocal of the limit of the ratios of coefficients).

On the other hand it is certainly possible that the ratios will 'settle down' and approach limit of, say, 1.

• The figure shows how, within the radius of convergence, the power series will tend to approximate the actual solution more and more accurately as the number of terms of the series increases, while still divergent outside the radius of convergence.

`04

One of the most common occurrences in the natural world is the tendency for objects to be restored to their equilibrium state.

• This is the case in a wide variety of physical, biological and chemical systems.

One of the easiest situations to understand is that of a pendulum, which for displacements x which are small compared to the length of the pendulum experiences a restoring force which is proportional to the distance from equilibrium.

• Most naturally occurring tendencies to restore equilibrium are, at least in first approximation, linear in nature, so this model has widespread applicability.

Specifically, for small displacements x, a freely swinging pendulum will experience a net force Fnet = - k x, where k is a constant of proportionality.

• The - sign in front of the k indicates that the force at position x is directed back toward the equilibrium position, in the direction opposite to the displacement x.
• Using Newton's Second Law, as indicated below, we obtain the differential equation in the third line.
• This differential equation is slightly simplified in the last line, so that the second derivative term has coefficient zero.

We cannot simply separate this differential equation, as we have been used to doing. This is because we have a second derivative.

• We could attempt a power-series solution, which is often what we do when we have no other method to use. However, to develop the techniques of power series solutions would be beyond the scope of this course.
• Before attempting a power series solution, in any case, we will usually attempt to find a solution of another type.

Common types of trial solutions include polynomial and exponential functions.

• Substituting this function for x(t) in the equation we obtain the equation in the third line, which we easily solve for c.
• The only problem with our solution is that, since k and m are both positive, c is imaginary.
• It turns out that imaginary solutions work just fine, the we need something called Euler's formula to make sense of such solutions.

`05

The problem we ran into the previous attempt was that we need a function whose second derivative is a negative multiple of the function.

• Exponential functions have second derivatives which are positive multiples of themselves.
• However, sine and cosine functions are easily see to have derivatives which are negative multiples of themselves.

We therefore attempt trial solution indicated in the second line below, using both sine and cosine functions.

• We obtain an equation which we easily solve for c, obtaining c = `sqrt( k / m ).
• Since c is a multiple of clock time t, it represents the angular frequency of the sine or cosine function.
• We therefore let `omega = c = `sqrt(k / m) and write the solution has in the last line below, making a clear that the solution has a consistent angular frequency.

The constant angular frequency tells us that in any system where the net force is the linear restoring force F = - k x acting on a mass m, we have an oscillatory solution with a constant angular frequency `omega = `sqrt( k / m ).

• If we think of the force F has the tendency to restore a system of any type to its equilibrium situation, and m as the 'inertia' which resists the restoring tendency, this model can be generalized to a wide variety of systems encompassing all the sciences.