Suppose that to maintain its weight, a certain type of whale requires 15 kg of plankton per ton of body weight daily.

• It is reasonable to assume that the rate at which the weight of the whale changes is proportional to the difference between the amount of food taken in and the amount needed.
• Based on a ball-park estimate of the nutritional composition of plankton and the energy composition of whale tissue, we assume a proportionality constant of 1 ton / 2000 kg (this is the proportionality between rate of weight change and excess food).

Representing the rate of weight change by dW / dt and the (constant) food intake by I, we write the proportionality as in the first line below.

• The equation is easily solved by separating variables as indicated below.
• The solution shows that the weight of the whale will exponentially approach the limit I / 15.
• Note that the time required for the weight of the whale to 'move' e times as close (i.e., 2.7 times as close) to its limiting value is 2000 / 15 = 133; since t is measured in days, it takes of whale over 4 months to move 2.7 times closer to its limiting weight.
• The time to move twice as close will be the solution to the equation e^(-15 t / 2000) = 1/2; the solution is t = -2000 / 15 ln(1/2), or about 93 days.

For example if the initial weight of the whale is 70 tons and the intake is 1200 kg / day, we could first note that the limiting weight should be 1200 kg / day / [ (15 kg / day) / ton ] = 80 tons; we expect an exponential approach from 70 tons to a limiting weight of 80 tons.

• Using the initial condition W(0) = 70 with I = 1200, we obtain the equation W = 80 - 10 e^(-.0075 t).

Video File #01

In the figure below we depict a curve whose points are (x,y).

• The curve has the property that when the distance BC from the point B where the tangent line at (x,y) meets the x axis to the point C on the x axis directly beneath (x,y) is equal to the square of the distance AC from the point A = (x,y) to C.

We wish to find the function y(x) that describes this curve.

• We observe that since the line is tangent to the curve, then the slope AC / BC must be equal to the derivative dy / dx at the point (x,y).
• Since BC = (AC)^2, we have slope = AC / BC = 1 / AC.
• Since y = AC, the slope is 1/y.
• We thus set dy / dx = 1 / y.

We can easily solve the equation:

• If we separate variables and integrate we easily obtain y^2 = (x + c), so that y = `sqrt(x + c).
• If we choose a curve thorough (0,0), we have c = 0 and y = `sqrt(x).

Video File #02

If Q stands for the quantity of salt in a reservoir from which salt is filtered out at a rate proportional to the amount salt present, with proportionality constant k = .0007, then if salt is added at 4 grams / hour, the differential equation representing the quantity of salt is dQ / dt = -.0007 Q + 4.

• The -.0007 Q expresses the proportionality; the - sign represents the fact that salt is being removed.
• The equation the says that the rate at which the quantity of salt changes is equal to the rate which salt is filtered out plus the rate which salt is added.
• This equation is easily solved by separation of variables.
• The limiting amount of salt will be 4 / .0007 = 5700 grams, and the exponential approach to this limit will be governed by C e^(-.0007 t).