Tangent Planes and Linear Approximations

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).

The slope of the tangent line at the point x=a

is given by 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)

be a point on a surface S,

and let C

be any curve passing through P0

and lying entirely in S.

If the tangent lines to all such curves C

at P0

lie in the same plane, then this plane is called the tangent plane to S

at P0

([link]).

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.

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. We define the term tangent plane here and then explore the idea intuitively.

Definition

Let S

be a surface defined by a differentiable function z=f(x,y),

and let P0=(x0,y0)

be a point in the domain of f.

Then, the equation of the tangent plane to S

at P0

is given by

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

To see why this formula is correct, let’s first find two tangent lines to the surface S.

The equation of the tangent line to the curve that is represented by the intersection of S

with the vertical trace given by x=x0

is 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 S

with the vertical trace given by y=y0

is z=f(x0,y0)+fx(x0,y0)(xx0).

A parallel vector to the first tangent line is a=j+fy(x0,y0)k;

a parallel vector to the second tangent line is 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.

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))

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.

Solving this equation for z

gives [link].

Finding a Tangent Plane

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

at point (2,−1).

First, we must calculate fx(x,y)

and fy(x,y),

then use [link] with x0=2

and 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.

Then [link] becomes

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).

Find the equation of the tangent plane to the surface defined by the function f(x,y)=x3x2y+y22x+3y2

at point (−1,3).

z=7x+8y3
Hint

First, calculate fx(x,y)

and fy(x,y),

then use [link].

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)

at the point (π/3,π/4).

First, calculate fx(x,y)

and fy(x,y),

then use [link] with x0=π/3

and 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.

Then [link] becomes

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).

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.

If either x=0

or y=0,

then 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,

so as either xory

approach zero, these partial derivatives stay equal to zero. Substituting them into [link] gives z=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.

If we put y=x

into the original function, it becomes

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

When x>0,

the slope of this curve is equal to 2/2;

when x<0,

the slope of this curve is equal to (2/2).

This presents a problem. In the definition of tangent plane, we presumed that all tangent lines through point P

(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)

at the point x=a

is given by

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.

The tangent line can be used as an approximation to the function f(x)

for values of x

reasonably close to 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)

with continuous partial derivatives that exist at the point (x0,y0),

the linear approximation of f

at the point (x0,y0)

is given by the equation

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

Notice that this equation also represents the tangent plane to the surface defined by z=f(x,y)

at the point (x0,y0).

The idea behind using a linear approximation is that, if there is a point (x0,y0)

at which the precise value of f(x,y)

is known, then for values of (x,y)

reasonably close to (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)

([link]). Furthermore the plane that is used to find the linear approximation is also the tangent plane to the surface at the point (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.

Using a Tangent Plane Approximation

Given the function f(x,y)=414x2y2,

approximate f(2.1,2.9)

using point (2,3)

for (x0,y0).

What is the approximate value of f(2.1,2.9)

to four decimal places?

To apply [link], we first must calculate f(x0,y0),

fx(x0,y0),

and fy(x0,y0)

using x0=2

and 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.

Now we substitute these values into [link]:

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

Last, we substitute x=2.1

and y=2.9

into L(x,y):

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)

to four decimal places is

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

which corresponds to a 0.2%

error in approximation.

Given the function f(x,y)=e52x+3y,

approximate f(4.1,0.9)

using point (4,1)

for (x0,y0).

What is the approximate value of f(4.1,0.9)

to four decimal places?

L(x,y)=62x+3y,

so L(4.1,0.9)=62(4.1)+3(0.9)=0.5

f(4.1,0.9)=e52(4.1)+3(0.9)=e−0.50.6065.
Hint

First calculate f(x0,y0),fx(x0,y0),andfy(x0,y0)

using x0=4

and y0=1,

then use [link].

Differentiability

When working with a function y=f(x)

of one variable, the function is said to be differentiable at a point x=a

if 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 P

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)

is given by

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

For a tangent plane to exist at the point (x0,y0),

the partial derivatives must therefore exist at that point. However, this is not a sufficient condition for smoothness, as was illustrated in [link]. 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)

is differentiable at a point P(x0,y0)

if, for all points (x,y)

in a δ

disk around P,

we can write

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

where the error term E

satisfies

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

The last term in [link] is referred to as the error term and it represents how closely the tangent plane comes to the surface in a small neighborhood (δ

disk) of point P.

For the function f

to be differentiable at P,

the function must be smooth—that is, the graph of f

must be close to the tangent plane for points near P.

Demonstrating Differentiability

Show that the function f(x,y)=2x24y

is differentiable at point (2,−3).

First, we calculate f(x0,y0),fx(x0,y0),andfy(x0,y0)

using x0=2

and y0=−3,

then we use [link]:

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

Therefore m1=8

and m2=−4,

and [link] 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.

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)(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)0

for any value of xory,

the original limit must be equal to zero. Therefore, f(x,y)=2x24y

is differentiable at point (2,−3).

Show that the function f(x,y)=3x4y2

is differentiable at point (−1,2).

f(−1,2)=−19,fx(−1,2)=3,fy(−1,2)=−16,E(x,y)=−4(y2)2. lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=lim(x,y)(−1,2)−4(y2)2(x+1)2+(y2)2lim(x,y)(−1,2)−4((x+1)2+(y2)2)(x+1)2+(y2)2=lim(x,y)(2,−3)4(x+1)2+(y2)2=0.
Hint

First, calculate f(x0,y0),fx(x0,y0),andfy(x0,y0)

using x0=−1

and y0=2,

then use [link] to find E(x,y).

Last, calculate the limit.

The function 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.

Substituting this information into [link] using x0=0

and 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.

Calculating lim(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.

Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and the function f

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.

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.

Differentiability Implies Continuity

Let z=f(x,y)

be a function of two variables with (x0,y0)

in the domain of f.

If f(x,y)

is differentiable at (x0,y0),

then f(x,y)

is continuous at (x0,y0).

[link] 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)

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.

[link] 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.

Continuity of First Partials Implies Differentiability

Let z=f(x,y)

be a function of two variables with (x0,y0)

in the domain of f.

If f(x,y),

fx(x,y),

and fy(x,y)

all exist in a neighborhood of (x0,y0)

and are continuous at (x0,y0),

then 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)

was not differentiable at the origin. Let’s calculate the partial derivatives fx

and fy:

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)

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

Let 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.

If y>0,

then this expression equals 1/(k2+1)3/2;

if y<0,

then it equals (1/(k2+1)3/2).

In either case, the value depends on k,

so the limit fails to exist.

Differentials

In Linear Approximations and Differentials we first studied the concept of differentials. The differential of y,

written dy,

is defined as f(x)dx.

The differential is used to approximate Δy=f(x+Δx)f(x),

where Δx=dx.

Extending this idea to the linear approximation of a function of two variables at the point (x0,y0)

yields the formula for the total differential for a function of two variables.

Definition

Let z=f(x,y)

be a function of two variables with (x0,y0)

in the domain of f,

and let Δx

and Δy

be chosen so that (x0+Δx,y0+Δy)

is also in the domain of f.

If f

is differentiable at the point (x0,y0),

then the differentials dx

and dy

are defined as

dx=Δxanddy=Δy.

The differential dz,

also called the total differential of z=f(x,y)

at (x0,y0),

is defined as

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

Notice that the symbol

is not used to denote the total differential; rather, d

appears in front of z.

Now, let’s define Δz=f(x+Δx,y+Δy)f(x,y).

We use dz

to approximate Δz,

so

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)

at the point (x0,y0)

for given values of Δx

and Δy.

Since 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,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.

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.

Approximation by Differentials

Find the differential dz

of the function f(x,y)=3x22xy+y2

and use it to approximate Δz

at point (2,−3).

Use Δx=0.1

and Δy=−0.05.

What is the exact value of Δz?

First, we must calculate f(x0,y0),fx(x0,y0),andfy(x0,y0)

using x0=2

and 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.

Then, we substitute these quantities into [link]:

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).

The exact value of Δz

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.

Find the differential dz

of the function f(x,y)=4y2+x2y2xy

and use it to approximate Δz

at point (1,−1).

Use Δx=0.03

and Δy=−0.02.

What is the exact value of Δz?

dz=0.18Δz=f(1.03,−1.02)f(1,−1)=0.180682
Hint

First, calculate fx(x0,y0)

and fy(x0,y0)

using x0=1

and y0=−1,

then use [link].

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)

is differentiable at a point P(x0,y0,z0)

if for all points (x,y,z)

in a δ

disk around P

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),

where the error term E satisfies

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),

then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.

Key Concepts

Key Equations


where the error term

E

satisfies


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

where the error term

E

satisfies


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

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

f(x,y)=x3,(2,−1,8)
(145145)(12ik)
ln(xyz)=0

when x=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.

x2+xy+y2=3,P(−1,−1)

Normal vector: i+j,

tangent vector: ij

(x2+y2)2=9(x2y2),P(2,1)
xy22x2+y+5x=6,P(4,2)

Normal vector: 7i17j,

tangent vector: 17i+7j

2x3x2y2=3xy7,P(1,−2)
zex2y23=0, P(2,2,3)
−1.094i0.18238j

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for z

in terms of x

and y.)

−8x3y7z=−19,P(1,−1,2)
z=−9x23y2,P(2,1,−39)
−36x6yz=−39
x2+10xyz+y2+8z2=0,P(−1,−1,−1)
z=ln(10x2+2y2+1),P(0,0,0)
z=0
z=e7x2+4y2, P(0,0,1)
xy+yz+zx=11,P(1,2,3)
5x+4y+3z22=0
x2+4y2=z2,P(3,2,5)
x3+y3=3xyz,P(1,2,32)
4x5y+4z=0
z=axy,P(1,1a,1)
z=sinx+siny+sin(x+y),P(0,0,0)
2x+2yz=0
h(x,y)=lnx2+y2,P(3,4)
z=x22xy+y2,P(1,2,1)
−2(x1)+2(y2)(z1)=0

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),

and a vector n=a,b,c

that is parallel to the line. Then the equation of the line is xx0=at,yy0=bt,zz0=ct.)

−3x+9y+4z=−4,P(1,−1,2)
z=5x22y2,P(2,1,18)
x=20t+2,y=−4t+1,z=t+18
x28xyz+y2+6z2=0,P(1,1,1)
z=ln(3x2+7y2+1),P(0,0,0)
x=0,y=0,z=t
z=e4x2+6y2,P(0,0,1)
z=x22xy+y2

at point P(1,2,1)

x1=2t;y2=−2t;z1=t

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).

The length of line segment AC

is equal to what mathematical expression?

The length of line segment BC

is equal to what mathematical expression?

The differential of the function z(x,y)=dz=fxdx+fydy

Using the figure, explain what the length of line segment AB

represents.

For the following exercises, complete each task.

Show that f(x,y)=exyx

is differentiable at point (1,0).

Using the definition of differentiability, we have exyxx+y.

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

Show that f(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,

where both ε1

and ε2

approach zero as (Δx,Δy)

approaches (0,0).

Δz=2xΔx+3Δy+(Δx)2. (Δx)20

for small Δx

and z

satisfies the definition of differentiability.

Find the total differential of the function z=xyy+x

where x

changes from 10to10.5

and y

changes from 15to13.

Let z=f(x,y)=xey.

Compute Δz

from P(1,2)

to Q(1.05,2.1)

and then find the approximate change in z

from point P

to point Q.

Recall Δz=f(x+Δx,y+Δy)f(x,y),

and dz

and Δz

are approximately equal.

Δz1.185422

and dz1.108.

They are relatively close.

The volume of a right circular cylinder is given by V(r,h)=πr2h.

Find the differential dV.

Interpret the formula geometrically.

See the preceding problem. Use differentials to estimate the amount of aluminum in an enclosed aluminum can with diameter 8.0cm

and height 12cm

if the aluminum is 0.04

cm thick.

16

cm3

Use the differential dz

to approximate the change in z=4x2y2

as (x,y)

moves from point (1,1)

to point (1.01,0.97).

Compare this approximation with the actual change in the function.

Let z=f(x,y)=x2+3xyy2.

Find the exact change in the function and the approximate change in the function as x

changes from 2.00to2.05

and y

changes from 3.00to2.96.

Δz=

exact change =0.6449,

approximate change is dz=0.65.

The two values are close.

The centripetal acceleration of a particle moving in a circle is given by a(r,v)=v2r,

where v

is the velocity and r

is the radius of the circle. Approximate the maximum percent error in measuring the acceleration resulting from errors of 3%

in v

and 2%

in 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 a

is given by daa.)

The radius r

and height h

of a right circular cylinder are measured with possible errors of 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 V

is given by dVV.)

13%or0.13

The base radius and height of a right circular cone are measured as 10

in. and 25

in., respectively, with a possible error in measurement of as much as 0.1

in. each. Use differentials to estimate the maximum error in the calculated volume of the cone.

The electrical resistance R

produced by wiring resistors R1

and R2

in parallel can be calculated from the formula 1R=1R1+1R2.

If R1

and R2

are measured to be 7Ω

and 6Ω,

respectively, and if these measurements are accurate to within 0.05Ω,

estimate the maximum possible error in computing R.

(The symbol Ω

represents an ohm, the unit of electrical resistance.)

0.025

The area of an ellipse with axes of length 2a

and 2b

is given by the formula

A=πab.

Approximate the percent change in the area when a

increases by 2%

and b

increases by 1.5%.

The period T

of a simple pendulum with small oscillations is calculated from the formula T=2πLg,

where L

is the length of the pendulum and g

is the acceleration resulting from gravity. Suppose that L

and g

have errors of, at most, 0.5%

and 0.1%,

respectively. Use differentials to approximate the maximum percentage error in the calculated value of T.

0.3%

Electrical power P

is given by P=V2R,

where V

is the voltage and R

is the resistance. Approximate the maximum percentage error in calculating power if 120

V

is applied to a 2000Ω

resistor and the possible percent errors in measuring V

and R

are 3%

and 4%,

respectively.

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

f(x,y)=xy,P(1,4)
2x+14y1
f(x,y)=excosy;P(0,0)
f(x,y)=arctan(x+2y),P(1,0)
12x+y+14π12
f(x,y)=20x27y2,P(2,1)
f(x,y,z)=x2+y2+z2,P(3,2,6)
37x+27y+67z

[T] Find the equation of the tangent plane to the surface f(x,y)=x2+y2

at point (1,2,5),

and graph the surface and the tangent plane at the point.

[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=0

A curved surface is shown with tangent plane at (0, 0, 0). The curved surface looks like the middle part of the bottom of a boat, and the tangent plane is z = 0.

[T] Find the equation of the tangent plane to the surface z=f(x,y)=sin(x+y2)

at point (π4,0,22),

and graph the surface and the tangent plane.

Glossary

differentiable
a function f(x,y)

is differentiable at

(x0,y0)

if

f(x,y)

can be expressed in the form

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

where the error term

E(x,y)

satisfies

lim(x,y)(x0,y0)E(x,y)(xx0)2+(yy0)2=0
linear approximation
given a function f(x,y)

and a tangent plane to the function at a point

(x0,y0),

we can approximate

f(x,y)

for points near

(x0,y0)

using the tangent plane formula

tangent plane
given a function f(x,y)

that is differentiable at a point

(x0,y0),

the equation of the tangent plane to the surface

z=f(x,y)

is given by

z=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)
total differential
the total differential of the function f(x,y)

at

(x0,y0)

is given by the formula

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

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