14.4: Tangent Planes and Linear Approximations

Julia Jackson

Department of MathematicsThe University of Oklahoma

Spring 2020

Overview

In the previous secction, we learned to calculate the instantaneous rate ofchange of a function f (x , y) at a point (a, b) in exactly two directions:parallel to the x-axis and parallel to the y -axis. Believe it or not, justthese two directions are enough to get us to our first application ofderivatives: tangent planes.

Recall that in single-variable calculus, you can use the derivative of afunction f at a point to give an equation of the tangent line to f at thatpoint. Given a two-variable function f (x , y), the partial derivatives at apoint can be used to specify a similar object: a plane tangent to thegraph of f . In this section we will define the tangent plane precisely andlearn how to give an equation of one. We will then turn to one of itsuses: estimating function values.

Table of Contents

Tangent Planes

Linear Approximations

Exercises

What is the Tangent Plane?

Suppose we have a two-variable function f (x , y) whose graph is thesurface S . Recall from the previous section that the partial derivativesfx(a, b) and fy (a, b) of f give the respective slopes of the lines T1 and T2

that lie tangent to S at the point P = (a, b, c) as in the following figure:

What is the Tangent Plane?, cont.

Note that the lines T1 and T2 generate a unique plane that containsthem both:

This is the plane tangent to S at the point P, i.e., the tangent plane atP, so called because it contains the two tangent lines. Note that it, toolies tangent to S .

Toward an Equation

This is a nice definition, but it tells us very little about how to give anequation for such a plane. That is our next goal.

Recall that any plane can be given in the form:

a(x − x0) + b(y − y0) + c(z − z0) = 0

where (x0, y0, z0) is a point in the plane. In particular, since the tangentplane passes through a given point on the surface S , it contains a pointof the form P = (x0, y0, f (x0, y0)). Rearranging slightly, we can give anequation of the tangent plane in the form:

z = A(x − x0) + B(y − y0) + f (x0, y0)

where A = −ac and B = −b

c (work this out to be sure you understand).Why this form? Hopefully the answer will become clear as we proceed.

The Constants

Our next goal is to work out the constants A and B in

z = A(x − x0) + B(y − y0) + f (x0, y0)

Let’s begin by working out A. To help us do so, we’ll examine thecross-section of the plane when y = y0. Plugging in, the equation of thiscross-section is:

z = A(x − x0) + f (x0, y0)

= Ax + (f (x0, y0)− Ax0)

This is the equation of a line with slope A! Why a line? Let’s return tothe picture.

The Constants, cont.

The cross-section cut from the tangent plane when y = y0 is the tangentline T1.

The Constants, cont.

Therefore,z = Ax + (f (x0, y0)− Ax0)

is an equation of the tangent line T1. From the previous section, weknow that the slope of T1 is fx(x0, y0). Therefore, A = fx(x0, y0).

If we fix x = x0 instead, we can repeat this argument to show thatB = fy (x0, y0) (this is a good mental exercise; try it!).

An Equation

Therefore, putting everything together, an equation of the plane tangentto the graph of f (x , y) at the point (x0, y0, f (x0, y0)) is:

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

Example

Find an equation of the plane tangent to the surface z = 2x2 + y2 at thepoint (1, 3, 11).

From the result above, we know that an equation of the plane tangent toz = f (x , y) at (1, 3, 11) is:

z = fx(1, 3)(x − 1) + fy (1, 3)(y − 3) + 11

Here we have z = f (x , y) = 2x2 + y2. Let’s calculate the derivatives:

fx(x , y) = 4x fx(1, 3) = 4

fy (x , y) = 2y fy (1, 3) = 6

Example, cont.

Therefore, we have:

z = 4(x − 1) + 6(y − 3) + 11

Or, simplifying, we have:

z = 4x + 6y − 11

Table of Contents

Tangent Planes

Linear Approximations

Exercises

Framing

We went to all this trouble to define the tangent plane and work out anequation for it, so a question now confronts us: what can we use this for?We turn back to single-variable calculus for inspiration.

Single-Variable Calculus

Recall that the derivative of a function f (x) at x = a can be used to givean equation for the line L(x) tangent to the graph of f (x) at the point(a, f (a)):

Note in particular that the values of L(x) are near the values of f (x)when x is near a, so, the values of L(x) can be used to approximate thevalues of f (x) near x = a.

Linear Approximation

In many cases, this observation can help us save time and energy.Suppose f is a computationally expensive function, like, say:

f (x) =

√cos(sin(x) + ln(x + π))−

√√√ln(cos(x + 7)−|x − 3|)

Depending on the level of accuracy needed, it may be worth it to insteadapproximate f with the computationally inexpensive tangent line:

L(x) = mx + b

The Picture

In R3 we have an analogous picture.

Below are three images of a surface and the plane tangent to that surfaceat a point. From left to right, we gradually zoom in on the point wherethe two meet:

The Picture, cont.

As we zoom in, the plane and the surface become almostindistinguishable from one another. Thus, if this surface is the graph of atwo-variable function f (x , y), we can use the tangent plane to estimatevalues of f near the point of intersection.

Linear Approximation

Suppose that the equation of the plane tangent to f (x , y) at(x0, y0, f (x0, y0)) has equation

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

Since the points on the plane are close to the points on the graph ofz = f (x , y) when (x , y) is near (x0, y0), we have:

f (x , y) ≈ fx(x0, y0)(x − x0) + fy (x0, y0)(y − y0) + f (x0, y0)

when (x , y) is near (x0, y0). This entire expression is called the linearapproximation or tangent plane approximation of f at (x0, y0). Theright-hand side alone is called the linearization of f at (x0, y0), oftenwritten:

L(x , y) = fx(x0, y0)(x − x0) + fy (x0, y0)(y − y0) + f (x0, y0)

Example

We saw above that the equation of the plane tangent to the graph off (x , y) = 2x2 + y2 at the point (1, 3, 11) is z = 4x + 6y − 11. Use thisto estimate f (1.1, 2.9).

From above, we have f (x , y) ≈ 4x + 6y − 11. Therefore, we have:

f (1.1, 2.9) ≈ 4(1.1) + 6(2.9)− 11 = 10.8

Example

Find the linearization L(x , y) of f (x , y) = xexy at (1, 0) and use it toapproximate f (1.1,−0.1).

Recall that the linearization of f an (1, 0) is simply the right side of anequation of the plane tangent to f at (1, 0). So, let’s find this first. Anequation for the plane tangent to f (x , y) at the point (1, 0) is:

z = fx(1, 0)(x − 1) + fy (1, 0)(y − 0) + f (1, 0)

We have:

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

fx(1, 0) = 1 fy (1, 0) = 1

f (1, 0) = 1

Example, cont.

Therefore, an equation of the plane tangent to f (x , y) at (1, 0) is:

z = 1(x − 1) + 1(y − 0) + 1

orz = x + y

Thus, the linearization of f at (1, 0) is

L(x , y) = x + y

and furthermore:f (1.1,−0.1) ≈ 1.1− 0.1 = 1

A Final Note

You can also create linear approximations for functions of more variables,and the equation is wholly analogous. For example, for a function ofthree variables, we can approximate it near (a, b, c) using:

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)

Table of Contents

Tangent Planes

Linear Approximations

Exercises

Exercises

1. Find an equation of the plane tangent to the graph ofz = (x + 2)2 − 2(y − 1)2 − 5 at (2, 3, 3).

2. Find the linearization L(x , y) of f (x , y) =√xy at (1, 4).

3. Use the linearization you found in the previous exercise to estimatef (1.1, 3.9).

Solutions

1. z = 8x − 8y + 11.

2. L(x , y) = x + y4

3. f (1.1, 3.9) ≈ L(1.1, 3.9) = 2.075