Partial Derivatives and Tangent Planes

Partial derivatives of the first order have an interesting geometric interpretation. Let z = f(x. y) be a function of two variables, with graph as shown in Fig. 15.14. Let us keep the value of y fixed at yo- The points (x. y) on the graph of / that have y = y0 are those that lie on the curve Ky indicated in the figure. The partial derivative f'x{x o, yo) is the derivative of z — fix, y0) with respect to x at the point x = xq, and is therefore the slope of the tangent line ly to the curve Ky at x = Join the same way, f.(xo, yo) is the slope of the tangent line lx to the curve Kx at >• = yo-

This geometric interpretation of the two partial derivatives can be explained in another way. Imagine that the graph of / describes a mountain, and suppose that we are standing at point P with coordinates (xo, yo, fix o, yo)) in three

FIGURE 15.14

dimensions, where the height is f(xo, yo) units above the xy-plane. The slope of the terrain at P depends on the direction in which we look. In particular, let us look in the direction parallel to the positive x-axis. Then fx(xo, y0) is a measure of the "steepness" in this direction. In the figure, fx{xo, yo) is negative, because moving from P in the direction given by the positive x-axis will take us downwards. In the same way, we see that f[.(xo, >'o) is a measure of the "steepness" in the direction parallel to the positive y-axis. We also see that f.(xo,yo) is positive, meaning that the slope is upward in this direction.

Let us now briefly consider the geometric interpretation of the "direct" second-order derivatives fxx and fyy. Consider the curve Ky on the graph of / in the figure. It seems that along this curve, fxx(x,yo) is negative, because fi(x, yo) decreases as x increases. In particular, fxx(xo,yo) <0. In the same way, we see that moving along Kx makes f[.(xo,y) decrease as >• increases, so fyy(xo, y) < 0 along Kx. In particular, fyy(x0, yo) < 0.

Example 15.17

Consider Fig. 15.15, showing some level curves of a function z = f(x, y).

On the basis of this figure, answer the following questions:

(a) What are the signs of fx(x, y) and f'y{x, y) at P and Q!

(b) What are the solutions of the two equations: (i) f(3,y) = 4 and (ii) fix, 4) = 6?

(c) What is the largest value that f(x,y) can attain when x = 2, and for which >• value does this maximum occur?

Solution

(a) If you stand at P, you are on the level curve f(x, >•) = 2. If you look in the direction of the positive x-axis (along the line y = 4). then you will see the terrain sloping upwards, because the (nearest) level curves

correspond to larger z values. Hence, f'x > 0. If you stand at P and look in the direction of the positive y-axis (along x = 2), the terrain will slope downwards. Thus, at P, we must have f[. < 0. At Q, we find similarly that f'x < 0 and f > 0.

(b) Equation (i) has the solutions y = 1 and y = 4, because the line x = 3 cuts the level curve f{x, y) = 4 at (3. 1) and at (3,4). Equation (ii) has no solutions, because the line y = 4 does not meet the level curve fix, y) = 6 at all.

(c) The highest value of c for which the level curve f(x, y) = c has a point in common with the line x = 2 is c = 6. The largest value of fix. y) when x = 2 is therefore 6, and we see that this maximum value is attained when y % 2.2.

Tangent Planes

Look back at Fig. 15.14. The two tangent lines lx and ly determine a unique plane through the point P = (xq, yo, f(xo, yo)). This plane is called the tangent plane to the surface at P. From [12.23] in Section 12.5, the general equation for a plane in three-dimensional space passing through a point (x0, yo, zo) is a(x — x0) + b(y — yo) 4- c(z — zo) =0. If c = 0, then this plane is parallel to the z-axis. If c ± 0 and we define A = —a/c, B — —b/c, then solving the equation for z — zo gives z - zo = A(x - x0) + B(y - y0) [1]

So the tangent plane to the surface at P must have this form. It remains to determine A and B. Now, line ly lies in the plane. Because the slope of the line is f[{xQ, y0), the points (x, y, z) that lie on ly are characterized by the two equations y = y0 and z - zo = /i'(*o, yoX* - *o). AH these points (x. y, z) lie in the plane

Tangent plane

FIGURE 15.16 THe graph of a function z = f(x, y) and its tangent plane at P.

Tangent plane fix.y)

FIGURE 15.16 THe graph of a function z = f(x, y) and its tangent plane at P.

[1] only if A = f{(xo, y0). In a similar way, we see that B = /2'0o, yo)- The conclusion is as follows:

The tangent plane to z = f(x,y) at the point Oo, yo, Zo), with zo = f(xo, yo), has the equation

Solution Because /( 1,1) = 5, the given point lies on the graph of /. We find that f[(x, y) = 2i + 2y, f±(x, y) = 2x + 4y Hence, /,'( 1,1) = 4 and /,'( 1, 1) = 6. Thus, [15.11] yields z-5 = 4(x- l) + 6(y- 1) or z = 4x + 6y-5

1. In Fig. 15.17, we have drawn some level curves for a function z = f(x, y)> together with the line 2x + 3y = 12.

The tangent plane is illustrated in Fig. 15.16. Example 15.18

Find the tangent plane at (xq, y0, Zo) = (1,1,5) to the graph of f(x,y)=x2 + 2xy + 2y2

Problems

a. What are the signs of f'x and f'y at P and <2?

b. Find possible solutions of the two equations (i) /(l, y) = 2 and (ii) / (*, 2) = 2.

c. What is the largest value of f(x,y) among those (x, y) that satisfy 2x + 3y = 12?

2. Suppose F(x, y) is a function about which all we know is that F(0,0) = 0, F[\x, y)> 2 for all (x, y), and Fiix, y) < 1 for all (x, y). What can be said about the relative sizes of F(0,0), F(1,0), F(2,0), F(0,1), and F(l, 1)? Write down the inequalities that have to hold between these numbers.

3. Find the tangent planes to the following surfaces at the indicated points:

a. z = x2 + r at (1,2,5) b. z = (y - x2)(y - 2x2) at (1,3,2)

4. Prove that all tangent planes to z = xf(y/x) pass through the origin.

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