274 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

3.5 Tangent Planes and Linear Approximations

In the same way that tangent lines played an important role for functions of onevariables, tangent planes play an important role for functions of two variables.We first learn how to derive them. We will then explain why they are important.

3.5.1 Tangent Planes

Goal: given a surface z = f (x, y) where f has continuous first partial deriva-tives, and a point P (x0, y0, z0) on the surface (see figure 3.14), we wish to findthe equation of the plane tangent to the surface at the given point. First, weneed to ask ourselves how this plane can be defined. For this, we let C1 be thecurve obtained by intersecting the surface and the plane y = y0, the blue curve.Similarly, let C2 be the curve obtained by intersecting the surface and the planex = x0, the red curve. Note that these two curves intersect at P (see figure). Let T1 be the line tangent to C1 at P . Let T2 be the line tangent to C2 atP . Then, the slope of T1 is fx (x0, y0) and the slope of T2 is fy (x0, y0). This isillustrated in figure 3.16. The plane tangent to the surface at the point wherethe two curves intersect is also shown.

Definition 3.5.1 The plane tangent to the surface at P is the plane containingT1 and T2.

Remark 3.5.2 At this point, it is important to understand that the techniqueto find tangent planes discussed in this section only applies to surfaces givenexplicitly, that is with an equation of the form z = f (x, y). Tangent planes oflevel surfaces, that is surfaces given by an equation of the form F (x, y, z) = Cwill be discussed in the next section.

Now, we want to derive the equation of this plane. The general form of thisequation is

A (x− x0) +B (y − y0) + C (z − z0) = 0 (3.7)

Dividing each side by C gives us

z − z0 = a (x− x0) + b (y − y0) (3.8)

where

a = −AC

and

b = −BC

The intersection of this plane with the plane y = y0 should be T1. To computeit, we simply set y = y0 in equation 3.8. This gives us{

z − z0 = a (x− x0)y = y0

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 275

Figure 3.14: Surface z = f (x, y) =−1

16

(x2 + y2

)+ 4 and the point P =(

1, 1,31

8

)

276 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

Figure 3.15:Blue curve: C1

Red curve: C2

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 277

Figure 3.16: Tangent plane

278 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

We should recognize the equation of a line (in point-slope form) in the xz-planewith slope a. But the slope of T1 is fx (x0, y0). Thus, a = fx (x0, y0). Similarly,we can see that b = fy (x0, y0). Therefore, we have:

Theorem 3.5.3 Suppose that a surface is given by z = f (x, y) where f hascontinuous first partials. Let P = (x0, y0, z0) be a point on the surface. Anequation of the tangent plane to the surface z = f (x, y) at P is

z − z0 = fx (x0, y0) (x− x0) + fy (x0, y0) (y − y0) (3.9)

Remark 3.5.4 We can also derive this result using a slightly different ap-proach. Recall that the equation of the plane through (x0, y0, z0) with normal〈a, b, c〉 is a (x− x0) + b (y − y0) + c (z − z0) = 0. We already have our pointon the tangent plane, it is (x0, y0, z0). To find a normal to the tangent plane,we find two vectors on the tangent plane and take their cross product. Usingthe notation above, A vector on the tangent plane will be a vector parallel toT1. Since the slope of T1 is fx (x0, y0) and T1 is in the xz-plane, a vector par-allel to T1 is 〈1, 0, fx (x0, y0)〉. Similarly, a vector on the tangent plane, parallelto T2 will be 〈0, 1, fy (x0, y0)〉. Hence, a vector normal to the tangent plane is〈1, 0, fx (x0, y0)〉 × 〈0, 1, fy (x0, y0)〉 = 〈−fx (x0y0) ,−fy (x0, y0) , 1〉. Hence, theequation of the tangent plane is

−fx (x0y0) (x− x0)− fy (x0, y0) (y − y0) + z − z0 = 0

orz − z0 = fx (x0, y0) (x− x0) + fy (x0, y0) (y − y0)

Example 3.5.5 Find the equation of the plane tangent to z = f (x, y) = x2+y2

at (.5, .5, .5).First, we compute the first order partials.

fx (x, y) = 2x

sofx (.5, .5) = 1

Alsofy (x, y) = 2y

sofy (.5, .5) = 1

Therefore, the equation of the tangent plane is

z − .5 = x− .5 + y − .5

orz = x+ y − .5

A picture of f (x, y) = x2 + y2 and its tangent plane are shown in figure 3.17.

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 279

­4 ­2

4

y ­10

­4

2 x

­2

0 00

2 4

z 5020

10

30

40

Figure 3.17: Graph of z = x2 + y2 and its tangent plane at (.5, .5, .5)

Example 3.5.6 Find the equation of the plane tangent to z = f (x, y) = ex2−y2

at (1,−1, 1).First, we compute the first partials.

fx (x, y) = 2xex2−y2

Sofx (1,−1) = 2

Alsofy (x, y) = −2yex

2−y2

Sofy (1,−1) = 2

It follows that the equation of the plane is

z − 1 = 2 (x− 1) + 2 (y + 1)

z = 2x+ 2y + 1

The graph of f and the tangent plane appear in figure 3.18

3.5.2 Linear Approximations

You will recall that one of the interpretations of the tangent line is that itapproximated a curve at the point of the tangent. More precisely, you remember

280 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

Figure 3.18: Graph of ex2−y2 and its tangent plane at (1,−1, 1)

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 281

that given a differentiable curve, if one zooms in close enough at one point, thecurve appears to be flat (like a line). The slope of that line is precisely thederivative of the curve at the given point. A similar analogy exists for tangentplanes. Using the notation of the previous section, if we zoom in close enoughto P , the surface z = f (x, y) will appear to be flat. It will be like the tangentplane at P . This means that as long as we are close enough to P , the functionz = f (x, y) can be approximated by the tangent plane whose equation is givenin 3.9.More precisely, we know from 3.9 that the equation of the tangent plane to

z = f (x, y) at (a, b, f (a, b)) is

z = f (a, b) + fx (a, b) (x− a) + fy (a, b) (y − b) (3.10)

Definition 3.5.7 The linear function defined by

L (x, y) = f (a, b) + fx (a, b) (x− a) + fy (a, b) (y − b) (3.11)

is called the linearization of f at (a, b).

Definition 3.5.8 The approximation

f (x, y) ≈ L (x, y)

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

Remark 3.5.9 The approximation f (x, y) ≈ L (x, y) is only valid for points(x, y) close to (a, b). For the linearization to be useful, we should be able tocompute f (a, b), fx (a, b) and fy (a, b) easily.

Example 3.5.10 Using example 3.5.6, approximate f (1.02,−.9). Compare itwith the exact answer.

f (1.02,−.9) = 2 (1.02) + 2 (−.9) + 1

= 1.24 (note we can do the computation by hand)

The exact value ise1.022−(−.9)2 = 1.259

Example 3.5.11 Find the linearization of f (x, y) = sinx+ cos y at (π, π) anduse it to approximate f (3.1, 2.9).We use the approximation f (3.1, 2.9) ≈ L (3.1, 2.9) where

L (x, y) = f (π, π) + fx (π, π) (x− π) + fy (π, π) (y − π)

We begin by computing first order partials.

fx (x, y) = cosx

282 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

so

fx (π, π) = cosπ

= −1

fy (x, y) = − sin y

so

fy (π, π) = − sinπ

= 0

Also

f (π, π) = sinπ + cosπ

= −1

HenceL (x, y) = −1− (x− π)

First, looking at figure , we can verify that at the point (π, π), f (x, y) ≈ L (x, y).

2

55

421

2 1 0

­2

­1z 0

1

0

x4 3

y

3

z = sinx+ cos y and its tangent plane at (π, π)

Now,

L (3.1, 2.9) = −1− (3.1− π)

= −0.958 41

Note that a calculator gives sin (3.1) + cos 2.9 = −0.929 38.

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 283

3.5.3 Differentiability

Loosely speaking, being differentiable means that locally (at each point), afunction can be linearized. In other words, its graphs near a point, if we zoomin close enough, will look like a plane. This means that the function can beapproximated by its tangent plane. This definition can be made much moreprecise, but we will not do it here. Instead, we give a theorem which provides asuffi cient condition for differentiability.

Theorem 3.5.12 If the partial derivatives fx and fy exist near a point (a, b)and are continuous at (a, b), then f is differentiable at (a, b).

Example 3.5.13 Show that f (x, y) = xexy is differentiable at (1, 0).We need to compute fx and fy and show that they exist near (1, 0) and arecontinuous at (1, 0).

fx (x, y) =∂x

∂xexy + x

∂exy

∂x

= exy + xexy∂ (xy)

∂x= exy + xyexy

This is continuous at (1, 0), in fact it is continuous everywhere (why?). It isalso defined at (1, 0), fx (1, 0) = 1.

fy (x, y) = x∂exy

∂y

= xexy∂ (xy)

∂y

= x2exy

This is also continuous at (1, 0), in fact it is continuous everywhere (why?). Itis also defined at (1, 0), fy (1, 0) = 1. Since both partial derivatives satisfy thecondition, it follows that f is differentiable at (1, 0).

3.5.4 The Differential

Given y = f (x), the differential of f , denoted dy or df was defined to be

dy = f ′ (x) dx

The equivalent for functions of several variables is:

Definition 3.5.14 Let z = f (x, y). the differential of f , denoted dz or df isdefined to be

dz = fx (x, y) dx+ fy (x, y) dy

=∂z

∂xdx+

∂z

∂ydy (3.12)

The differential tells us how small changes in x and y produce changes in z.

284 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

Example 3.5.15 Find the differential of the function z = f (x, y) = x2 + xy

dz =∂z

∂xdx+

∂z

∂ydy

= (2x+ y) dx+ xdy

We have similar formulas for functions of three or more variables. For ex-ample, if w = f (x, y, z), then

dw = fx (x, y, z) dx+ fy (x, y, z) dy + fz (x, y, z) dz

=∂w

∂xdx+

∂w

∂ydy +

∂w

∂zdz

3.5.5 Problems

Make sure you have read, studied and understood what was done above beforeattempting the problems.

1. Find an equation of the plane tangent to z = 4x2 − y2 + 2y at the point(−1, 2, 4).

2. Find an equation of the plane tangent to z = y cos (x− y) at the point(2, 2, 2).

3. Find an equation of the plane tangent to z = ln(x2 + y2

)at the point

(1, 0, 0).

4. Graph both z = x2 + xy + 3y2 and its tangent plane at (1, 1, 5). Zoomin close enough until the surface and the tangent plane become indistin-guishable.

5. Explain why f (x, y) = x√y is differentiable at the point (1, 4), then find

the linearization L (x, y) at the given point.

6. Explain why f (x, y) = tan−1 (x+ 2y) is differentiable at the point (1, 0),then find the linearization L (x, y) at the given point.

7. Find the linear approximation of f (x, y) =√

20− x2 − 7y2 at (2, 1) anduse it to approximate f (1.95, 1.08).

8. Find the differential dz of z = x3 ln y2.

9. Find the differential dR of R = αβ2 cos γ.

10. What is a normal to the surface given by z = f (x, y) at the point (a, b, c)where c = f (a, b) (hint: use the equation of the plane tangent to thesame surface at the given point, write it in the standard equation of aplane Ax+By + Cz +D = 0 remembering the meaning of the constantsA,B,C)?

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 285

3.5.6 Answers

1. Find an equation of the plane tangent to z = 4x2 − y2 + 2y at the point(−1, 2, 4).

z = −8x− 2y

2. Find an equation of the plane tangent to z = y cos (x− y) at the point(2, 2, 2).

z = y

3. Find an equation of the plane tangent to z = ln(x2 + y2

)at the point

(1, 0, 0).z = 2x− 2

4. Graph both z = x2 + xy + 3y2 and its tangent plane at (1, 1, 5). Zoomin close enough until the surface and the tangent plane become indistin-guishable.

z = 3x+ 7y − 5

The graphs are shown below.

­4

­50

­2

y4

­4

x200

0­2

2

4

50z

100

Graph of x2 + xy + 3y2 and its tangent plane at (1, 1, 5)

5. Explain why f (x, y) = x√y is differentiable at the point (1, 4), then find

the linearization L (x, y) at the given point.

We first compute the partials.∂f

∂x=√y hence

∂f (1, 4)

∂x= 2 and

∂f

∂y=

286 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES

x

2√yhence

∂f (1, 4)

∂y=

1

4. We see that f is differentiable at (1, 4) because

both partials exist and are continuous at (1, 4). The linearization is

L (x, y) = f (1, 4) +∂f (1, 4)

∂x(x− 1) +

∂f (1, 4)

∂y(y − 4)

= 2 + 2 (x− 1) +1

4(y − 4)

= 2 + 2x− 2 +1

4y − 1

= 2x+1

4y − 1

6. Explain why f (x, y) = tan−1 (x+ 2y) is differentiable at the point (1, 0),then find the linearization L (x, y) at the given point.

L (x, y) =1

2x+ y +

π

4− 1

2

7. Find the linear approximation of f (x, y) =√

20− x2 − 7y2 at (2, 1) anduse it to approximate f (1.95, 1.08).

L (x, y) =−2x− 7y + 20

3

Therefore

f (1.95, 1.08) ≈ L (1.95, 1.08)

=−2 (1.95)− 7 (1.08) + 20

3= 2. 846 7

Note that the exact value is√

20− (1.95)2 − 7 (1.08)

2= 2. 834 2.

8. Find the differential of z = x3 ln y2

dz =∂z

∂xdx+

∂z

∂ydy

= x2 ln y2dx+2x3

ydy

9. Find the differential of R = αβ2 cos γ

dR = β2 cos γdα+ 2αβ cos γdβ − αβ2 sin γdγ

10. What is a normal to the surface given by z = f (x, y) at the point (a, b, c)where c = f (a, b) (hint: use the equation of the plane tangent to the

3.5. TANGENT PLANES AND LINEAR APPROXIMATIONS 287

same surface at the given point, write it in the standard equation of aplane Ax+By + Cz +D = 0 remembering the meaning of the constantsA,B,C)?

A vector normal to the surface at (a, b, c) is⟨−∂f (a, b)

∂x,−∂f (a, b)

∂y, 1

⟩.

Bibliography

[1] Joel Hass, Maurice D. Weir, and George B. Thomas, University calculus:Early transcendentals, Pearson Addison-Wesley, 2012.

[2] James Stewart, Calculus, Cengage Learning, 2011.

[3] Michael Sullivan and Kathleen Miranda, Calculus: Early transcendentals,Macmillan Higher Education, 2014.

615