12.4 Tangent Planes and Linear Approximations

Recall that one of the features of the tangent line to the curve y = f (x) at x = a is that it stays close to the curve near the point of tangency. This enables us to use the tangent line to approximate values of the function close to the point of tangency (see Figure 12.24a). Recall that the equation of the tangent line is given by
y = f (a)+f ' (a)(x-a).
(4.1)
Notice that the function on the right side of (4.1) is a linear function of x. When we discussed this in section 3.1, we called this the linear approximation to f (x) at x = a.



Figure 12.24a
Linear approximation.

In much the same way, information about the behavior of a function of two variables near a given point can be obtained from the tangent plane to the surface at that point. For instance, the graph of z = 6-x2-y2 and its tangent plane at the point (1, 2, 1) are shown in Figure 12.24b. Notice that near the point (1, 2, 1), the surface and the tangent plane are very close together. We will develop and exploit this approximation in this section.



Figure 12.24b
z = 6-x2-y2 and the tangent plane at (1, 2, 1).

Our development of the tangent plane will parallel the development of the tangent line in section 2.1. For functions of two variables, we will see that the tangent plane is determined by two slopes, given by the partial derivatives.

We want to find a general equation for the tangent plane to z = f (x, y) at the point (a, b, f (a, b)). You should refer to Figures 12.25a and 12.25b to visualize the process. Starting from a standard graphing window (Figure 12.25a shows z = 6-x2-y2 with -3 x 3 and -3 y 3 ), zoom in on the point (a, b, f (a, b)), as in Figure 12.25b (showing z = 6-x2-y2 with 0.9 x 1.1 and 1.9 y 2.1 ). The surface in Figure12.25b looks like a plane. What has happened is that we have zoomed in far enough that the surface and its tangent plane are difficult to distinguish visually. This suggests that for points (x, y) close to the point of tangency, we can use the corresponding z - value on the tangent plane as an approximation to the value of the function at that point. First, we must find an equation of the tangent plane. Recall that an equation of a plane can be constructed from a point in the plane and any vector normal to the plane. One point lying in the tangent plane is, of course, the point of tangency (a, b, f (a, b)). To find a normal vector, we will find two vectors lying in the plane and then take their cross product to find a vector orthogonal to both (and thus, orthogonal to the plane).



Figure 12.25a
z = 6-x2-y2, with -3 x 3 and -3 y 3.



Figure 12.25b
z = 6-x2-y2, with 0.9 x 1.1 and 1.9 y 2.1.

Imagine intersecting the surface z = f (x, y) with the plane y = b, as shown in Figure 12.26a. As we observed in section 12.3, the result is a curve in the plane y = b whose slope at x = a is given by f x (a, b) . Along the tangent line at x = a, a change of 1 unit in x corresponds to a change of f x (a, b) in z. Since we're looking at a curve that lies in the plane y = b, the value of y doesn't change at all along the curve. A vector with the same direction as the tangent line is then 1, 0, f x(a, b). Because of the way in which we constructed it, this vector must lie in the tangent plane (think about this some). Now, intersecting the surface z = f (x, y) with the plane x = a, as shown in Figure 12.26b, we get a curve lying in the plane x = a, whose slope at y = b is given by f y(a, b). A vector with the same direction as the tangent line at y = b is then 0, 1, f y (a, b).



Figure 12.26a
The intersection of the surface z = f (x, y) with the plane y = b.



Figure 12.26b
The intersection of the surface z = f (x, y) with the plane x = a.

We have now found two vectors in the tangent plane: 1, 0, f x(a, b) and 0, 1, f y (a, b). A vector normal to the plane is then given by the cross product:

0, 1, f y(a, b) 1, 0, f x(a, b) = f x(a, b), f y(a, b), -1,
where we have left the details of the calculation as an exercise. We indicate the tangent plane and normal vector at a point in Figure 12.26c. We have now derived the following result.



Figure 12.26c
Tangent plane and normal vector.

4.1
4.1   
 
Suppose that f (x, y) has continuous partial derivatives at (a, b). A normal vector to the tangent plane to z = f (x, y) at (a, b) is then f x(a, b), f y(a, b), -1. Further, an equation of the tangent plane is given by
f x(a, b)(x -a) + f y(a, b)(y-b)-[z-f (a, b)] = 0
or
z = f (a, b)+f x(a, b)(x-a)+f y(a, b)(y-b).
(4.2)

Observe that since we now know a normal vector to the tangent plane, a line orthogonal to the tangent plane is given by
x = a + f x (a, b)t,        y = b + f y(a, b)t,        z = f (a, b) - t.
(4.3)
This line is called the normal line to the surface at the point (a, b, f (a, b)).

It's now a simple matter to use Theorem 4.1 to construct the equations of a tangent plane and normal line to nearly any surface, as we illustrate in the following examples.

4.1   
Finding Equations of the Tangent Plane and the Normal Line
 
Find equations of the tangent plane and the normal line to z = 6-x2-y2 at the point (1, 2, 1).
 
 
For f (x, y) = 6-x2-y2, we have f x = -2x and f y = -2y. This gives us f x(1, 2) = -2 and f y(1, 2) = -4. A normal vector is then -2, -4, -1 and from (4.2), an equation of the tangent plane is
z-1 = -2(x-1)-4(y-2).
From (4.3), equations of the normal line are
x = 1 - 2t,       y = 2 - 4t,        z = 1 - t.
A sketch of the surface, the tangent plane and the normal line is shown in Figure 12.27.



Figure 12.27
Surface, tangent plane and normal line at the point (1, 2, 1).
 

4.2   
Finding Equations of the Tangent Plane and the Normal Line
 
Find equations of the tangent plane and the normal line to at (2, 1, 13).
 
 
First, notice that here, and so that f x(2, 1) = 12 + 4 = 16 and f y(2, 1) = 3-4 = -1. A normal vector is then 16, -1, -1 and from (4.2), an equation of the tangent plane is
z-13 = 16(x-2)-(y-1).
From (4.3), equations of the normal line are
x = 2 + 16t,       y = 1 - t,        z = 13 - t.
A sketch of the surface, the tangent plane and the normal line is shown in Figure 12.28.



Figure 12.28
Surface, tangent plane and normal line at the point (2, 1, 13).
 



 Notice that in each of Figures 12.27 and 12.28, the tangent plane appears to stay close to the surface near the point of tangency. This says that the z - values on the tangent plane should be close to the corresponding z - values on the surface, which are given by the function values f (x, y), at least for (x, y) close to the point of tangency. Further, the simple form of the equation for the tangent plane makes it ideal for approximating the value of complicated functions. From Theorem 4.1, we have that the tangent plane to z = f (x, y) at the point (a, b) is given by
z = f (a, b)+f x(a, b)(x-a)+f y(a, b)(y-b).
We define the linear approximation L(x, y) of f (x, y) at the point (a, b) to be the function defining the z - values on the tangent plane, namely,
L(x, y) = f (a, b)+f x(a, b)(x-a)+f y(a, b)(y-b).
(4.4)
We illustrate this with the following example.

4.3   
Finding a Linear Approximation
 
Compute the linear approximation of at (0, 0). Compare the linear approximation to the actual function values for (a) x = 0 and y near 0; (b) y = 0 and x near 0; (c) y = x, with both x and y near 0 and (d) y = 2x, with both x and y near 0.
 
 
First, notice that and so that f x(0, 0) = 2 and f y(0, 0) = -1. Also, f (0, 0) = 1. From (4.4), the linear approximation is then given by
L(x, y) = 1+2(x-0)-(y-0) = 1+2x-y.
The following table compares values of L(x, y) and f (x, y) for a number of points of the form (0, y), (x, 0), (x, x) and (x, 2x).

(x, y) f (x, y) L(x, y) (x, y) f (x, y) L(x, y)
(0, 0.1) 0.905 0.9 (0.1, 0.1) 1.11393 1.1
(0, 0.01) 0.99005 0.99 (0.01, 0.01) 1.01015 1.01
(0, - 0.1) 1.105 1.1 ( - 0.1, - 0.1) 0.91628 0.9
(0, - 0.01) 1.01005 1.01 ( - 0.01, - 0.01) 0.99015 0.99
           
(0.1, 0) 1.21005 1.2 (0.1, 0.2) 1.02696 1.0
(0.01, 0) 1.02010 1.02 (0.01, 0.02) 1.00030 1.0
( - 0.1, 0) 0.81005 0.8 ( - 0.1, - 0.2) 1.03368 1.0
( - 0.01, 0) 0.98010 0.98 ( - 0.01, - 0.02) 1.00030 1.0

Notice that the closer a given point is to the point of tangency, the more accurate the linear approximation is at that point. This is typical of this type of approximation and was also the case for linear approximations to functions of a single variable. We will explore this further in the exercises.
 

Increments and Differentials

Now that we have examined linear approximations from a graphical perspective, we will examine these in a symbolic fashion. In the course of doing so, we will gain some insight into the behavior of functions of several variables. First, we remind you of the notation and some alternative language that we have already used in section 3.1 for functions of a single variable. We defined the increment y to be

y = f (a+ x)-f (a).
Referring to Figure 12.29, notice that for x small,
y dy = f ' (a) x,
where we referred to dy as the differential of y. Further, observe that if f  is differentiable at x = a and then we have
 
as x 0. (You'll need to recognize the definition of derivative here!) Finally, solving



Figure 12.29
Increments and differentials for a function of one variable.

for y in terms of we have
where as x 0. We can make a similar observation for functions of several variables, as follows.

For z = f (x, y), we define the increment of f  to be

z = f (a+ x, b+ y)-f (a, b).
That is, z is the change in z that occurs when a is incremented by x and b is incremented by y, as illustrated in Figure 12.30.



Figure 12.30
Linear approximation.

Notice that as long as f  is continuous in some open region containing (a, b) and f  has first partial derivatives on that region, we can write
z = f (a + x, b + y) - f (a, b)
  = [f (a+ x, b+ y)-f (a, b+ y)]+[f (a, b+ y)-f (a, b)] Adding and subtracting f (a, b+ y).
  = f x(u, b+ y)[(a+ x)-a]+f y(a, v)[(b+ y)-b] Applying the Mean Value Theorem to both terms.
  = f x(u, b+ y) x+f y(a, v) y,
by the Mean Value Theorem. Here, u is some value between a and a+ x and v is some value between b and b+ y (see Figure 12.31). This gives us
z = f x(u, b + y) x + f y(a, v) y
  = {f x(a, b)+[f x(u, b+ y)-f x(a, b)]} x
     + { f y(a, b)+[ f y(a, v)-f y(a, b)]} y.
(4.5)



Figure 12.31
Intermediate points from the Mean Value Theorem.

Notice that we can rewrite (4.5) as
where
Finally, observe that if f x and f y are both continuous in some open region containing (a, b), then and will both tend to 0, as ( x, y) (0, 0). In fact, you should recognize that since as ( x, y) (0, 0), the products and both tend to 0 even faster than do x or y individually. (Think about this!)

We have now established the following result.

4.2   
 
Suppose that z = f (x, y) is defined on the rectangle R = {(x, y)|x0 < x < x1, y0 < y < y1} and f x and f y are defined on R and are continuous at (a, b) R. Then for (a+ x, b+ y) R,
(4.6)
where and are functions of x and y that both tend to zero, as ( x, y) (0, 0).
 

For some very simple functions, we can compute z by hand, as illustrated in the following example.

4.4   
Computing the Increment z
 
For z = f (x, y) = x2-5xy, find z and write it in the form indicated in Theorem 4.2.
 
 
We have
z = f (x + x, y + y) - f (x, y)
  = [(x+ x)2-5(x+ x)(y+ y)] -(x2-5xy)
  = x2+2x x+( x)2-5(xy+x y+y x+ x y)-x2+5xy
 
 
where and both tend to zero, as ( x, y) (0, 0), as indicated in Theorem 4.2.
 

Look closely at the first two terms in the expansion of the increment z given in (4.6). If we take x = x-a and y = y-b, then they correspond to the linear approximation of f (x, y). In this context, we give this a special name. If we increment x by the amount dx = x and increment y by dy = y, then we define the differential of z to be

dz = f x(x, y) dx+f y(x, y) dy.

This is sometimes referred to as a total differential. Notice that for dx and dy small, we have from (4.6) that

z dz.
You should recognize that this is the same approximation as the linear approximation developed in the beginning of this section. In this case though, we have developed this from an analytical perspective, rather than the geometrical one used in the beginning of the section.

Functions that can be approximated linearly in the above fashion we give a special name, as in the following definition.

4.1   
 
Let z = f (x, y). We say that f is differentiable at (a, b) if we can write
where and are both functions of x and y and as ( x, y) (0, 0). We say that f is differentiable on a region R whenever f is differentiable at every point in R.
 

Note that from Theorem 4.2, if f x and f y are defined on some open rectangle R containing the point (a, b) and if f x and f y are continuous at (a, b), then f will be differentiable at (a, b). Just as with functions of a single variable, it can be shown that if f is differentiable at a point (a, b), then it is also continuous at (a, b). Further, owing to Theorem 4.2, if a function is differentiable at a point, then the linear approximation (differential) at that point provides a good approximation to the function near that point. Be very careful of what this does not say, however. If a function has partial derivatives at a point, it need not be differentiable or even continuous at that point. (In the exercises, you will see an example of a function with partial derivatives defined everywhere, but that is not continuous at a point.)

The idea of the linear approximation extends easily to three or more dimensions. We lose the graphical interpretation of a tangent plane approximating a surface, but the definition should make sense.

4.2   
 
The linear approximation to f (x, y, z) at the point (a, b, c) is given by
 

We can write the linear approximation in the context of increments and differentials, as follows. If we increment x by x, y by y and z by z, then the increment of w = f (x, y, z) is given by
w = f (x + x, y + y, z + z) - f (x, y, z)
  dw = f x(x, y, z) x+f y(x, y, z) y+f z(x, y, z) z.



A good way to interpret (and remember!) the linear approximation is that each partial derivative represents the change in the function relative to the change in that variable. The linear approximation starts with the function value at the known point and adds in the approximate changes corresponding to each of the independent variables.

4.5   
Approximating the Sag in a Beam
 
Suppose that the sag in a beam of length L, width w and height h is given by with all lengths measured in inches. We illustrate the beam in Figure 12.32. A beam is supposed to measure L = 36, w = 2 and h = 6 with a



Figure 12.32
A typical beam.

corresponding sag of 1.5552 inches. Due to weathering and other factors, the manufacturer only guarantees measurements with error tolerances L = 36 1, w = 2 0.4 and h = 6 0.8 Use a linear approximation to estimate the possible range of sags in the beam.
 
 
We first compute and . At the point (36, 2, 6), we then have , and . From Definition 4.2, the linear approximation of the sag is then given by
S 1.5552+0.1728(L-36)-0.7776 (w-2)-0.7776(h-6).
From the stated tolerances, L-36 must be between -1 and 1, w-2 must be between -0.4 and 0.4 and h-6 must be between -0.8 and 0.8. Notice that the maximum sag then occurs with L-36 = 1, w-2 = -0.4 and h-6 = -0.8. The linear approximation predicts that
S-1.5552 0.1728+0.31104+0.62208 = 1.10592.
Similarly, the minimum sag occurs with L-36 = -1, w-2 = 0.4 and h-6 = 0.8. The linear approximation predicts that
S-1.5552 -0.1728-0.31104-0.62208 = -1.10592.
Based on the linear approximation, the sag is 1.5552 1.10592, or between 0.44928 and 2.66112. As you can see, in this case, the uncertainty in the sag is substantial.
 

In many real-world situations, we do not have a formula for the quantity we are interested in computing. Even so, given sufficient information, we can still use linear approximations to estimate the desired quantity.

4.6   
Estimating the Gauge of a Sheet of Metal
 
Manufacturing plants create rolls of metal of a desired gauge (thickness) by feeding the metal through very large rollers. The thickness of the resulting metal depends on the gap between the working rollers, the speed at which the rollers turn and the temperature of the metal. Suppose that for a certain metal, a gauge of 4 mm is produced by a gap of 4 mm, a speed of 10 m/s and a temperature of 900°. Experiments show that an increase in speed of 0.2 m/s increases the gauge by 0.06 mm and an increase in temperature of 10° decreases the gauge by 0.04 mm. Use a linear approximation to estimate the gauge at 10.1 m/s and 880°.
 
 
With no change in gap, we assume that the gauge is a function g(s, t ) of the speed s and the temperature t. Based on our data, and From Definition 4.2, the linear approximation of g(s, t ) is given by
g(s, t ) 4+0.3(s-10)-0.004(t-900).
With s = 10.1 and t = 880, we get the estimate
g(10.1, 880) 4+0.3(0.1)-0.004(-20) = 4.11.

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