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Section 2.4 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 does it mean to say that a multivariable function is differentiable?

  • 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

\begin{equation*} f(x,y) = 6 - \frac{x^2}2 - y^2, \end{equation*}

whose graph is shown in Figure 2.4.1.

Figure 2.4.1. 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 2.4.2

Figure 2.4.2. 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 2.4.3. In the following preview activity, we explore how to find the equation of this plane.

Figure 2.4.3. The graph of \(f(x,y)=6-x^2/2 - y^2\text{.}\)

In what follows, we will also use the important fact 1 As we saw in Section 1.5, 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 \amp = z_0-\frac AC(x-x_0) - \frac BC(y-y_0)\\ \amp = z_0 + a(x-x_0) + b(y-y_0) \end{align*}
where \(a=-A/C\) and \(b=-B/C\text{,}\) respectively. 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.

Preview Activity 2.4.1.

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

  1. 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 \(f_y(1,1)\text{.}\)

  2. 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 \(z = z_0 + a(x-x_0) + b(y-y_0)\text{.}\)

  3. 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 2.4.4.

    Figure 2.4.4. The traces of \(f(x,y)\) with \(y=y_0=1\) and \(x=x_0=1\text{.}\)

  4. 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 2.4.5. The traces of \(f(x,y)\) and the tangent plane.

  5. Figure 2.4.5 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 \(z_x(x_0,y_0)\text{?}\))

  6. 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 \(z_y(x_0,y_0)\text{?}\))

  7. 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 \((x_0,y_0)=(1,1)\text{.}\)

Subsection 2.4.1 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 2.4.5.11 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 2.4.12). 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 dicussion of differentiability of functions of more than one variable (see Exercise 2.4.5.15) for a little more detail), so we will be content to just state 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 differentiable at \((x_0,y_0)\text{.}\)

As a consequence, whenever a function \(z = f(x,y)\) is 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. In addition, as in Preview Activity 2.4.1, 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

\begin{equation} 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{2.4.1} \end{equation}

Important Note: As can be seen in Exercise 2.4.5.11, 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 2.4.2.
  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 \((1,2)\text{.}\)

Subsection 2.4.2 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 2.4.6.

Figure 2.4.6. 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

\begin{equation*} L(x) = f(x_0) + f'(x_0)(x-x_0) \end{equation*}

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

\begin{equation*} 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). \end{equation*}

Finally, note that \(f(x,y)\approx L(x,y)\) for points near \((x_0, y_0)\text{.}\) This is illustrated in Figure 2.4.7.

Figure 2.4.7. The linearization of \(f(x,y)\text{.}\)
Activity 2.4.3.

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

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

    \begin{equation*} g(x,y) = \frac{x}{x^2+y^2} \end{equation*}

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

  2. Table 2.4.8 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.

    \(v \backslash T\) \(-30\) \(-25\) \(-20\) \(-15\) \(-10\) \(-5\) \(0\) \(5\) \(10\) \(15\) \(20\)
    \(5\) \(-46\) \(-40\) \(-34\) \(-28\) \(-22\) \(-16\) \(-11\) \(-5\) \(1\) \(7\) \(13\)
    \(10\) \(-53\) \(-47\) \(-41\) \(-35\) \(-28\) \(-22\) \(-16\) \(-10\) \(-4\) \(3\) \(9\)
    \(15\) \(-58\) \(-51\) \(-45\) \(-39\) \(-32\) \(-26\) \(-19\) \(-13\) \(-7\) \(0\) \(6\)
    \(20\) \(-61\) \(-55\) \(-48\) \(-42\) \(-35\) \(-29\) \(-22\) \(-15\) \(-9\) \(-2\) \(4\)
    \(25\) \(-64\) \(-58\) \(-51\) \(-44\) \(-37\) \(-31\) \(-24\) \(-17\) \(-11\) \(-4\) \(3\)
    \(30\) \(-67\) \(-60\) \(-53\) \(-46\) \(-39\) \(-33\) \(-26\) \(-19\) \(-12\) \(-5\) \(1\)
    \(35\) \(-69\) \(-62\) \(-55\) \(-48\) \(-41\) \(-34\) \(-27\) \(-21\) \(-14\) \(-7\) \(0\)
    \(40\) \(-71\) \(-64\) \(-57\) \(-50\) \(-43\) \(-36\) \(-29\) \(-22\) \(-15\) \(-8\) \(-1\)
    Table 2.4.8. Wind chill as a function of wind speed and temperature.
    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 2.2.5

  3. Figure 2.4.9 gives a contour plot of a differentiable function \(f\text{.}\)

    Figure 2.4.9. 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 \(f(2.2, 0.8)\text{.}\)

Subsection 2.4.3 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 2.4.10. The differential \(df\) approximates the change in \(f(x,y)\text{.}\)

Figure 2.4.10 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

\begin{equation*} \Delta f = f(x,y) - f(x_0, y_0). \end{equation*}

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 \amp = L(x,y)-f(x_0, y_0)\\ \amp = [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)\\ \amp = 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

\begin{equation} \Delta f \approx df = f_x(x_0,y_0)~dx + f_y(x_0,y_0)~dy.\label{E_10_4_differential}\tag{2.4.2} \end{equation}

Expressed equivalently in Leibniz notation, we have

\begin{equation} df = \frac{\partial f}{\partial x}~dx + \frac{\partial f}{\partial y}~dy.\label{E_10_4_differential_leib}\tag{2.4.3} \end{equation}

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 (2.4.2) and (2.4.3) express the relationship between these changes. Equation (2.4.3) resembles an important idea from single-variable calculus: when \(y\) depends on \(x\text{,}\) it follows in the notation of differentials that

\begin{equation*} dy = y'~dx = \frac{dy}{dx}~dx. \end{equation*}

We will illustrate the use of differentials with an example.

Example 2.4.11.

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

\begin{equation*} A(x,y) = xy, \end{equation*}

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 (2.4.2), and find that

\begin{equation*} \Delta A \approx dA = A_x(20, 10)~dx + A_y(20,10)~dy. \end{equation*}

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

\begin{equation*} \Delta A \approx dA = 10~dx + 20~dy = 10\cdot0.2 + 20\cdot0.4 = 10. \end{equation*}

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

Activity 2.4.4.

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

    \begin{equation*} P= P(T,V) = 8.31 T/V, \end{equation*}

    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 2.4.8, 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.

Subsection 2.4.4 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

    \begin{equation*} z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0). \end{equation*}
  • 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

    \begin{equation*} df = f_x(x_0,y_0) dx + f_y(x_0,y_0)dy. \end{equation*}

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

Exercises 2.4.5 Exercises

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

Let \(f\) be the function defined by \(f(x,y) = 2x^2+3y^3\text{.}\)

  1. Find the equation of the tangent plane to \(f\) at the point \((1,1)\text{.}\)

  2. Use the linearization to approximate the values of \(f\) at the points \((1.1, 2.05)\) and \((1.3,2.2)\text{.}\)

  3. Compare the approximations form part (b) to the exact values of \(f(1.1, 2.05)\) and \(f(1.3, 2.2)\text{.}\) Which approximation is more accurate. Explain why this should be expected.

11.

Let \(f\) be the function defined by \(f(x,y) = x^{1/3}y^{1/3}\text{,}\) whose graph is shown in Figure 2.4.12.

Figure 2.4.12. The surface for \(f(x,y) = x^{1/3}y^{1/3}\text{.}\)
  1. Determine

    \begin{equation*} \lim_{h \to 0} \frac{f(0+h,0)-f(0,0)}{h}. \end{equation*}

    What does this limit tell us about \(f_x(0,0)\text{?}\)

  2. Note that \(f(x,y)=f(y,x)\text{,}\) and this symmetry implies that \(f_x(0,0) = f_y(0,0)\text{.}\) So both partial derivatives of \(f\) exist at \((0,0)\text{.}\) A picture of the surface defined by \(f\) near \((0,0)\) is shown in Figure 2.4.12. Based on this picture, do you think \(f\) is locally linear at \((0,0)\text{?}\) Why?

  3. Show that the curve where \(x=y\) on the surface defined by \(f\) is not differentiable at 0. What does this tell us about the local linearity of \(f\) at \((0,0)\text{?}\)

  4. Is the function \(f\) defined by \(f(x,y) = \frac{x^2}{y^2+1}\) locally linear at \((0,0)\text{?}\) Why or why not?

12.

Let \(g\) be a function that is differentiable at \((-2,5)\) and suppose that its tangent plane at this point is given by \(z = -7 + 4(x+2) - 3(y-5)\text{.}\)

  1. Determine the values of \(g(-2,5)\text{,}\) \(g_x(-2,5)\text{,}\) and \(g_y(-2,5)\text{.}\) Write one sentence to explain your thinking.

  2. Estimate the value of \(g(-1.8, 4.7)\text{.}\) Clearly show your work and thinking.

  3. Given changes of \(dx = -0.34\) and \(dy = 0.21\text{,}\) estimate the corresponding change in \(g\) that is given by its differential, \(dg\text{.}\)

  4. Suppose that another function \(h\) is also differentiable at \((-2,5)\text{,}\) but that its tangent plane at \((-2,5)\) is given by \(3x + 2y - 4z = 9.\) Determine the values of \(h(-2,5)\text{,}\) \(h_x(-2,5)\text{,}\) and \(h_y(-2,5)\text{,}\) and then estimate the value of \(h(-1.8, 4.7)\text{.}\) Clearly show your work and thinking.

13.

In the following questions, we determine and apply the linearization for several different functions.

  1. Find the linearization \(L(x,y)\) for the function \(f\) defined by \(f(x,y) = \cos(x)(2e^{2y}+e^{-2y})\) at the point \((x_0,y_0) = (0,0)\text{.}\) Hence use the linearization to estimate the value of \(f(0.1, 0.2)\text{.}\) Compare your estimate to the actual value of \(f(0.1, 0.2)\text{.}\)

  2. The Heat Index, \(I\text{,}\) (measured in apparent degrees F) is a function of the actual temperature \(T\) outside (in degrees F) and the relative humidity \(H\) (measured as a percentage). A portion of the table which gives values for this function, \(I=I(T,H)\text{,}\) is provided in Table 2.4.13.

    T \(\downarrow \backslash\) H \(\rightarrow\) \(70\) \(75\) \(80\) \(85\)
    \(90\) \(106\) \(109\) \(112\) \(115\)
    \(92\) \(112\) \(115\) \(119\) \(123\)
    \(94\) \(118\) \(122\) \(127\) \(132\)
    \(96\) \(125\) \(130\) \(135\) \(141\)
    Table 2.4.13. Heat index.
    Suppose you are given that \(I_T(94,75) = 3.75\) and \(I_H(94,75) = 0.9\text{.}\) Use this given information and one other value from the table to estimate the value of \(I(93.1,77)\) using the linearization at \((94,75)\text{.}\) Using proper terminology and notation, explain your work and thinking.

  3. Just as we can find a local linearization for a differentiable function of two variables, we can do so for functions of three or more variables. By extending the concept of the local linearization from two to three variables, find the linearization of the function \(h(x,y,z) = e^{2x}(y+z^2)\) at the point \((x_0,y_0,z_0) = (0, 1, -2)\text{.}\) Then, use the linearization to estimate the value of \(h(-0.1, 0.9, -1.8)\text{.}\)

14.

In the following questions, we investigate two different applied settings using the differential.

  1. Let \(f\) represent the vertical displacement in centimeters from the rest position of a string (like a guitar string) as a function of the distance \(x\) in centimeters from the fixed left end of the string and \(y\) the time in seconds after the string has been plucked. (An interesting video of this can be seen at https://www.youtube.com/watch?v=TKF6nFzpHBUA.) A simple model for \(f\) could be

    \begin{equation*} f(x,y) = \cos(x)\sin(2y). \end{equation*}

    Use the differential to approximate how much more this vibrating string is vertically displaced from its position at \((a,b) = \left(\frac{\pi}{4}, \frac{\pi}{3} \right)\) if we decrease \(a\) by \(0.01\) cm and increase the time by \(0.1\) seconds. Compare to the value of \(f\) at the point \(\left(\frac{\pi}{4}-0.01, \frac{\pi}{3}+0.1\right)\text{.}\)

  2. Resistors used in electrical circuits have colored bands painted on them to indicate the amount of resistance and the possible error in the resistance. When three resistors, whose resistances are \(R_1\text{,}\) \(R_2\text{,}\) and \(R_3\text{,}\) are connected in parallel, the total resistance \(R\) is given by

    \begin{equation*} \frac1R = \frac1{R_1} + \frac1{R_2} + \frac1{R_3}. \end{equation*}

    Suppose that the resistances are \(R_1=25\Omega\text{,}\) \(R_2=40\Omega\text{,}\) and \(R_3=50\Omega\text{.}\) Find the total resistance \(R\text{.}\) If you know each of \(R_1\text{,}\) \(R_2\text{,}\) and \(R_3\) with a possible error of \(0.5\)%, estimate the maximum error in your calculation of \(R\text{.}\)

15.

In this exercise we explore the concept of differetiability of a function of two variables in more detail. We will consider the function \(f\) defined by \(f(x,y) = |x|+|y|\text{.}\)

  1. Use appropriate technology to plot the graph of \(f\) on the domain \([-1,1] \times [-1,1]\text{.}\) Based on the graph, do you think that \(f\) is locally linear at \((0,0)\text{?}\) Explain your reasoning.

  2. Show that both \(f_x(0,0)\) and \(f_y(0,0)\) exist. If \(f\) is locally linear at \((0,0)\text{,}\) what must be the equation of the tangent plane \(L\) to \(f\) at \((0,0)\text{?}\)

  3. In general, if \(g = g(x,y)\) is a function of two variables, and \(g\) is differentiable at a point \((x_0,y_0)\text{,}\) let

    \begin{equation*} L(x,y) = g(x_0,y_0) + g_x(x_0,y_0)(x-x_0) + g_y(x_0,y_0)(y-y_0) \end{equation*}

    be the linearization of \(g\) at \((x_0,y_0)\text{.}\) The error in approximating \(g(x,y)\) by \(L(x,y)\) for points near \((x_0,y_0)\) is given by

    \begin{equation*} E(x,y) = g(x,y) - L(x,y). \end{equation*}

    It might be reasonable to think that if the error term goes to \(0\) as \((x,y)\) approaches \((x_0,y_0)\text{,}\) then \(g\) is locally linear at \((x_0,y_0)\text{.}\) Assume that \(f(x,y) = |x| + |y|\) is differentiable at \((0,0)\) and use what must be its linearization at the origin that you found in (b) and demonstrate that the limit of \(E(x,y)\) for \(f\) is \(0\) at \((0,0)\text{.}\) This shows that just because an error term goes to \(0\) as \((x,y)\) approaches \((x_0,y_0)\text{,}\) we cannot conclude that a function is locally linear at \((x_0,y_0)\text{.}\)

  4. To help understand the condition we need for differentiability (local linearity) at a point, let us recall the absolute value function \(a\) of a single variable, that is \(a(x) = |x|\text{.}\) The reason that \(a\) is not differentiable at \(0\) is that to the right of \(0\) \(a\) is the linear function \(y=x\) and to the left of \(0\) \(a\) is the linear function \(y=-x\text{.}\) In other words, the slope of \(a\) is different on the two sides of \(0\text{.}\) The slope is the change in \(a\) divided by the change in \(x\text{,}\) or how far the values of \(a\) are from the origin in relation to how far the values of \(x\) are from the origin. This is a relative error instead of an actual error. We can apply the same idea for functions of two variables and measure how well the error term approximates a function relative to how far the point in question is from the base point.

    For a differentiable function \(g\) of two variables \(x\) and \(y\text{,}\) we define the relative error in approximating \(g(x,y)\) with \(L(x,y)\) as

    \begin{equation*} \frac{E(x_0+h,y_0+k)}{\sqrt{h^2+k^2}}, \end{equation*}

    where \(h=x-x_0\) and \(k = y-y_0\text{.}\) Notice that \(\sqrt{h^2+k^2}\) measures how far the point \((x,y)\) is from \((x_0,y_0)\text{.}\) It is this relative error that we want to go to \(0\) in order for our function to be differentiable at \((x_0,y_0)\text{.}\) We can use this idea to more formally define differentiability of a function of two variables.

    Definition 2.4.14.

    A function \(f = f(x,y)\) is differentiable at a point \((x_0,y_0)\) if there is a linear function \(L = L(x,y) = f(x_0,y_0) + m(x-x_0) + n(y-y_0)\) such that the relative error

    \begin{equation*} \frac{E(x_0+h,y_0+k)}{\sqrt{h^2+k^2}}, \end{equation*}

    has at limit of \(0\) at \((h,k) = (0,0)\text{,}\) where \(E(x,y) = f(x,y) - L(x,y)\text{,}\) \(h=x-x_0\text{,}\) and \(k = y-y_0\text{.}\)

    Show that for \(f(x,y) = |x| +|y|\text{,}\) the relative error at \((x,y) = (0,0)\) does not have a limit at \((h,k) = (0,0)\text{,}\) using \(L(x,y)\) as in part (b). So in this case the relative error does show that \(f\) is not differentiable at the origin. As we have seen, it is often difficult to verify a limit of a function at a point, so this definition of differentibility can be hard to use.

  5. We conclude this exercise by showing that the linear function in Definition 2.4.14 is in fact our tangent plane. So assume that a function \(g\) is differentiable at a point \((x_0,y_0)\) and that \(L = L(x,y) = f(x_0,y_0) + m(x-x_0) + n(y-y_0)\) satisfies the conditions of Definition 2.4.14. Show that \(m = g_x(x_0,y_0)\) and \(n = g_y(x_0,y_0)\text{.}\) (Hint: Calculate the limits of the relative errors when \(h = 0\) and \(k = 0\text{.}\))