The Calculus Primer (2011)

Part X. Partial Differentiation

Chapter 37. SIGNIFICANCE OF PARTIAL AND TOTAL DERIVATIVES

10—6.Partial Differentials and Partial Derivatives. Some notion of the significance of partial differentials will be obtained from the following examples. Consider the changes in the area of a rectangle caused by variation in the lengths of the sides separately, and when varying simultaneously. Let the rectangle ABCD have a variable base x and a variable altitude y; let its area equal z = xy. Now if we consider x constant while y increases by an increment BP = dy, the corresponding change in z is PQCB = dy; if we consider y constant while x increases by DS = dx, the corresponding change in z is CBSD = dx; and the total differential of z is given by

Moreover, from the equation z = xy, we see that

It should also be carefully noted that the total increment Δz due to increments dx and dy is given by

Δz = x dy + y dx + dx dy;

the small rectangle RQ, whose area equals dx dy, represents the difference between Δz and dz. Thus the total increment and the total differential of a function of two or more variables are not, in general, equal.

EXAMPLE. The total area A of a right circular cone is given by A = πrs + πr², where s = slant height and r = radius of base. Find the rate of change of A with respect to r when s remains constant; the rate of change of A with respect to s when r is constant.

Solution.

A = πrs + πr²;

10—7.The Total Derivative. As we have already learned, the total differential of u = f(x,y) is

if now x and y, and therefore u,are functions of another variable t, we may divide (1) by dt:

This, as we have seen, is the total derivative of u with respect to t; here and may be regarded as derivatives which represent the rates of change of x and y, respectively, with respect to t. In practical applications, the variable toften represents time.

EXAMPLE 1.The base of a rectangle equals 8 inches, and its altitude is 4 inches. At a certain instant the base is increasing at the rate of 3 in./sec, and the altitude at the rate of 2 in./sec. At what rate is the area changing at the same instant?

Solution. Let x = base, y = altitude, u = area.

By equation (2) above:

But x = 8 in., y = 4 in., = 3 in./sec, = 2 in./sec.;

hence = (4) (3) + (8) (2) = 28 sq. in./sec.

EXAMPLE 2.Assume that the formula for a gas is pv = kT, where p = pressure in lb./sq. unit, v = the volume in corresponding cubic units, T = absolute temperature, and k = a constant. If k = 60, and at a certain instant v = 12 cu. ft. and T = 200°, find p at this instant. If at the same instant v is increasing at the rate of 0.5 cu. ft./min., and T is increasing at 0.6 degree/min., at what rate is p changing at that instant?

Solution. Substituting k = 60, v = 12, T = 200 in the formula pv = kT, we find p = 1000. At the instant in question,

Now, since , we note that

In other words, the pressure is decreasing at the rate of 38.7 lb./sq. ft./minute.