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Calculus Volume 3

4.4 Tangent Planes and Linear Approximations

Calculus Volume 34.4 Tangent Planes and Linear Approximations
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 4.4.1. Determine the equation of a plane tangent to a given surface at a point.
  • 4.4.2. Use the tangent plane to approximate a function of two variables at a point.
  • 4.4.3. Explain when a function of two variables is differentiable.
  • 4.4.4. Use the total differential to approximate the change in a function of two variables.

In this section, we consider the problem of finding the tangent plane to a surface, which is analogous to finding the equation of a tangent line to a curve when the curve is defined by the graph of a function of one variable, y=f(x).y=f(x). The slope of the tangent line at the point x=ax=a is given by m=f(a);m=f(a); what is the slope of a tangent plane? We learned about the equation of a plane in Equations of Lines and Planes in Space; in this section, we see how it can be applied to the problem at hand.

Tangent Planes

Intuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensional space, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at that point. A tangent plane at a regular point contains all of the lines tangent to that point. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.

Definition

Let P0=(x0,y0,z0)P0=(x0,y0,z0) be a point on a surface S,S, and let CC be any curve passing through P0P0 and lying entirely in S.S. If the tangent lines to all such curves CC at P0P0 lie in the same plane, then this plane is called the tangent plane to SS at P0P0 (Figure 4.27).

A surface S is shown with a point P0 = (x0, y0, z0). There are two intersecting curves shown on S that pass through P0. There are tangents drawn for each of these curves at P0, and these tangent lines create a plane, namely, the tangent plane at P0.
Figure 4.27 The tangent plane to a surface SS at a point P0P0 contains all the tangent lines to curves in SS that pass through P0.P0.

For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface to be differentiable at that point, defined later in this section. We define the term tangent plane here and then explore the idea intuitively.

Definition

Let SS be a surface defined by a differentiable function z=f(x,y),z=f(x,y), and let P0=(x0,y0)P0=(x0,y0) be a point in the domain of f.f. Then, the equation of the tangent plane to SS at P0P0 is given by

z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0).z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0).
4.24

To see why this formula is correct, let’s first find two tangent lines to the surface S.S. The equation of the tangent line to the curve that is represented by the intersection of SS with the vertical trace given by x=x0x=x0 is z=f(x0,y0)+fy(x0,y0)(yy0).z=f(x0,y0)+fy(x0,y0)(yy0). Similarly, the equation of the tangent line to the curve that is represented by the intersection of SS with the vertical trace given by y=y0y=y0 is z=f(x0,y0)+fx(x0,y0)(xx0).z=f(x0,y0)+fx(x0,y0)(xx0). A parallel vector to the first tangent line is a=j+fy(x0,y0)k;a=j+fy(x0,y0)k; a parallel vector to the second tangent line is b=i+fx(x0,y0)k.b=i+fx(x0,y0)k. We can take the cross product of these two vectors:

a×b=(j+fy(x0,y0)k)×(i+fx(x0,y0)k)=|ijk01fy(x0,y0)10fx(x0,y0)|=fx(x0,y0)i+fy(x0,y0)jk.a×b=(j+fy(x0,y0)k)×(i+fx(x0,y0)k)=|ijk01fy(x0,y0)10fx(x0,y0)|=fx(x0,y0)i+fy(x0,y0)jk.

This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as a normal vector to the tangent plane, along with the point P0=(x0,y0,f(x0,y0))P0=(x0,y0,f(x0,y0)) in the equation for a plane:

n·((xx0)i+(yy0)j+(zf(x0,y0))k)=0(fx(x0,y0)i+fy(x0,y0)j-k)·((xx0)i+(yy0)j+(zf(x0,y0))k)=0fx(x0,y0)(xx0)+fy(x0,y0)(yy0)(zf(x0,y0))=0.n·((xx0)i+(yy0)j+(zf(x0,y0))k)=0(fx(x0,y0)i+fy(x0,y0)j-k)·((xx0)i+(yy0)j+(zf(x0,y0))k)=0fx(x0,y0)(xx0)+fy(x0,y0)(yy0)(zf(x0,y0))=0.

Solving this equation for zz gives Equation 4.24.

Example 4.21

Finding a Tangent Plane

Find the equation of the tangent plane to the surface defined by the function f(x,y)=2x23xy+8y2+2x4y+4f(x,y)=2x23xy+8y2+2x4y+4 at point (2,−1).(2,−1).

Solution

First, we must calculate fx(x,y)fx(x,y) and fy(x,y),fy(x,y), then use Equation 4.24 with x0=2x0=2 and y0=−1:y0=−1:

fx(x,y)=4x3y+2fy(x,y)=−3x+16y4f(2,−1)=2(2)23(2)(−1)+8(−1)2+2(2)4(−1)+4=34.fx(2,−1)=4(2)3(−1)+2=13fy(2,−1)=−3(2)+16(−1)4=−26.fx(x,y)=4x3y+2fy(x,y)=−3x+16y4f(2,−1)=2(2)23(2)(−1)+8(−1)2+2(2)4(−1)+4=34.fx(2,−1)=4(2)3(−1)+2=13fy(2,−1)=−3(2)+16(−1)4=−26.

Then Equation 4.24 becomes

z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)z=34+13(x2)26(y(−1))z=34+13x2626y26z=13x26y18.z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)z=34+13(x2)26(y(−1))z=34+13x2626y26z=13x26y18.

(See the following figure).

A curved surface f(x, y) = 2x2 – 3xy + 8y2 + 2x + 4y + 4 with tangent plane z = 13x – 26y – 18 at point (2, –1, 34).
Figure 4.28 Calculating the equation of a tangent plane to a given surface at a given point.
Checkpoint 4.19

Find the equation of the tangent plane to the surface defined by the function f(x,y)=x3x2y+y22x+3y2f(x,y)=x3x2y+y22x+3y2 at point (−1,3).(−1,3).

Example 4.22

Finding Another Tangent Plane

Find the equation of the tangent plane to the surface defined by the function f(x,y)=sin(2x)cos(3y)f(x,y)=sin(2x)cos(3y) at the point (π/3,π/4).(π/3,π/4).

Solution

First, calculate fx(x,y)fx(x,y) and fy(x,y),fy(x,y), then use Equation 4.24 with x0=π/3x0=π/3 and y0=π/4:y0=π/4:

fx(x,y)=2cos(2x)cos(3y)fy(x,y)=−3sin(2x)sin(3y)f(π3,π4)=sin(2(π3))cos(3(π4))=(32)(22)=64fx(π3,π4)=2cos(2(π3))cos(3(π4))=2(12)(22)=22fy(π3,π4)=−3sin(2(π3))sin(3(π4))=−3(32)(22)=364.fx(x,y)=2cos(2x)cos(3y)fy(x,y)=−3sin(2x)sin(3y)f(π3,π4)=sin(2(π3))cos(3(π4))=(32)(22)=64fx(π3,π4)=2cos(2(π3))cos(3(π4))=2(12)(22)=22fy(π3,π4)=−3sin(2(π3))sin(3(π4))=−3(32)(22)=364.

Then Equation 4.24 becomes

z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)z=64+22(xπ3)364(yπ4)z=22x364y64π26+3π616.z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)z=64+22(xπ3)364(yπ4)z=22x364y64π26+3π616.

A tangent plane to a surface does not always exist at every point on the surface. Consider the function

f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0).f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0).

The graph of this function follows.

A curved surface that passes through (0, 0, 0) and that folds up on either side of the z axis.
Figure 4.29 Graph of a function that does not have a tangent plane at the origin.

If either x=0x=0 or y=0,y=0, then f(x,y)=0,f(x,y)=0, so the value of the function does not change on either the x- or y-axis. Therefore, fx(x,0)=fy(0,y)=0,fx(x,0)=fy(0,y)=0, so as either xoryxory approach zero, these partial derivatives stay equal to zero. Substituting them into Equation 4.24 gives z=0z=0 as the equation of the tangent line. However, if we approach the origin from a different direction, we get a different story. For example, suppose we approach the origin along the line y=x.y=x. If we put y=xy=x into the original function, it becomes

f(x,x)=x(x)x2+(x)2=x22x2=|x|2.f(x,x)=x(x)x2+(x)2=x22x2=|x|2.

When x>0,x>0, the slope of this curve is equal to 2/2;2/2; when x<0,x<0, the slope of this curve is equal to (2/2).(2/2). This presents a problem. In the definition of tangent plane, we presumed that all tangent lines through point PP (in this case, the origin) lay in the same plane. This is clearly not the case here. When we study differentiable functions, we will see that this function is not differentiable at the origin.

Linear Approximations

Recall from Linear Approximations and Differentials that the formula for the linear approximation of a function f(x)f(x) at the point x=ax=a is given by

yf(a)+f(a)(xa).yf(a)+f(a)(xa).

The diagram for the linear approximation of a function of one variable appears in the following graph.

A curve in the xy plane with a point and a tangent to that point. The figure is marked tangent line approximation.
Figure 4.30 Linear approximation of a function in one variable.

The tangent line can be used as an approximation to the function f(x)f(x) for values of xx reasonably close to x=a.x=a. When working with a function of two variables, the tangent line is replaced by a tangent plane, but the approximation idea is much the same.

Definition

Given a function z=f(x,y)z=f(x,y) with continuous partial derivatives that exist at the point (x0,y0),(x0,y0), the linear approximation of ff at the point (x0,y0)(x0,y0) is given by the equation

L(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0).L(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0).
4.25

Notice that this equation also represents the tangent plane to the surface defined by z=f(x,y)z=f(x,y) at the point (x0,y0).(x0,y0). The idea behind using a linear approximation is that, if there is a point (x0,y0)(x0,y0) at which the precise value of f(x,y)f(x,y) is known, then for values of (x,y)(x,y) reasonably close to (x0,y0),(x0,y0), the linear approximation (i.e., tangent plane) yields a value that is also reasonably close to the exact value of f(x,y)f(x,y) (Figure 4.31). Furthermore the plane that is used to find the linear approximation is also the tangent plane to the surface at the point (x0,y0).(x0,y0).

A paraboloid with surface z = f(x, y). There is a point given on the paraboloid P (x0, y0) with a tangent plane at that point. There is a point on the plane which is marked as the linear approximation L(x, y) to f(x0, y0), which is close to the corresponding point on the paraboloid.
Figure 4.31 Using a tangent plane for linear approximation at a point.

Example 4.23

Using a Tangent Plane Approximation

Given the function f(x,y)=414x2y2,f(x,y)=414x2y2, approximate f(2.1,2.9)f(2.1,2.9) using point (2,3)(2,3) for (x0,y0).(x0,y0). What is the approximate value of f(2.1,2.9)f(2.1,2.9) to four decimal places?

Solution

To apply Equation 4.25, we first must calculate f(x0,y0),f(x0,y0), fx(x0,y0),fx(x0,y0), and fy(x0,y0)fy(x0,y0) using x0=2x0=2 and y0=3:y0=3:

f(x0,y0)=f(2,3)=414(2)2(3)2=41169=16=4fx(x,y)=4x414x2y2sofx(x0,y0)=4(2)414(2)2(3)2=−2fy(x,y)=y414x2y2sofy(x0,y0)=3414(2)2(3)2=34.f(x0,y0)=f(2,3)=414(2)2(3)2=41169=16=4fx(x,y)=4x414x2y2sofx(x0,y0)=4(2)414(2)2(3)2=−2fy(x,y)=y414x2y2sofy(x0,y0)=3414(2)2(3)2=34.

Now we substitute these values into Equation 4.25:

L(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)=42(x2)34(y3)=4142x34y.L(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)=42(x2)34(y3)=4142x34y.

Last, we substitute x=2.1x=2.1 and y=2.9y=2.9 into L(x,y):L(x,y):

L(2.1,2.9)=4142(2.1)34(2.9)=10.254.22.175=3.875.L(2.1,2.9)=4142(2.1)34(2.9)=10.254.22.175=3.875.

The approximate value of f(2.1,2.9)f(2.1,2.9) to four decimal places is

f(2.1,2.9)=414(2.1)2(2.9)2=14.953.8665,f(2.1,2.9)=414(2.1)2(2.9)2=14.953.8665,

which corresponds to a 0.2%0.2% error in approximation.

Checkpoint 4.20

Given the function f(x,y)=e52x+3y,f(x,y)=e52x+3y, approximate f(4.1,0.9)f(4.1,0.9) using point (4,1)(4,1) for (x0,y0).(x0,y0). What is the approximate value of f(4.1,0.9)f(4.1,0.9) to four decimal places?

Differentiability

When working with a function y=f(x)y=f(x) of one variable, the function is said to be differentiable at a point x=ax=a if f(a)f(a) exists. Furthermore, if a function of one variable is differentiable at a point, the graph is “smooth” at that point (i.e., no corners exist) and a tangent line is well-defined at that point.

The idea behind differentiability of a function of two variables is connected to the idea of smoothness at that point. In this case, a surface is considered to be smooth at point PP if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point. Recall the formula for a tangent plane at a point (x0,y0)(x0,y0) is given by

z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0),z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0),

For a tangent plane to exist at the point (x0,y0),(x0,y0), the partial derivatives must therefore exist at that point. However, this is not a sufficient condition for smoothness, as was illustrated in Figure 4.29. In that case, the partial derivatives existed at the origin, but the function also had a corner on the graph at the origin.

Definition

A function f(x,y)f(x,y) is differentiable at a point P(x0,y0)P(x0,y0) if, for all points (x,y)(x,y) in a δδ disk around P,P, we can write

f(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)+E(x,y),f(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)+E(x,y),
4.26

where the error term EE satisfies

lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=0.lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=0.

The last term in Equation 4.26 is referred to as the error term and it represents how closely the tangent plane comes to the surface in a small neighborhood (δ(δ disk) of point P.P. For the function ff to be differentiable at P,P, the function must be smooth—that is, the graph of ff must be close to the tangent plane for points near P.P.

Example 4.24

Demonstrating Differentiability

Show that the function f(x,y)=2x24yf(x,y)=2x24y is differentiable at point (2,−3).(2,−3).

Solution

First, we calculate f(x0,y0),fx(x0,y0),andfy(x0,y0)f(x0,y0),fx(x0,y0),andfy(x0,y0) using x0=2x0=2 and y0=−3,y0=−3, then we use Equation 4.26:

f(2,−3)=2(2)24(−3)=8+12=20fx(2,−3)=4(2)=8fy(2,−3)=−4.f(2,−3)=2(2)24(−3)=8+12=20fx(2,−3)=4(2)=8fy(2,−3)=−4.

Therefore m1=8m1=8 and m2=−4,m2=−4, and Equation 4.26 becomes

f(x,y)=f(2,−3)+fx(2,−3)(x2)+fy(2,−3)(y+3)+E(x,y)2x24y=20+8(x2)4(y+3)+E(x,y)2x24y=20+8x164y12+E(x,y)2x24y=8x4y8+E(x,y)E(x,y)=2x28x+8.f(x,y)=f(2,−3)+fx(2,−3)(x2)+fy(2,−3)(y+3)+E(x,y)2x24y=20+8(x2)4(y+3)+E(x,y)2x24y=20+8x164y12+E(x,y)2x24y=8x4y8+E(x,y)E(x,y)=2x28x+8.

Next, we calculate lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2:lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2:

lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=lim(x,y)(2,−3)2x28x+8(x2)2+(y+3)2=lim(x,y)(2,−3)2(x24x+4)(x2)2+(y+3)2=lim(x,y)(2,−3)2(x2)2(x2)2+(y+3)2lim(x,y)(2,−3)2((x2)2+(y+3)2)(x2)2+(y+3)2=lim(x,y)(2,−3)2(x2)2+(y+3)2=0.lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=lim(x,y)(2,−3)2x28x+8(x2)2+(y+3)2=lim(x,y)(2,−3)2(x24x+4)(x2)2+(y+3)2=lim(x,y)(2,−3)2(x2)2(x2)2+(y+3)2lim(x,y)(2,−3)2((x2)2+(y+3)2)(x2)2+(y+3)2=lim(x,y)(2,−3)2(x2)2+(y+3)2=0.

Since E(x,y)0E(x,y)0 for any value of xory,xory, the original limit must be equal to zero. Therefore, f(x,y)=2x24yf(x,y)=2x24y is differentiable at point (2,−3).(2,−3).

Checkpoint 4.21

Show that the function f(x,y)=3x4y2f(x,y)=3x4y2 is differentiable at point (−1,2).(−1,2).

The function f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0)f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0) is not differentiable at the origin. We can see this by calculating the partial derivatives. This function appeared earlier in the section, where we showed that fx(0,0)=fy(0,0)=0.fx(0,0)=fy(0,0)=0. Substituting this information into Equation 4.26 using x0=0x0=0 and y0=0,y0=0, we get

f(x,y)=f(0,0)+fx(0,0)(x0)+fy(0,0)(y0)+E(x,y)E(x,y)=xyx2+y2.f(x,y)=f(0,0)+fx(0,0)(x0)+fy(0,0)(y0)+E(x,y)E(x,y)=xyx2+y2.

Calculating lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2 gives

lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=lim(x,y)(0,0)xyx2+y2x2+y2=lim(x,y)(0,0)xyx2+y2.lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=lim(x,y)(0,0)xyx2+y2x2+y2=lim(x,y)(0,0)xyx2+y2.

Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and the function ff is not differentiable at the origin as shown in the following figure.

A curved surface in xyz space that remains constant along the positive x axis and curves downward along the line y = –x in the second quadrant.
Figure 4.32 This function f(x,y)f(x,y) is not differentiable at the origin.

Differentiability and continuity for functions of two or more variables are connected, the same as for functions of one variable. In fact, with some adjustments of notation, the basic theorem is the same.

Theorem 4.6

Differentiability Implies Continuity

Let z=f(x,y)z=f(x,y) be a function of two variables with (x0,y0)(x0,y0) in the domain of f.f. If f(x,y)f(x,y) is differentiable at (x0,y0),(x0,y0), then f(x,y)f(x,y) is continuous at (x0,y0).(x0,y0).

Differentiability Implies Continuity shows that if a function is differentiable at a point, then it is continuous there. However, if a function is continuous at a point, then it is not necessarily differentiable at that point. For example,

f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0)f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0)

is continuous at the origin, but it is not differentiable at the origin. This observation is also similar to the situation in single-variable calculus.

Continuity of First Partials Implies Differentiability further explores the connection between continuity and differentiability at a point. This theorem says that if the function and its partial derivatives are continuous at a point, the function is differentiable.

Theorem 4.7

Continuity of First Partials Implies Differentiability

Let z=f(x,y)z=f(x,y) be a function of two variables with (x0,y0)(x0,y0) in the domain of f.f. If f(x,y),f(x,y), fx(x,y),fx(x,y), and fy(x,y)fy(x,y) all exist in a neighborhood of (x0,y0)(x0,y0) and are continuous at (x0,y0),(x0,y0), then f(x,y)f(x,y) is differentiable there.

Recall that earlier we showed that the function

f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0)f(x,y)={xyx2+y2(x,y)(0,0)0(x,y)=(0,0)

was not differentiable at the origin. Let’s calculate the partial derivatives fxfx and fy:fy:

fx=y3(x2+y2)3/2andfy=x3(x2+y2)3/2.fx=y3(x2+y2)3/2andfy=x3(x2+y2)3/2.

The contrapositive of the preceding theorem states that if a function is not differentiable, then at least one of the hypotheses must be false. Let’s explore the condition that fx(0,0)fx(0,0) must be continuous. For this to be true, it must be true that lim(x,y)(0,0)fx(0,0)=fx(0,0):lim(x,y)(0,0)fx(0,0)=fx(0,0):

lim(x,y)(0,0)fx(x,y)=lim(x,y)(0,0)y3(x2+y2)3/2.lim(x,y)(0,0)fx(x,y)=lim(x,y)(0,0)y3(x2+y2)3/2.

Let x=ky.x=ky. Then

lim(x,y)(0,0)y3(x2+y2)3/2=limy0y3((ky)2+y2)3/2=limy0y3(k2y2+y2)3/2=limy0y3|y|3(k2+1)3/2=1(k2+1)3/2limy0|y|y.lim(x,y)(0,0)y3(x2+y2)3/2=limy0y3((ky)2+y2)3/2=limy0y3(k2y2+y2)3/2=limy0y3|y|3(k2+1)3/2=1(k2+1)3/2limy0|y|y.

If y>0,y>0, then this expression equals 1/(k2+1)3/2;1/(k2+1)3/2; if y<0,y<0, then it equals (1/(k2+1)3/2).(1/(k2+1)3/2). In either case, the value depends on k,k, so the limit fails to exist.

Differentials

In Linear Approximations and Differentials we first studied the concept of differentials. The differential of y,y, written dy,dy, is defined as f(x)dx.f(x)dx. The differential is used to approximate Δy=f(x+Δx)f(x),Δy=f(x+Δx)f(x), where Δx=dx.Δx=dx. Extending this idea to the linear approximation of a function of two variables at the point (x0,y0)(x0,y0) yields the formula for the total differential for a function of two variables.

Definition

Let z=f(x,y)z=f(x,y) be a function of two variables with (x0,y0)(x0,y0) in the domain of f,f, and let ΔxΔx and ΔyΔy be chosen so that (x0+Δx,y0+Δy)(x0+Δx,y0+Δy) is also in the domain of f.f. If ff is differentiable at the point (x0,y0),(x0,y0), then the differentials dxdx and dydy are defined as

dx=Δxanddy=Δy.dx=Δxanddy=Δy.

The differential dz,dz, also called the total differential of z=f(x,y)z=f(x,y) at (x0,y0),(x0,y0), is defined as

dz=fx(x0,y0)dx+fy(x0,y0)dy.dz=fx(x0,y0)dx+fy(x0,y0)dy.
4.27

Notice that the symbol is not used to denote the total differential; rather, dd appears in front of z.z. Now, let’s define Δz=f(x+Δx,y+Δy)f(x,y).Δz=f(x+Δx,y+Δy)f(x,y). We use dzdz to approximate Δz,Δz, so

Δzdz=fx(x0,y0)dx+fy(x0,y0)dy.Δzdz=fx(x0,y0)dx+fy(x0,y0)dy.

Therefore, the differential is used to approximate the change in the function z=f(x0,y0)z=f(x0,y0) at the point (x0,y0)(x0,y0) for given values of ΔxΔx and Δy.Δy. Since Δz=f(x+Δx,y+Δy)f(x,y),Δz=f(x+Δx,y+Δy)f(x,y), this can be used further to approximate f(x+Δx,y+Δy):f(x+Δx,y+Δy):

f(x+Δx,y+Δy)=f(x,y)+Δzf(x,y)+fx(x0,y0)Δx+fy(x0,y0)Δy.f(x+Δx,y+Δy)=f(x,y)+Δzf(x,y)+fx(x0,y0)Δx+fy(x0,y0)Δy.

See the following figure.

A surface f in the xyz plane, with a tangent plane at the point (x, y, f(x, y)). On the (x, y) plane, there is a point marked (x + Δx, y + Δy). There is a dashed line to the corresponding point on the graph of f and the line then continues to the tangent plane; the distance to the graph of f is marked f(x + + Δx, y + Δy), and the distance to the tangent plane is marked as the linear approximation.
Figure 4.33 The linear approximation is calculated via the formula f(x+Δx,y+Δy)f(x,y)+fx(x0,y0)Δx+fy(x0,y0)Δy.f(x+Δx,y+Δy)f(x,y)+fx(x0,y0)Δx+fy(x0,y0)Δy.

One such application of this idea is to determine error propagation. For example, if we are manufacturing a gadget and are off by a certain amount in measuring a given quantity, the differential can be used to estimate the error in the total volume of the gadget.

Example 4.25

Approximation by Differentials

Find the differential dzdz of the function f(x,y)=3x22xy+y2f(x,y)=3x22xy+y2 and use it to approximate ΔzΔz at point (2,−3).(2,−3). Use Δx=0.1Δx=0.1 and Δy=−0.05.Δy=−0.05. What is the exact value of Δz?Δz?

Solution

First, we must calculate f(x0,y0),fx(x0,y0),andfy(x0,y0)f(x0,y0),fx(x0,y0),andfy(x0,y0) using x0=2x0=2 and y0=−3:y0=−3:

f(x0,y0)=f(2,−3)=3(2)22(2)(−3)+(−3)2=12+12+9=33fx(x,y)=6x2yfy(x,y)=−2x+2yfx(x0,y0)=fx(2,−3)=6(2)2(−3)=12+6=18fy(x0,y0)=fy(2,−3)=−2(2)+2(−3)=−46=−10.f(x0,y0)=f(2,−3)=3(2)22(2)(−3)+(−3)2=12+12+9=33fx(x,y)=6x2yfy(x,y)=−2x+2yfx(x0,y0)=fx(2,−3)=6(2)2(−3)=12+6=18fy(x0,y0)=fy(2,−3)=−2(2)+2(−3)=−46=−10.

Then, we substitute these quantities into Equation 4.27:

dz=fx(x0,y0)dx+fy(x0,y0)dydz=18(0.1)10(−0.05)=1.8+0.5=2.3.dz=fx(x0,y0)dx+fy(x0,y0)dydz=18(0.1)10(−0.05)=1.8+0.5=2.3.

This is the approximation to Δz=f(x0+Δx,y0+Δy)f(x0,y0).Δz=f(x0+Δx,y0+Δy)f(x0,y0). The exact value of ΔzΔz is given by

Δz=f(x0+Δx,y0+Δy)f(x0,y0)=f(2+0.1,−30.05)f(2,−3)=f(2.1,−3.05)f(2,−3)=2.3425.Δz=f(x0+Δx,y0+Δy)f(x0,y0)=f(2+0.1,−30.05)f(2,−3)=f(2.1,−3.05)f(2,−3)=2.3425.
Checkpoint 4.22

Find the differential dzdz of the function f(x,y)=4y2+x2y2xyf(x,y)=4y2+x2y2xy and use it to approximate ΔzΔz at point (1,−1).(1,−1). Use Δx=0.03Δx=0.03 and Δy=−0.02.Δy=−0.02. What is the exact value of Δz?Δz?

Differentiability of a Function of Three Variables

All of the preceding results for differentiability of functions of two variables can be generalized to functions of three variables. First, the definition:

Definition

A function f(x,y,z)f(x,y,z) is differentiable at a point P(x0,y0,z0)P(x0,y0,z0) if for all points (x,y,z)(x,y,z) in a δδ disk around PP we can write

f(x,y)=f(x0,y0,z0)+fx(x0,y0,z0)(xx0)+fy(x0,y0,z0)(yy0)+fz(x0,y0,z0)(zz0)+E(x,y,z),f(x,y)=f(x0,y0,z0)+fx(x0,y0,z0)(xx0)+fy(x0,y0,z0)(yy0)+fz(x0,y0,z0)(zz0)+E(x,y,z),
4.28

where the error term E satisfies

lim(x,y,z)(x0,y0,z0)E(x,y,z)(xx0)2+(yy0)2+(zz0)2=0.lim(x,y,z)(x0,y0,z0)E(x,y,z)(xx0)2+(yy0)2+(zz0)2=0.

If a function of three variables is differentiable at a point (x0,y0,z0),(x0,y0,z0), then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.

Section 4.4 Exercises

For the following exercises, find a unit normal vector to the surface at the indicated point.

163.

f(x,y)=x3,(2,−1,8)f(x,y)=x3,(2,−1,8)

164.

ln(xyz)=0ln(xyz)=0 when x=y=1x=y=1

For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point P.P.

165.

x2+xy+y2=3,P(−1,−1)x2+xy+y2=3,P(−1,−1)

166.

(x2+y2)2=9(x2y2),P(2,1)(x2+y2)2=9(x2y2),P(2,1)

167.

xy22x2+y+5x=6,P(4,2)xy22x2+y+5x=6,P(4,2)

168.

2x3x2y2=3xy7,P(1,−2)2x3x2y2=3xy7,P(1,−2)

169.

zex2y23=0,zex2y23=0, P(2,2,3)P(2,2,3)

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for zz in terms of xx and y.)y.)

170.

−8x3y7z=−19,P(1,−1,2)−8x3y7z=−19,P(1,−1,2)

171.

z=−9x23y2,P(2,1,−39)z=−9x23y2,P(2,1,−39)

172.

x2+10xyz+y2+8z2=0,P(−1,−1,−1)x2+10xyz+y2+8z2=0,P(−1,−1,−1)

173.

z=ln(10x2+2y2+1),P(0,0,0)z=ln(10x2+2y2+1),P(0,0,0)

174.

z=e7x2+4y2,z=e7x2+4y2, P(0,0,1)P(0,0,1)

175.

xy+yz+zx=11,P(1,2,3)xy+yz+zx=11,P(1,2,3)

176.

x2+4y2=z2,P(3,2,5)x2+4y2=z2,P(3,2,5)

177.

x3+y3=3xyz,P(1,2,32)x3+y3=3xyz,P(1,2,32)

178.

z=axy,P(1,1a,1)z=axy,P(1,1a,1)

179.

z=sinx+siny+sin(x+y),P(0,0,0)z=sinx+siny+sin(x+y),P(0,0,0)

180.

h(x,y)=lnx2+y2,P(3,4)h(x,y)=lnx2+y2,P(3,4)

181.

z=x22xy+y2,P(1,2,1)z=x22xy+y2,P(1,2,1)

For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, P0(x0,y0,z0),P0(x0,y0,z0), and a vector n=a,b,cn=a,b,c that is parallel to the line. Then the equation of the line is xx0=at,yy0=bt,zz0=ct.)xx0=at,yy0=bt,zz0=ct.)

182.

−3x+9y+4z=−4,P(1,−1,2)−3x+9y+4z=−4,P(1,−1,2)

183.

z=5x22y2,P(2,1,18)z=5x22y2,P(2,1,18)

184.

x28xyz+y2+6z2=0,P(1,1,1)x28xyz+y2+6z2=0,P(1,1,1)

185.

z=ln(3x2+7y2+1),P(0,0,0)z=ln(3x2+7y2+1),P(0,0,0)

186.

z=e4x2+6y2,P(0,0,1)z=e4x2+6y2,P(0,0,1)

187.

z=x22xy+y2z=x22xy+y2 at point P(1,2,1)P(1,2,1)

For the following exercises, use the figure shown here.

A surface in the xyz plane is marked as z = f(x, y). This surface has a tangent plane at (x0, y0, z0), with the corresponding point (x0, y0) marked on the xy plane. Also marked on the xy plane is the point (x0 + Δx, y0 + Δy). From this point, a line is drawn to the surface and three points are marked. The first point is C, which is (x0 + Δx, y0 + Δy, z0), then there is B, which is on the tangent plane, and then there is A, which is on the surface. The distance between B and C is marked df(x0, y0).
188.

The length of line segment ACAC is equal to what mathematical expression?

189.

The length of line segment BCBC is equal to what mathematical expression?

190.

Using the figure, explain what the length of line segment ABAB represents.

For the following exercises, complete each task.

191.

Show that f(x,y)=exyxf(x,y)=exyx is differentiable at point (1,0).(1,0).

192.

Find the total differential of the function w=eycos(x)+z2.w=eycos(x)+z2.

193.

Show that f(x,y)=x2+3yf(x,y)=x2+3y is differentiable at every point. In other words, show that Δz=f(x+Δx,y+Δy)f(x,y)=fxΔx+fyΔy+ε1Δx+ε2Δy,Δz=f(x+Δx,y+Δy)f(x,y)=fxΔx+fyΔy+ε1Δx+ε2Δy, where both ε1ε1 and ε2ε2 approach zero as (Δx,Δy)(Δx,Δy) approaches (0,0).(0,0).

194.

Find the total differential of the function z=xyy+xz=xyy+x where xx changes from 10to10.510to10.5 and yy changes from 15to13.15to13.

195.

Let z=f(x,y)=xey.z=f(x,y)=xey. Compute ΔzΔz from P(1,2)P(1,2) to Q(1.05,2.1)Q(1.05,2.1) and then find the approximate change in zz from point PP to point Q.Q. Recall Δz=f(x+Δx,y+Δy)f(x,y),Δz=f(x+Δx,y+Δy)f(x,y), and dzdz and ΔzΔz are approximately equal.

196.

The volume of a right circular cylinder is given by V(r,h)=πr2h.V(r,h)=πr2h. Find the differential dV.dV. Interpret the formula geometrically.

197.

See the preceding problem. Use differentials to estimate the volume of aluminum in an enclosed aluminum can with diameter 8.0cm8.0cm and height 12cm12cm if the aluminum is 0.040.04 cm thick.

198.

Use the differential dzdz to approximate the change in z=4x2y2z=4x2y2 as (x,y)(x,y) moves from point (1,1)(1,1) to point (1.01,0.97).(1.01,0.97). Compare this approximation with the actual change in the function.

199.

Let z=f(x,y)=x2+3xyy2.z=f(x,y)=x2+3xyy2. Find the exact change in the function and the approximate change in the function as xx changes from 2.00to2.052.00to2.05 and yy changes from 3.00to2.96.3.00to2.96.

200.

The centripetal acceleration of a particle moving in a circle is given by a(r,v)=v2r,a(r,v)=v2r, where vv is the velocity and rr is the radius of the circle. Approximate the maximum percent error in measuring the acceleration resulting from errors of 3%3% in vv and 2%2% in r.r. (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in aa is given by daa.)daa.)

201.

The radius rr and height hh of a right circular cylinder are measured with possible errors of 4%and5%,4%and5%, respectively. Approximate the maximum possible percentage error in measuring the volume (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in VV is given by dVV.)dVV.)

202.

The base radius and height of a right circular cone are measured as 1010 in. and 2525 in., respectively, with a possible error in measurement of as much as 0.10.1 in. each. Use differentials to estimate the maximum error in the calculated volume of the cone.

203.

The electrical resistance RR produced by wiring resistors R1R1 and R2R2 in parallel can be calculated from the formula 1R=1R1+1R2.1R=1R1+1R2. If R1R1 and R2R2 are measured to be 7Ω7Ω and 6Ω,6Ω, respectively, and if these measurements are accurate to within 0.05Ω,0.05Ω, estimate the maximum possible error in computing R.R. (The symbol ΩΩ represents an ohm, the unit of electrical resistance.)

204.

The area of an ellipse with axes of length 2a2a and 2b2b is given by the formula

A=πab.A=πab. Approximate the percent change in the area when aa increases by 2%2% and bb increases by 1.5%.1.5%.

205.

The period TT of a simple pendulum with small oscillations is calculated from the formula T=2πLg,T=2πLg, where LL is the length of the pendulum and gg is the acceleration resulting from gravity. Suppose that LL and gg have errors of, at most, 0.5%0.5% and 0.1%,0.1%, respectively. Use differentials to approximate the maximum percentage error in the calculated value of T.T.

206.

Electrical power PP is given by P=V2R,P=V2R, where VV is the voltage and RR is the resistance. Approximate the maximum percentage error in calculating power if 120120 VV is applied to a 2000Ω2000Ω resistor and the possible percent errors in measuring VV and RR are 3%3% and 4%,4%, respectively.

For the following exercises, find the linear approximation of each function at the indicated point.

207.

f(x,y)=xy,P(1,4)f(x,y)=xy,P(1,4)

208.

f(x,y)=excosy;P(0,0)f(x,y)=excosy;P(0,0)

209.

f(x,y)=arctan(x+2y),P(1,0)f(x,y)=arctan(x+2y),P(1,0)

210.

f(x,y)=20x27y2,P(2,1)f(x,y)=20x27y2,P(2,1)

211.

f(x,y,z)=x2+y2+z2,P(3,2,6)f(x,y,z)=x2+y2+z2,P(3,2,6)

212.

[T] Find the equation of the tangent plane to the surface f(x,y)=x2+y2f(x,y)=x2+y2 at point (1,2,5),(1,2,5), and graph the surface and the tangent plane at the point.

213.

[T] Find the equation for the tangent plane to the surface at the indicated point, and graph the surface and the tangent plane: z=ln(10x2+2y2+1),P(0,0,0).z=ln(10x2+2y2+1),P(0,0,0).

214.

[T] Find the equation of the tangent plane to the surface z=f(x,y)=sin(x+y2)z=f(x,y)=sin(x+y2) at point (π4,0,22),(π4,0,22), and graph the surface and the tangent plane.

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