10.4: Linearization- Tangent Planes and Differentials

Motivating Questions

• What does it mean for a function of two variables to be locally linear at a point?
• How do we find the equation of the plane tangent to a locally linear function at a point?
• What is the differential of a multivariable function of two variables and what are its uses?

One of the central concepts in single variable calculus is that the graph of a differentiable function, when viewed on a very small scale, looks like a line. We call this line the tangent line and measure its slope with the derivative. In this section, we will extend this concept to functions of several variables.

Let's see what happens when we look at the graph of a two-variable function on a small scale. To begin, let's consider the function \(f\) defined by

\[ f(x,y) = 6 - \frac{x^2}2 - y^2, \nonumber \]

whose graph is shown in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\). The graph of \(f(x,y)=6-x^2/2 - y^2\text{.}\)

We choose to study the behavior of this function near the point \((x_0, y_0) = (1,1)\text{.}\) In particular, we wish to view the graph on an increasingly small scale around this point, as shown in the two plots in Figure \(\PageIndex{2}\)

Figure \(\PageIndex{2}\). The graph of \(f(x,y)=6-x^2/2 - y^2\text{.}\)

Just as the graph of a differentiable single-variable function looks like a line when viewed on a small scale, we see that the graph of this particular two-variable function looks like a plane, as seen in Figure \(\PageIndex{3}\). In the following preview activity, we explore how to find the equation of this plane.

Figure \(\PageIndex{3}\). The graph of \(f(x,y)=6-x^2/2 - y^2\text{.}\)

In what follows, we will also use the important fact¹ that the plane passing through \((x_0, y_0, z_0)\) may be expressed in the form \(z = z_0 + a(x-x_0) + b(y-y_0)\text{,}\) where \(a\) and \(b\) are constants.

As we saw in Section 9.5, the equation of a plane passing through the point \((x_0, y_0, z_0)\) may be written in the form \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\text{.}\) If the plane is not vertical, then \(C\neq 0\text{,}\) and we can rearrange this and hence write \(C(z-z_0) = -A(x-x_0) - B(y-y_0)\) and thus

\begin{align*} z & = z_0-\frac AC(x-x_0) - \frac BC(y-y_0)\\[4pt] & = z_0 + a(x-x_0) + b(y-y_0) \end{align*}

where \(a=-A/C\) and \(b=-B/C\text{,}\) respectively.

Preview Activity \(\PageIndex{1}\)

Let \(f(x,y) = 6 - \frac{x^2}2 - y^2\text{,}\) and let \((x_0,y_0) = (1,1)\text{.}\)

a. Evaluate \(f(x,y) = 6 - \frac{x^2}2 - y^2\) and its partial derivatives at \((x_0,y_0)\text{;}\) that is, find \(f(1,1)\text{,}\) \(f_x(1,1)\text{,}\) and \(f_y(1,1)\text{.}\)

b. We know one point on the tangent plane; namely, the \(z\)-value of the tangent plane agrees with the \(z\)-value on the graph of \(f(x,y) = 6 - \frac{x^2}2 - y^2\) at the point \((x_0, y_0)\text{.}\) In other words, both the tangent plane and the graph of the function \(f\) contain the point \((x_0, y_0, z_0)\text{.}\) Use this observation to determine \(z_0\) in the expression \(z = z_0 + a(x-x_0) + b(y-y_0)\text{.}\)

c. Sketch the traces of \(f(x,y) = 6 - \frac{x^2}2 - y^2\) for \(y=y_0=1\) and \(x=x_0=1\) below in Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\). The traces of \(f(x,y)\) with \(y=y_0=1\) and \(x=x_0=1\text{.}\)

d. Determine the equation of the tangent line of the trace that you sketched in the previous part with \(y=1\) (in the \(x\) direction) at the point \(x_0=1\text{.}\)

Figure \(\PageIndex{5}\). The traces of \(f(x,y)\) and the tangent plane.

e. Figure \(\PageIndex{5}\) shows the traces of the function and the traces of the tangent plane. Explain how the tangent line of the trace of \(f\text{,}\) whose equation you found in the last part of this activity, is related to the tangent plane. How does this observation help you determine the constant \(a\) in the equation for the tangent plane \(z = z_0+a(x-x_0) + b(y-y_0)\text{?}\) (Hint: How do you think \(f_x(x_0,y_0)\) should be related to \(z_x(x_0,y_0)\text{?}\))

f. In a similar way to what you did in (d), determine the equation of the tangent line of the trace with \(x=1\) at the point \(y_0=1\text{.}\) Explain how this tangent line is related to the tangent plane, and use this observation to determine the constant \(b\) in the equation for the tangent plane \(z=z_0+a(x-x_0) + b(y-y_0)\text{.}\) (Hint: How do you think \(f_y(x_0,y_0)\) should be related to \(z_y(x_0,y_0)\text{?}\))

g. Finally, write the equation \(z=z_0 + a(x-x_0) + b(y-y_0)\) of the tangent plane to the graph of \(f(x,y)=6-x^2/2 - y^2\) at the point \((x_0,y_0)=(1,1)\text{.}\)

The Tangent Plane

Before stating the formula for the equation of the tangent plane at a point for a general function \(f=f(x,y)\text{,}\) we need to discuss a technical condition. As we have noted, when we look at the graph of a single-variable function on a small scale near a point \(x_0\text{,}\) we expect to see a line; in this case, we say that \(f\) is locally linear near \(x_0\) since the graph looks like a linear function locally around \(x_0\text{.}\) Of course, there are functions, such as the absolute value function given by \(f(x)=|x|\text{,}\) that are not locally linear at every point. In single-variable calculus, we learn that if the derivative of a function exists at a point, then the function is guaranteed to be locally linear there.

In a similar way, we say that a two-variable function \(f\) is locally linear near \((x_0,y_0)\) provided that the graph of \(f\) looks like a plane (its tangent plane) when viewed on a small scale near \((x_0,y_0)\text{.}\) How can we tell when a function of two variables is locally linear at a point?

It is not unreasonable to expect that if \(f_x(a,b)\) and \(f_y(a,b)\) exist for some function \(f\) at a point \((a,b)\text{,}\) then \(f\) is locally linear at \((a,b)\text{.}\) This is not sufficient, however. As an example, consider the function \(f\) defined by \(f(x,y) = x^{1/3} y^{1/3}\text{.}\) In Exercise \(\PageIndex{15}\) you are asked to show that \(f_x(0,0)\) and \(f_y(0,0)\) both exist, but that \(f\) is not locally linear at \((0,0)\) (see Figure \(\PageIndex{1}\)2). So the existence of the two first order partial derivatives at a point does not guarantee local linearity at that point.

It would take us too far afield to provide a rigorous discussion of differentiability of functions of more than one variable (see Exercise \(\PageIndex{15}\) for a little more detail), so we will be content to define stronger, but more easily verified, conditions that ensure local linearity.

Differentiablity

If \(f\) is a function of the independent variables \(x\) and \(y\) and both \(f_x\) and \(f_y\) exist and are continuous in an open disk containing the point \((x_0,y_0)\text{,}\) then \(f\) is continuously differentiable at \((x_0,y_0)\text{.}\)

As a consequence, whenever a function \(z = f(x,y)\) is continuously differentiable at a point \((x_0,y_0)\text{,}\) it follows that the function has a tangent plane at \((x_0,y_0)\text{.}\) Viewed up close, the tangent plane and the function are then virtually indistinguishable. (We won't formally define differentiability of multivariable functions here, and for our purposes continuous differentiability is the only condition we will ever need to use. It is important to note that continuous differentiability is a stronger condition than differentiability. All of the results we encounter will apply to differentiable functions, and so also apply to continuously differentiable functions.) In addition, as in Preview Activity \(\PageIndex{1}\), we find the following general formula for the tangent plane.

The tangent plane

If \(f(x,y)\) has continuous first-order partial derivatives, then the equation of the plane tangent to the graph of \(f\) at the point \((x_0,y_0,f(x_0,y_0))\) is

\[ z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0).\label{eq_10_4_tan_plane}\tag{\(\PageIndex{1}\)} \]

Important Note: As can be seen in Exercise \(\PageIndex{5}\).11, it is possible that \(f_x(x_0,y_0)\) and \(f_y(x_0,y_0)\) can exist for a function \(f\text{,}\) and so the plane \(z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\) exists even though \(f\) is not locally linear at \((x_0,y_0)\) (because the graph of \(f\) does not look linear when we zoom in around the point \((x_0,y_0)\)). In such a case this plane is not tangent to the graph. Differentiability for a function of two variables implies the existence of a tangent plane, but the existence of the two first order partial derivatives of a function at a point does not imply differentiaility. This is quite different than what happens in single variable calculus.

Finally, one important note about the form of the equation for the tangent plane, \(z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\text{.}\) Say, for example, that we have the particular tangent plane \(z = 7 - 2(x-3) + 4(y+1)\text{.}\) Observe that we can immediately read from this form that \(f_x(3,-1) = -2\) and \(f_y(3,-1) = 4\text{;}\) furthermore, \(f_x(3,-1)=-2\) is the slope of the trace to both \(f\) and the tangent plane in the \(x\)-direction at \((3,-1)\text{.}\) In the same way, \(f_y(3,-1) = 4\) is the slope of the trace of both \(f\) and the tangent plane in the \(y\)-direction at \((3,-1)\text{.}\)

Activity \(\PageIndex{2}\)

1. Find the equation of the tangent plane to \(f(x,y) = 2 + 4x - 3y\) at the point \((1,2)\text{.}\) Simplify as much as possible. Does the result surprise you? Explain.
2. Find the equation of the tangent plane to \(f(x,y) = x^2y\) at the point \((1,2)\text{.}\)

Linearization

In single variable calculus, an important use of the tangent line is to approximate the value of a differentiable function. Near the point \(x_0\text{,}\) the tangent line to the graph of \(f\) at \(x_0\) is close to the graph of \(f\) near \(x_0\text{,}\) as shown in Figure \(\PageIndex{6}\).

Figure \(\PageIndex{6}\). The linearization of the single-variable function \(f(x)\text{.}\)

In this single-variable setting, we let \(L\) denote the function whose graph is the tangent line, and thus

\[ L(x) = f(x_0) + f'(x_0)(x-x_0) \nonumber \]

Furthermore, observe that \(f(x) \approx L(x)\) near \(x_0\text{.}\) We call \(L\) the linearization of \(f\text{.}\)

In the same way, the tangent plane to the graph of a differentiable function \(z = f(x,y)\) at a point \((x_0,y_0)\) provides a good approximation of \(f(x,y)\) near \((x_0, y_0)\text{.}\) Here, we define the linearization, \(L\text{,}\) to be the two-variable function whose graph is the tangent plane, and thus

\[ L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0). \nonumber \]

Finally, note that \(f(x,y)\approx L(x,y)\) for points near \((x_0, y_0)\text{.}\) This is illustrated in Figure \(\PageIndex{7}\).

Figure \(\PageIndex{7}\). The linearization of \(f(x,y)\text{.}\)

Activity \(\PageIndex{3}\)

In what follows, we find the linearization of several different functions that are given in algebraic, tabular, or graphical form.

a. Find the linearization \(L(x,y)\) for the function \(g\) defined by

\[ g(x,y) = \frac{x}{x^2+y^2} \nonumber \]

at the point \((1,2)\text{.}\) Then use the linearization to estimate the value of \(g(0.8, 2.3)\text{.}\)

b. Table \(\PageIndex{1}\) provides a collection of values of the wind chill \(w(v,T)\text{,}\) in degrees Fahrenheit, as a function of wind speed, in miles per hour, and temperature, also in degrees Fahrenheit.

Table \(\PageIndex{1}\). Wind chill as a function of wind speed and temperature.

\(v \backslash T\)	\(-30\)	\(-25\)	\(-20\)	\(-15\)	\(-10\)	\(-5\)	\(0\)	\(5\)	\(10\)	\(15\)	\(20\)
\(5\)	\(-46\)	\(-40\)	\(-34\)	\(-28\)	\(-22\)	\(-16\)	\(-11\)	\(-5\)	\(1\)	\(7\)	\(13\)
\(10\)	\(-53\)	\(-47\)	\(-41\)	\(-35\)	\(-28\)	\(-22\)	\(-16\)	\(-10\)	\(-4\)	\(3\)	\(9\)
\(15\)	\(-58\)	\(-51\)	\(-45\)	\(-39\)	\(-32\)	\(-26\)	\(-19\)	\(-13\)	\(-7\)	\(0\)	\(6\)
\(20\)	\(-61\)	\(-55\)	\(-48\)	\(-42\)	\(-35\)	\(-29\)	\(-22\)	\(-15\)	\(-9\)	\(-2\)	\(4\)
\(25\)	\(-64\)	\(-58\)	\(-51\)	\(-44\)	\(-37\)	\(-31\)	\(-24\)	\(-17\)	\(-11\)	\(-4\)	\(3\)
\(30\)	\(-67\)	\(-60\)	\(-53\)	\(-46\)	\(-39\)	\(-33\)	\(-26\)	\(-19\)	\(-12\)	\(-5\)	\(1\)
\(35\)	\(-69\)	\(-62\)	\(-55\)	\(-48\)	\(-41\)	\(-34\)	\(-27\)	\(-21\)	\(-14\)	\(-7\)	\(0\)
\(40\)	\(-71\)	\(-64\)	\(-57\)	\(-50\)	\(-43\)	\(-36\)	\(-29\)	\(-22\)	\(-15\)	\(-8\)	\(-1\)

Use the data to first estimate the appropriate partial derivatives, and then find the linearization \(L(v,T)\) at the point \((20,-10)\text{.}\) Finally, use the linearization to estimate \(w(10,-10)\text{,}\) \(w(20,-12)\text{,}\) and \(w(18,-12)\text{.}\) Compare your results to what you obtained in Activity 10.2.5

c. Figure \(\PageIndex{9}\) gives a contour plot of a continuously differentiable function \(f\text{.}\)

Figure \(\PageIndex{9}\). A contour plot of \(f(x,y)\text{.}\)

After estimating appropriate partial derivatives, determine the linearization \(L(x,y)\) at the point \((2,1)\text{,}\) and use it to estimate \(f(2.2, 1)\text{,}\) \(f(2, 0.8)\text{,}\) and \(f(2.2, 0.8)\text{.}\)

Differentials

As we have seen, the linearization \(L(x,y)\) enables us to estimate the value of \(f(x,y)\) for points \((x,y)\) near the base point \((x_0, y_0)\text{.}\) Sometimes, however, we are more interested in the change in \(f\) as we move from the base point \((x_0,y_0)\) to another point \((x,y)\text{.}\)

Figure \(\PageIndex{10}\). The differential \(df\) approximates the change in \(f(x,y)\text{.}\)

Figure \(\PageIndex{10}\ illustrates this situation. Suppose we are at the point \((x_0,y_0)\text{,}\) and we know the value \(f(x_0,y_0)\) of \(f\) at \((x_0,y_0)\text{.}\) If we consider the displacement \(\langle \Delta x, \Delta y\rangle\) to a new point \((x,y) = (x_0+\Delta x, y_0 + \Delta y)\text{,}\) we would like to know how much the function has changed. We denote this change by \(\Delta f\text{,}\) where

\[ \Delta f = f(x,y) - f(x_0, y_0). \nonumber \]

A simple way to estimate the change \(\Delta f\) is to approximate it by \(df\text{,}\) which represents the change in the linearization \(L(x,y)\) as we move from \((x_0,y_0)\) to \((x,y)\text{.}\) This gives

\begin{align*} \Delta f \approx df & = L(x,y)-f(x_0, y_0)\\[4pt] & = [f(x_0,y_0)+ f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)] - f(x_0, y_0)\\[4pt] & = f_x(x_0,y_0)\Delta x + f_y(x_0, y_0)\Delta y. \end{align*}

For consistency, we will denote the change in the independent variables as \(dx = \Delta x\) and \(dy = \Delta y\text{,}\) and thus

\[ \Delta f \approx df = f_x(x_0,y_0)~dx + f_y(x_0,y_0)~dy.\label{E_10_4_differential}\tag{\(\PageIndex{2}\)} \]

Expressed equivalently in Leibniz notation, we have

\[ df = \frac{\partial f}{\partial x}~dx + \frac{\partial f}{\partial y}~dy.\label{E_10_4_differential_leib}\tag{\(\PageIndex{3}\)} \]

We call the quantities \(dx\text{,}\) \(dy\text{,}\) and \(df\) differentials, and we think of them as measuring small changes in the quantities \(x\text{,}\) \(y\text{,}\) and \(f\text{.}\) Equations (\(\PageIndex{2}\)) and (\(\PageIndex{3}\)) express the relationship between these changes. Equation (\(\PageIndex{3}\)) resembles an important idea from single-variable calculus: when \(y\) depends on \(x\text{,}\) it follows in the notation of differentials that

\[ dy = y'~dx = \frac{dy}{dx}~dx. \nonumber \]

We will illustrate the use of differentials with an example.

Example \(\PageIndex{1}\)

Suppose we have a machine that manufactures rectangles of width \(x=20\) cm and height \(y=10\) cm. However, the machine isn't perfect, and therefore the width could be off by \(dx = \Delta x = 0.2\) cm and the height could be off by \(dy = \Delta y = 0.4\) cm.

The area of the rectangle is

\[ A(x,y) = xy, \nonumber \]

so that the area of a perfectly manufactured rectangle is \(A(20, 10) = 200\) square centimeters. Since the machine isn't perfect, we would like to know how much the area of a given manufactured rectangle could differ from the perfect rectangle. We will estimate the uncertainty in the area using (\(\PageIndex{2}\)), and find that

\[ \Delta A \approx dA = A_x(20, 10)~dx + A_y(20,10)~dy. \nonumber \]

Since \(A_x = y\) and \(A_y = x\text{,}\) we have

\[ \Delta A \approx dA = 10~dx + 20~dy = 10\cdot0.2 + 20\cdot0.4 = 10. \nonumber \]

That is, we estimate that the area in our rectangles could be off by as much as 10 square centimeters.

Activity \(\PageIndex{4}\)

The questions in this activity explore the differential in several different contexts.

1. Suppose that the elevation of a landscape is given by the function \(h\text{,}\) where we additionally know that \(h(3,1) = 4.35\text{,}\) \(h_x(3,1) = 0.27\text{,}\) and \(h_y(3,1) = -0.19\text{.}\) Assume that \(x\) and \(y\) are measured in miles in the easterly and northerly directions, respectively, from some base point \((0,0)\text{.}\) Your GPS device says that you are currently at the point \((3,1)\text{.}\) However, you know that the coordinates are only accurate to within \(0.2\) units; that is, \(dx = \Delta x = 0.2\) and \(dy= \Delta y = 0.2\text{.}\) Estimate the uncertainty in your elevation using differentials.
2. The pressure, volume, and temperature of an ideal gas are related by the equation
   \[ P= P(T,V) = 8.31 T/V, \nonumber \]

   where \(P\) is measured in kilopascals, \(V\) in liters, and \(T\) in kelvin. Find the pressure when the volume is 12 liters and the temperature is 310 K. Use differentials to estimate the change in the pressure when the volume increases to 12.3 liters and the temperature decreases to 305 K.

3. Refer to Table \(\PageIndex{8}\), the table of values of the wind chill \(w(v,T)\text{,}\) in degrees Fahrenheit, as a function of temperature, also in degrees Fahrenheit, and wind speed, in miles per hour. Suppose your anemometer says the wind is blowing at \(25\) miles per hour and your thermometer shows a reading of \(-15^\circ\) degrees. However, you know your thermometer is only accurate to within \(2^\circ\) degrees and your anemometer is only accurate to within \(3\) miles per hour. What is the wind chill based on your measurements? Estimate the uncertainty in your measurement of the wind chill.

Summary

• A function \(f\) of two independent variables is locally linear at a point \((x_0,y_0)\) if the graph of \(f\) looks like a plane as we zoom in on the graph around the point \((x_0,y_0)\text{.}\) In this case, the equation of the tangent plane is given by
  \[ z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0). \nonumber \]
• The tangent plane \(L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\text{,}\) when considered as a function, is called the linearization of a differentiable function \(f\) at \((x_0,y_0)\) and may be used to estimate values of \(f(x,y)\text{;}\) that is, \(f(x,y) \approx L(x,y)\) for points \((x,y)\) near \((x_0,y_0)\text{.}\)
• A function \(f\) of two independent variables is differentiable at \((x_0,y_0)\) provided that both \(f_x\) and \(f_y\) exist and are continuous in an open disk containing the point \((x_0,y_0)\text{.}\)
• The differential \(df\) of a function \(f= f(x,y)\) is related to the differentials \(dx\) and \(dy\) by
  \[ df = f_x(x_0,y_0) dx + f_y(x_0,y_0)dy. \nonumber \]

  We can use this relationship to approximate small changes in \(f\) that result from small changes in \(x\) and \(y\text{.}\)