Partial Differentiation

We sometimes come across quantities that are functions of more than one independent variable. Thus, we may find a case where depends on two other variable quantities, one of which we will call and the other . In symbols Take the simplest concrete case.

Let What are we to do? If we were to treat as a constant, and differentiate with respect to , we should get or if we treat as a constant, and differentiate with respect to , we should have:

The little letters here put as subscripts are to show which quantity has been taken as constant in the operation.

Another way of indicating that the differentiation has been performed only partially, that is, has been performed only with respect to one of the independent variables, is to write the derivatives with a “curved dee” , instead of the regular letter . In this way

If we put in these values for and respectively, we shall have

But, if you think of it, you will observe that the total variation of depends on both these things at the same time. That is to say, if both are varying, the real ought to be written and this is called a total differential.

Example 16.1. Find the partial derivatives of the expression .

Solution. The answers are:

The first is obtained by supposing constant, the second is obtained by supposing constant; then

Example 16.2. Let . Then, treating first and then as constant, we get in the usual way so that .

Example 16.3. A cone having height and radius of base has volume . If its height remains constant, while changes, the ratio of change of volume, with respect to radius, is different from ratio of change of volume with respect to height which would occur if the height were varied and the radius kept constant, for

The variation when both the radius and the height change is given by .

Example 16.4. In the following example and denote two arbitrary functions of any form whatsoever. For example, they may be sine-functions, or exponentials, or mere algebraic functions of the two independent variables, and . This being understood, let us take the expression or where , and .
Then (where the figure is simply the coefficient of in and );
and Also and whence

This differential equation is of immense importance in mathematical physics.

Maxima and Minima of Functions of two Independent Variables

Let us take up again the following exercise from the chapter on Maxima and Minima:

Example 16.5. A piece of string inches long has its two ends joined together and is stretched by pegs so as to form a triangle. What is the largest triangular area that can be enclosed by the string?

Solution. Let and be the length of two of the portions of the string. The third is , and the area of the triangle is , where is the half perimeter, , so that , where \begin{align} P &= (15-x)(15-y)(x+y-15) \\ &= xy^2 + x^2y - 15x^2 - 15y^2 - 45xy + 450x + 450y - 3375. \end{align}

Clearly is maximum when is maximum. For a maximum (clearly it will not be a minimum in this case), one must have simultaneously

that is,

An immediate solution is .

If we now introduce this condition in the value of , we find Now is a function of alone. For maximum or minimum, , which gives or .

Clearly gives minimum area; gives the maximum, for , which is for and for (see the Second Derivative Test).

Example 16.6. Find the dimensions of an ordinary railway coal truck with rectangular ends, so that, for a given volume the area of sides and floor together is as small as possible.

Solution. The truck is a rectangular box open at the top. Let be the length and be the width; then the depth is . The surface area is . For minimum (clearly it won’t be a maximum here),

Here also, an immediate solution is , so that , for minimum, and

Exercises

Exercise 16.1. Differentiate the expression with respect to alone, and with respect to alone.

Answer

Solution

Exercise 16.2. Find the partial derivatives with respect to , and , of the expression

Answer

;
;
.

Solution

Let Then

Exercise 16.3. Let .

Find the value of . Also find the value of .

Answer

; .

Solution

Similarly

Therefore,

Using the Quotient Rule

Similarly and

Hence

Exercise 16.4. Find the total differential of .

Answer

Solution

The total differential of is

Exercise 16.5. Find the total differential of ; of ; and of .

Answer

,
,
.

Solution

(a)

(b)

(c)

Exercise 16.6. Verify that the sum of three quantities , , , whose product is a constant , is maximum when these three quantities are equal.

Solution

We want to maximize

provided that .

We find from the constraint and then substitute it in :

The maximum occurs when If we divide by , we get Simplifying the left side, we get: Therefore and

Exercise 16.7. Find the maximum or minimum of the function

Answer

Minimum for .

Solution

has a maximum or minimum where :

By examining nearby points (such as ), we can say that is a minimum when .

Exercise 16.8. The post-office regulations state that no parcel is to be of such a size that its length plus its girth exceeds feet. What is the greatest volume that can be sent by post (a) in the case of a package of rectangular cross section; (b) in the case of a package of circular cross section?

Answer

(a) Length feet, width = depth = foot, vol. = cubic feet.
(b) Radius = feet = in., length = feet, vol. = .

Solution

(a) Let Then

We want to maximize provided that or

Using this constraint, we can write as

The maximum occurs where and : Subtracting the second equation from the first one or

The expression between brackets is and since , the only solution is or .

Substituting in yields or Since Therefore, [Alternatively we could substitute in .]

In this case and .

(b) radius of the cross section

We want to maximize

Since :

Now is a function of alone:

Since , the only solution is .

When :

and the maximum volume is obtained when and

Exercise 16.9. Divide into parts such that the continued product of their sines may be a maximum or minimum.

Answer

All three parts equal; the product is maximum.

Solution

We want to maximize

provided that . Also none of them can be zero because will be zero if or or of occurs if . By symmetry we can imagine the maximum of occurs if

But here we want to use the calculus: and thus Recall that . Hence

To maximize :

Subtracting (A) from (B), we obtain Since , we can write the expression between brackets as . Hence

since , only is acceptable:

Using in (A), we get

Note :

Since ; therefore

This means is a maximum when and the maximum value of is

Exercise 16.10. Find the maximum or minimum of .

Answer

Minimum for .

Solution

Since , we must have

When .

Let’s examine some nearby points:

Therefore, has a minimum of .

Exercise 16.11. Find maximum and minimum of

Answer

Min.: and .

Solution

When and ,

We can examine a couple of points near

When

Hence, is a minimum when and .

Exercise 16.12. A telpherage bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open. Find its dimensions in order that the least amount of iron sheet may be used in its construction.

Answer

Angle at apex ; equal sides = length = .

Solution

To minimize, we set and

Since and , we multiply both sides of the first equation by and divide them by :

So the second equation can be written as

Since , we must have or

Since , from the first equation we obtain and because

Maxima and Minima of Functions of two Independent Variables

Exercises

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