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# 14.4

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## 14.4 Tangent Planes and

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### Tangent Planes

Suppose a surface S has equation z = f(x, y), where f has continuous first partial derivatives, and let P(x0, y0, z0) be a point on S.

Let C1 and C2 be the curves obtained by intersecting the vertical planes y = y0 and x = x0 with the surface S. Then the point P lies on both C1 and C2.

Let T1 and T2 be the tangent lines to the curves C1 and C2 at the point P.

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### Tangent Planes

Then the tangent plane to the surface S at the point P is defined to be the plane that contains both tangent lines T1 and T2. (See Figure 1.)

The tangent plane contains the tangent lines T1 and T2.

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### Tangent Planes

If C is any other curve that lies on the surface S and

passes through P, then its tangent line at P also lies in the tangent plane.

Therefore you can think of the tangent plane to S at P as consisting of all possible tangent lines at P to curves that lie on S and pass through P. The tangent plane at P is the

plane that most closely approximates the surface S near the point P. We know that any plane passing through the point P(x0, y0, z0) has an equation of the form

A(x – x0) + B(y – y0) + C(z – z0) = 0

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### Tangent Planes

By dividing this equation by C and letting a = –A/C and b = –B/C, we can write it in the form

z – z0 = a(x – x0) + b(y – y0)

If Equation 1 represents the tangent plane at P, then its intersection with the plane y = y0 must be the tangent line T1. Setting y = y0 in Equation 1 gives

z – z0 = a(x – x0) where y = y0

and we recognize this as the equation (in point-slope form) of a line with slope a.

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### Tangent Planes

But we know that the slope of the tangent T1 is fx(x0, y0).

Therefore a = fx(x0, y0).

Similarly, putting x = x0 in Equation 1, we get

z – z0 = b(y – y0), which must represent the tangent line T2, so b = fy(x0, y0).

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### Example 1

Find the tangent plane to the elliptic paraboloid z = 2x2 + y2 at the point (1, 1, 3).

Solution:

Let f(x, y) = 2x2 + y2. Then

fx(x, y) = 4x fy(x, y) = 2y fx(1, 1) = 4 fy(1, 1) = 2

Then (2) gives the equation of the tangent plane at (1, 1, 3) as

z – 3 = 4(x – 1) + 2(y – 1) or z = 4x + 2y – 3

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### Tangent Planes

Figure 2(a) shows the elliptic paraboloid and its tangent plane at (1, 1, 3) that we found in Example 1. In parts (b) and (c) we zoom in toward the point (1, 1, 3) by restricting the domain of the function f(x, y) = 2x2 + y2.

Figure 2

The elliptic paraboloid z = 2x2+ y2 appears to coincide with its tangent plane as we zoom in toward (1, 1, 3).

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### Tangent Planes

Notice that the more we zoom in, the flatter the graph appears and the more it resembles its tangent plane.

In Figure 3 we corroborate this impression by zooming in toward the point (1, 1) on a contour map of the function f(x, y) = 2x2 + y2.

Figure 3

Zooming in toward (1, 1) on a contour map of f(x, y) = 2x2 + y2

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### Tangent Planes

Notice that the more we zoom in, the more the level curves look like equally spaced parallel lines, which is

characteristic of a plane.

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### Linear Approximations

In Example 1 we found that an equation of the tangent plane to the graph of the function f(x, y) = 2x2 + y2 at the point (1, 1, 3) is z = 4x + 2y – 3. Therefore, the linear

function of two variables

L(x, y) = 4x + 2y – 3

is a good approximation to f(x, y) when (x, y) is near (1, 1).

The function L is called the linearization of f at (1, 1) and the approximation

f(x, y) ≈ 4x + 2y – 3

is called the linear approximation or tangent plane approximation of f at (1, 1).

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### Linear Approximations

For instance, at the point (1.1, 0.95) the linear approximation gives

f(1.1, 0.95) ≈ 4(1.1) + 2(0.95) – 3 = 3.3 which is quite close to the true value of

f(1.1, 0.95) = 2(1.1)2 + (0.95)2 = 3.3225.

But if we take a point farther away from (1, 1), such as (2, 3), we no longer get a good approximation.

In fact, L(2, 3) = 11 whereas f(2, 3) = 17.

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### Linear Approximations

In general, we know from (2) that an equation of the

tangent plane to the graph of a function f of two variables at the point (a, b, f(a, b)) is

z = f(a, b) + fx(a, b)(x – a) + fy(a, b)(y – b) The linear function whose graph is this tangent plane, namely

L(x, y) = f(a, b) + fx(a, b)(x – a) + fy(a, b)(y – b) is called the linearization of f at (a, b).

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### Linear Approximations

The approximation

f(x, y) ≈ f(a, b) + fx(a, b)(x – a) + fy(a, b)(y – b)

is called the linear approximation or the tangent plane approximation of f at (a, b).

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### Linear Approximations

We have defined tangent planes for surfaces z = f(x, y), where f has continuous first partial derivatives. What

happens if fx and fy are not continuous? Figure 4 pictures such a function; its equation is

You can verify that its partial derivatives exist at the origin and, in fact, fx(0, 0) = 0 and fy(0, 0) = 0, but fx and fy are not continuous.

Figure 4

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### Linear Approximations

The linear approximation would be f(x, y) ≈ 0, but f(x, y) = at all points on the line y = x.

So a function of two variables can behave badly even though both of its partial derivatives exist.

To rule out such behavior, we formulate the idea of a differentiable function of two variables.

Recall that for a function of one variable, y = f(x), if x

changes from a to a + ∆x, we defined the increment of y as

∆y = f(a + ∆x) – f(a)

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### Linear Approximations

If f is differentiable at a, then

∆y = f′(a) ∆x + ε ∆x where ε → 0 as ∆x → 0 Now consider a function of two variables, z = f(x, y), and suppose x changes from a to a + ∆x and y changes from b to b + ∆y. Then the corresponding increment of z is

∆z = f(a + ∆x, b + ∆y) – f(a, b)

Thus the increment ∆z represents the change in the value of f when (x, y) changes from (a, b) to (a + ∆x, b + ∆y).

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### Linear Approximations

By analogy with (5) we define the differentiability of a function of two variables as follows.

Definition 7 says that a differentiable function is one for

which the linear approximation (4) is a good approximation when (x, y) is near (a, b). In other words, the tangent plane approximates the graph of f well near the point of tangency.

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### Linear Approximations

It’s sometimes hard to use Definition 7 directly to check the differentiability of a function, but the next theorem provides a convenient sufficient condition for differentiability.

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### Example 2

Show that f (x, y) = xexy is differentiable at (1, 0) and find its linearization there. Then use it to approximate f(1.1,– 0.1).

Solution:

The partial derivatives are

fx(x, y) = exy + xyexy fy(x, y) = x2exy fx(1, 0) = 1 fy(1, 0) = 1

Both fx and fy are continuous functions, so f is differentiable by Theorem 8. The linearization is

L(x, y) = f (1, 0) + fx(1, 0)(x – 1) + fy(1, 0)(y – 0)

= 1 + 1(x – 1) + 1  y = x + y

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### Example 2 – Solution

The corresponding linear approximation is xexy ≈ x + y

so f(1.1,– 0.1) ≈ 1.1 – 0.1 = 1 Compare this with the actual value of

f(1.1,– 0.1) = 1.1e –0.11

≈ 0.98542.

cont’d

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### Differentials

For a differentiable function of one variable, y = f(x), we

define the differential dx to be an independent variable; that is, dx can be given the value of any real number.

The differential of y is then defined as dy = f′(x) dx

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### Differentials

Figure 6 shows the relationship between the increment ∆y and the differential dy: ∆y represents the change in height of the curve y = f(x) and dy represents the change in height of the tangent line when x changes by an amount dx = ∆x.

Figure 6

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### Differentials

For a differentiable function of two variables, z = f(x, y), we define the differentials dx and dy to be independent

variables; that is, they can be given any values. Then the differential dz, also called the total differential, is

defined by

Sometimes the notation df is used in place of dz.

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### Differentials

If we take dx = ∆x = x – a and dy = ∆y = y – b in Equation 10, then the differential of z is

dz = fx(a, b)(x – a) + fy(a, b)(y – b)

So, in the notation of differentials, the linear approximation (4) can be written as

f(x, y) ≈ f(a, b) + dz

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### Differentials

Figure 7 is the three-dimensional counterpart of Figure 6 and shows the geometric interpretation of the differential dz and the increment ∆z: dz represents the change in height of the tangent plane, whereas ∆z represents the change in height of the surface z = f(x, y) when (x, y) changes from (a, b) to (a + ∆x, b + ∆y).

Figure 6 Figure 7

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### Example 4

(a) If z = f(x, y) = x2 + 3xy – y2, find the differential dz.

(b) If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values of ∆z and dz.

Solution:

(a) Definition 10 gives

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### Example 4 – Solution

(b) Putting x = 2, dx = ∆x = 0.05, y = 3, and dy = ∆y = –0.04, we get

dz = [2(2) + 3(3)]0.05 + [3(2) – 2(3)](–0.04)

The increment of z is

∆z = f(2.05, 2.96) – f(2, 3)

= [(2.05)2 + 3(2.05)(2.96) – (2.96)2] – [22 + 3(2)(3) – 32]

= 0.6449

Notice that ∆z ≈ dz but dz is easier to compute.

cont’d

= 0.65

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### Functions of Three or More Variables

Linear approximations, differentiability, and differentials can be defined in a similar manner for functions of more than two variables. A differentiable function is defined by an expression similar to the one in Definition 7. For such functions the linear approximation is

f(x, y, z) ≈ f(a, b, c) + fx(a, b, c)(x – a) + fy(a, b, c)(y – b) + fz(a, b, c)(z – c)

and the linearization L(x, y, z) is the right side of this expression.

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### Functions of Three or More Variables

If w = f(x, y, z), then the increment of w is

∆w = f(x + ∆x, y + ∆y, z + ∆z) – f(x, y, z)

The differential dw is defined in terms of the differentials dx, dy, and dz of the independent variables by

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### Example 6

The dimensions of a rectangular box are measured to be 75 cm, 60 cm, and 40 cm, and each measurement is

correct to within 0.2 cm. Use differentials to estimate the largest possible error when the volume of the box is

calculated from these measurements.

Solution:

If the dimensions of the box are x, y, and z, its volume is V = xyz and so

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### Example 6 – Solution

We are given that |∆x| ≤ 0.2, |∆y| ≤ 0.2, and |∆z| ≤ 0.2.

To estimate the largest error in the volume, we therefore

use dx = 0.2, dy = 0.2, and dz = 0.2 together with x = 75, y = 60, and z = 40:

∆V ≈ dV = (60)(40)(0.2) + (75)(40)(0.2) + (75)(60)(0.2)

= 1980

cont’d

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### Example 6 – Solution

Thus an error of only 0.2 cm in measuring each dimension could lead to an error of approximately 1980 cm3 in the

calculated volume! This may seem like a large error, but it’s only about 1% of the volume of the box.

cont’d

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