14.3

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14.3 Partial Derivatives

Partial Derivatives

On a hot day, extreme humidity makes us think the

temperature is higher than it really is, whereas in very dry air we perceive the temperature to be lower than the

thermometer indicates.

The National Weather Service has devised the heat index (also called the temperature-humidity index, or humidex, in some countries) to describe the combined effects of

temperature and humidity.

The heat index I is the perceived air temperature when the actual temperature is T and the relative humidity is H.

So I is a function of T and H and we can write I = f(T, H).

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Partial Derivatives

The following table of values of I is an excerpt from a table compiled by the National Weather Service.

Table 1

Heat index I as a function of temperature and humidity

Partial Derivatives

If we concentrate on the highlighted column of the table, which corresponds to a relative humidity of H = 70%, we are considering the heat index as a function of the single variable T for a fixed value of H. Let’s write g(T ) = f(T, 70).

Then g(T ) describes how the heat index I increases as the actual temperature T increases when the relative humidity is 70%.

The derivative of g when T = 96°F is the rate of change of I with respect to T when T = 96°F:

5

Partial Derivatives

We can approximate g′(96) using the values in Table 1 by taking h = 2 and –2:

Averaging these values, we can say that the derivative g′(96) is approximately 3.75.

Partial Derivatives

This means that, when the actual temperature is 96°F and the relative humidity is 70%, the apparent temperature

(heat index) rises by about 3.75°F for every degree that the actual temperature rises!

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Partial Derivatives

Now let’s look at the highlighted row in Table 1, which corresponds to a fixed temperature of T = 96°F.

Table 1

Heat index I as a function of temperature and humidity

Partial Derivatives

The numbers in this row are values of the function G(H) = f(96, H), which describes how the heat index

increases as the relative humidity H increases when the actual temperature is T = 96°F.

The derivative of this function when H = 70% is the rate of change of I with respect to H when H = 70%:

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Partial Derivatives

By taking h = 5 and –5, we approximate G′(70) using the tabular values:

By averaging these values we get the estimate

G′(70) ≈ 0.9. This says that, when the temperature is 96°F and the relative humidity is 70%, the heat index rises about 0.9°F for every percent that the relative humidity rises.

Partial Derivatives

In general, if f is a function of two variables x and y,

suppose we let only x vary while keeping y fixed, say y = b, where b is a constant.

Then we are really considering a function of a single

variable x, namely, g(x) = f(x, b). If g has a derivative at a, then we call it the partial derivative of f with respect to x at (a, b) and denote it by f_x(a, b). Thus

11

Partial Derivatives

By the definition of a derivative, we have

and so Equation 1 becomes

Partial Derivatives

Similarly, the partial derivative of f with respect to y at (a, b), denoted by f_y(a, b), is obtained by keeping x fixed (x = a) and finding the ordinary derivative at b of the

function G(y) = f(a, y):

With this notation for partial derivatives, we can write the rates of change of the heat index I with respect to the

actual temperature T and relative humidity H when T = 96°F and H = 70% as follows:

f_T(96, 70) ≈ 3.75 f_H(96, 70) ≈ 0.9

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Partial Derivatives

If we now let the point (a, b) vary in Equations 2 and 3, f_x and f_y become functions of two variables.

Partial Derivatives

There are many alternative notations for partial derivatives.

For instance, instead of f_x we can write f₁ or D₁f (to indicate differentiation with respect to the first variable) or ∂f/∂x.

But here ∂f/∂x can’t be interpreted as a ratio of differentials.

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Partial Derivatives

To compute partial derivatives, all we have to do is

remember from Equation 1 that the partial derivative with respect to x is just the ordinary derivative of the function g of a single variable that we get by keeping y fixed.

Thus we have the following rule.

If f(x, y) = x³ + x²y³ – 2y², find f_x(2, 1) and f_y(2, 1).

Solution:

Holding y constant and differentiating with respect to x, we get

f_x(x, y) = 3x² + 2xy³

and so f_x(2, 1) = 3  2² + 2  2  1³

Holding x constant and differentiating with respect to y, we get

f_y(x, y) = 3x²y² – 4y

f_y(2, 1) = 3  2²  1² – 4  1

Example 1

= 16

= 8

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Interpretations of Partial Derivatives

Interpretations of Partial Derivatives

To give a geometric interpretation of partial derivatives, we recall that the equation z = f(x, y) represents a surface

S (the graph of f ). If f(a, b) = c, then the point P(a, b, c) lies on S.

By fixing y = b, we are restricting our attention to the curve C₁ in which the vertical plane y = b intersects S. (In other words, C₁ is the trace of S in the plane y = b.)

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Interpretations of Partial Derivatives

Likewise, the vertical plane x = a intersects S in a curve C₂. Both of the curves C₁ and C₂ pass through the point P.

(See Figure 1.)

Note that the curve C₁ is the graph of the function

g(x) = f(x, b), so the slope of its tangent T₁ at P is g′(a) = f_x(a, b).

The curve C₂ is the graph of the function G(y) = f(a, y), so the slope of its tangent T₂ at P is G′(b) = f_y(a, b).

Figure 1

The partial derivatives of f at (a, b) are the slopes of the tangents to C₁and C₂.

Interpretations of Partial Derivatives

Thus the partial derivatives f_x(a, b) and f_y(a, b) can be

interpreted geometrically as the slopes of the tangent lines at P(a, b, c) to the traces C₁ and C₂ of S in the planes y = b and x = a.

As we have seen in the case of the heat index function, partial derivatives can also be interpreted as rates of change.

If z = f(x, y), then ∂z/∂x represents the rate of change of z with respect to x when y is fixed. Similarly, ∂z/∂y

represents the rate of change of z with respect to y when x is fixed.

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Example 2

If f(x, y) = 4 – x² – 2y², find f_x(1, 1) and f_y(1, 1) and interpret these numbers as slopes.

Solution:

We have

f_x(x, y) = –2x f_y(x, y) = –4y f_x(1, 1) = –2 f_y(1, 1) = –4 .

Example 2 – Solution

The graph of f is the paraboloid z = 4 – x² – 2y² and the

vertical plane y = 1 intersects it in the parabola z = 2 – x², y = 1. (As in the preceding discussion, we label it C₁ in

Figure 2.)

The slope of the tangent line to this parabola at the point

(1, 1, 1) is f_x(1, 1) = –2.

Figure 2

cont’d

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Example 2 – Solution

Similarly, the curve C₂ in which the plane x = 1 intersects the paraboloid is the parabola z = 3 – 2y², x = 1, and the slope of the tangent line at (1, 1, 1) is f_y(1, 1) = –4. (See Figure 3.)

Figure 3

cont’d

Functions of More Than Two

Variables

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Functions of More Than Two Variables

Partial derivatives can also be defined for functions of three or more variables. For example, if f is a function of three

variables x, y, and z, then its partial derivative with respect to x is defined as

and it is found by regarding y and z as constants and differentiating f(x, y, z) with respect to x.

Functions of More Than Two Variables

If w = f(x, y, z), then f_x = ∂w/∂x can be interpreted as the rate of change of w with respect to x when y and z are held fixed. But we can’t interpret it geometrically because the graph of f lies in four-dimensional space.

In general, if u is a function of n variables,

u = f(x₁, x₂,…, x_n), its partial derivative with respect to the i th variable x_i is

and we also write

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Example 6

Find f_x, f_y, and f_z if f(x, y, z) = e^xy ln z.

Solution:

Holding y and z constant and differentiating with respect to x, we have

f_x = ye^xy ln z Similarly,

f_y = xe^xy ln z and f_z =

Higher Derivatives

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Higher Derivatives

If f is a function of two variables, then its partial derivatives f_x and f_y are also functions of two variables, so we can

consider their partial derivatives (f_x)_x, (f_x)_y, (f_y)_x, and (f_y)_y, which are called the second partial derivatives of f.

If z = f(x, y), we use the following notation:

Higher Derivatives

Thus the notation f_xy (or ∂²f/∂y ∂x) means that we first differentiate with respect to x and then with respect to y, whereas in computing f_yx the order is reversed.

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Example 7

Find the second partial derivatives of f(x, y) = x³ + x²y³ – 2y²

Solution:

In Example 1 we found that

f_x(x, y) = 3x² + 2xy³ f_y(x, y) = 3x²y² – 4y Therefore

f_xx = (3x² + 2xy³)

= 6x + 2y³

Example 7 – Solution

f_xy = (3x² + 2xy³)

= 6xy²

f_yx = (3x²y² – 4y)

= 6xy²

f_yy = (3x²y² – 4y)

= 6x²y – 4

cont’d

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Higher Derivatives

Notice that f_xy = f_yx in Example 7. This is not just a

coincidence. It turns out that the mixed partial derivatives f_xy and f_yx are equal for most functions that one meets in practice.

The following theorem, which was discovered by the

French mathematician Alexis Clairaut (1713–1765), gives conditions under which we can assert that f_xy = f_yx.

Higher Derivatives

Partial derivatives of order 3 or higher can also be defined.

For instance,

and using Clairaut’s Theorem it can be shown that f_xyy = f_yxy = f_yyx if these functions are continuous.

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Partial Differential Equations

Partial Differential Equations

Partial derivatives occur in partial differential equations that express certain physical laws.

For instance, the partial differential equation

is called Laplace’s equation after Pierre Laplace (1749–1827).

Solutions of this equation are called harmonic functions;

they play a role in problems of heat conduction, fluid flow, and electric potential.

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Example 9

Show that the function u(x, y) = e^x sin y is a solution of Laplace’s equation.

Solution:

We first compute the needed second-order partial derivatives:

u_x = e^x sin y u_y = e^x cos y u_xx = e^x sin y u_yy = –e^x sin y So u_xx + u_yy = e^x sin y – e^x sin y = 0 Therefore u satisfies Laplace’s equation.

Partial Differential Equations

The wave equation

describes the motion of a waveform, which could be an ocean wave, a sound wave, a light wave, or a wave

traveling along a vibrating string.

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Partial Differential Equations

For instance, if u(x, t) represents the displacement of a

vibrating violin string at time t and at a distance x from one end of the string (as in Figure 8), then u(x, t) satisfies the wave equation.

Here the constant a depends on the density of the string and on the tension in the string.

Figure 8

Partial Differential Equations

Partial differential equations involving functions of three variables are also very important in science and

engineering. The three-dimensional Laplace equation is

and one place it occurs is in geophysics. If u(x, y, z)

represents magnetic field strength at position (x, y, z), then it satisfies Equation 5. The strength of the magnetic field indicates the distribution of iron-rich minerals and reflects different rock types and the location of faults.

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Partial Differential Equations

Figure 9 shows a contour map of the earth’s magnetic field as recorded from an aircraft carrying a magnetometer and flying 200 m above the surface of the ground. The contour map is enhanced by color-coding of the regions between the level curves.

Magnetic field strength of the earth

Figure 9

Partial Differential Equations

Figure 10 shows a contour map for the second-order partial derivative of u in the vertical direction, that is, u_zz. It turns out that the values of the partial derivatives u_xx and u_yy are relatively easily measured from a map of the magnetic field.

Then values of u_zz can be calculated from Laplace’s equation (5).

Second vertical derivative of the magnetic field

Figure 10

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The Cobb-Douglas Production

Function

The Cobb-Douglas Production Function

We have described the work of Cobb and Douglas in

modeling the total production P of an economic system as a function of the amount of labor L and the capital

investment K.

Here we use partial derivatives to show how the particular form of their model follows from certain assumptions they made about the economy.

If the production function is denoted by P = P(L, K), then the partial derivative ∂P/∂L is the rate at which production changes with respect to the amount of labor. Economists call it the marginal production with respect to labor or the marginal productivity of labor.

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The Cobb-Douglas Production Function

Likewise, the partial derivative ∂P/∂K is the rate of change of production with respect to capital and is called the

marginal productivity of capital.

In these terms, the assumptions made by Cobb and Douglas can be stated as follows.

(i) If either labor or capital vanishes, then so will production.

(ii) The marginal productivity of labor is proportional to the amount of production per unit of labor.

(iii) The marginal productivity of capital is proportional to the amount of production per unit of capital.

The Cobb-Douglas Production Function

Because the production per unit of labor is P/L, assumption (ii) says that

for some constant α.

If we keep K constant (K = K₀), then this partial differential equation becomes an ordinary differential equation:

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The Cobb-Douglas Production Function

If we solve this separable differential equation, we get P(L, K₀) = C₁(K₀)L^α

Notice that we have written the constant C₁ as a function of K₀ because it could depend on the value of K₀.

Similarly, assumption (iii) says that

and we can solve this differential equation to get P(L₀, K) = C₂(L₀)K^β

The Cobb-Douglas Production Function

Comparing Equations 7 and 8, we have P(L, K) = bL^αK^β

where b is a constant that is independent of both L and K.

Assumption (i) shows that α > 0 and β > 0.

Notice from Equation 9 that if labor and capital are both increased by a factor m, then

P(mL, mK) = b(mL)^α(mK)^β = m^α+βbL^αK^β = m^α+βP(L, K)

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The Cobb-Douglas Production Function

If α + β = 1, then P(mL, mK) = mP(L, K), which means that production is also increased by a factor of m. That is why Cobb and Douglas assumed that α + β = 1 and therefore

P(L, K) = bL^αK^{1 –α}

This is the Cobb-Douglas production function.