Linearization: Tangent Planes and Differentials

Section 11.5 Linearization: Tangent Planes and Differentials

Motivating Questions

What does it mean for a function of two variables to be locally linear at a point?
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How do we find the equation of the plane tangent to a locally linear function at a point?
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What is the differential of a multivariable function of two variables and what are its uses?
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Subsection 11.5.1 Introduction

Throughout the study of in single-variable calculus, the tangent line serves as a geometric and conceptual tool to understand the derivative at a point. While the value of a derivative at a point measures the instantaneous rate of change, we visualize the value of the derivative of a function as the slope of the line tangent to its graph. This allowed use to “see” the value of the derivative in terms of an important behavior of the function’s graph.

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In Subsection 10.2.4, we saw that a curve in space, as the graph of a vector-valued function of one variable, is locally linear and approximated well by a tangent line provided that the function is differentiable. So a curve in space/vector-valued function will be differentiable if the graph looks like a line on a small scale.

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In this section, we will explore the various ways in which we can extend the ideas described above to a function of two variables. Geometrically, this means we will use a tangent plane as a “flat” analog for surfaces of the form \(z=f(x,y)\text{.}\)

For graphs in the cartesian plane, there was only one type of flat graph: a line. When we generalized to higher dimensions, we saw how both lines and planes can be considered “flat” graphs in three dimensions. However, we noted that lines are 1-dimensional flat graphs because there is only direction to move along the graph. On the other hand, planes are 2-dimensional flat graphs.

We will also introduce a couple of new tools to help us describe change in the output of \(f(x,y)\) algebraically and numerically.

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To start our investigation, let’s consider the graph of a “nice” two-variable function on a small scale. We will look at

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

whose graph is shown in Figure 11.5.1.

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Examine the behavior of \(f\) near the input \((x_0,y_0) = (1,1)\text{,}\) which has output of \(f(1,1)=4.5\text{.}\) Adjust the Zoom slider in Figure 11.5.1 to view the graph of \(z=f(x,y)\) on smaller and smaller scales around the point \((1,1,4.5)\text{.}\) With zoom set to 0, the plot clearly reveals that the surface is curved. However, once you increase the zoom level to 3 or higher, the surface looks more like a tilted plane than a curved surface.

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Figure 11.5.1. A plot of \(f(x,y)=6-\frac{x^2}{2}-y^2\) with the point \((1,1,4.5)\) highlighted and the option to zoom in around the highlighted point.

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Just as the graph of a differentiable single-variable function or a differentiable vector-valued function of one variable looks like a line when viewed on a small scale, the graph of this two-variable function looks like a plane when viewed on a small scale. The Preview Activity guides you through finding an equation for this plane, but we first state a key idea that will be helpful algebraic form to use when thinking about tangent planes to surfaces of the form \(z=f(x,y)\text{.}\)

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Key Idea 11.5.2.

A non-vertical plane that goes through the point \((x_0,y_0,z_0)\) can be expressed in the form \(z = z_0 + a(x-x_0) + b(y-y_0)\text{,}\) where \(a\) and \(b\) are constants.

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Proof.

As we saw in Section 9.6, 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 equation and write \(C(z-z_0) = -A(x-x_0) - B(y-y_0)\) and thus

\begin{align*} z \amp = z_0-\frac{A}{C}(x-x_0) - \frac{B}{C}(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.

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After our work in the Preview Activity, we will look at how to obtain this equation in terms of the normal vector of this plane that best describes the surface at a particular point.

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Preview Activity 11.5.1.

We want to find the equation of the plane, using the form given in Key Idea 11.5.2, that best describes the surface given by \(z=f(x,y)=6-\frac{x^2}2 - y^2\) for input values around \((x_0,y_0) = (1,1)\text{.}\) In particular, we will need to find how the values of \(z_0\text{,}\) \(a\text{,}\) and \(b\) are related to \(f(x,y)\text{.}\)

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(a)

Find \(f(1,1)\text{,}\) \(f_x(1,1)\text{,}\) and \(f_y(1,1)\text{.}\)

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(b)

We want the plane to match the height of the surface at \((1,1)\text{.}\) For this to happen, the value \(z_0\) from Key Idea 11.5.2 must be the \(z\)-coordinate value where the plane intersects the surface. What is the \(z\)-coordinate of the point where the tangent plane and the surface intersect?

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(c)

For the plane to have the same behavior as the surface \(z=f(x,y)\) near \((1,1)\text{,}\) the plane must match the behavior of the traces \(x=1\) and \(y=1\) near this point.

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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. Draw the tangent lines to the each of the traces when the appropriate input is 1.

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(d)

What are the slopes of the tangent lines to the traces that you drew in the previous part? Write a couple of sentences to explain why the tilt of the tangent plane in the \(x\)-direction is given by the partial derivative \(f_x\) and the tilt of the tangent plane in the \(y\)-direction is given by the partial derivative \(f_y\text{.}\) Your answer should discuss how each of these slopes/partial derivatives relates to the traces of the surface and the plane illustrated below. You should identify whether the red/orange traces along the plane and the black/blue traces along the surface correspond to constant values of \(x\) or \(y\text{.}\)

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(e)

Fill in the blanks below with the proper values to give the tangent plane to the graph of \(f(x,y)=6-x^2/2 - y^2\) at the point \((x_0,y_0)=(1,1)\text{.}\)

\begin{equation*} z=z_0 + a(x-x_0) + b(y-y_0)= \fillinmath{X} + \fillinmath{X}(x-\fillinmath{X}) +\fillinmath{X}(y-\fillinmath{X})\text{.} \end{equation*}

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Subsection 11.5.2 The Tangent Plane

As the preview activity suggests, the graph of most functions we have encountered will be approximated well by a plane tangent to the graph at a point of interest. In Subsection 11.5.5, we will talk more about the technical conditions for the tangent plane to be a good approximation to the surface, but for now we will use the following general formula for the tangent plane from our final result of Preview Activity 11.5.1.

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The Tangent Plane.

If \(f\) is a function of \(x\) and \(y\) for which both \(f_x\) and \(f_y\) exist and are continuous in an open disk containing the point \((x_0,y_0)\text{,}\) 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)\text{.}\tag{11.5.1} \end{equation}

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Having the equation for a tangent plane in the form given by \(z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\) allows us to quickly identify important information about the function \(f\) at the point \((x_0,y_0)\text{.}\) For example, if a function \(f\) has a tangent plane given by \(z = 7 - 2(x-3) + 4(y+1)\text{,}\) then we can immediately read from this form of the tangent plane equation the following:

\(f_x(3,-1) = -2\text{,}\)
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\(f_y(3,-1) = 4\text{,}\)
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\(f_x(3,-1)=-2\) is the slope of the trace to both \(f\) and the tangent plane in the \(x\)-direction at \((3,-1)\text{,}\) and
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\(f_y(3,-1) = 4\) is the slope of the trace of both \(f\) and the tangent plane in the \(y\)-direction at \((3,-1)\text{.}\)
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Activity 11.5.2.

(a)

Find the equation of the tangent plane to \(f(x,y) = x^2y\) at the point \((1,2)\text{.}\)

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(b)

Suppose that the tangent plane to the graph of a continuously differentiable function \(z=g(x,y)\) is given in the form

\begin{equation*} z = 5 - 3(x+2) + (y-3)\text{.} \end{equation*}

Use the equation of the tangent plane to identify a point on the graph as well as a value of \(g_x\) and a value of \(g_y\text{.}\) Be sure to identify at what point(s) you have found the values of the partial derivatives.

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In single-variable calculus, an important use of the tangent line is to approximate 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 \(f\) for input values close to \(x_0\text{,}\) as shown in Figure 11.5.3. Adjust the Zoom slider to examine a small region around the highlighted point to see how the tangent line is a good approximation for a small region around that point. In fact, the smaller the scale around the point, the better the approximation is.

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Figure 11.5.3. A plot of a locally linear function \(f\)

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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 \((x_0,y_0)\text{.}\) Determining when a function of two variables is locally linear at a point involves more nuance than in the single-variable setting. In Subsection 11.5.5 we will discuss the technical details and present some examples of functions that are not locally linear.

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If we consider a locally linear function, the tangent plane is a geometric version of the approximation to a surface given by \(z=f(x,y)\text{,}\) as shown in Subsection 11.5.2. In the next subsection, we will discuss the linearization, a functional form of the tangent plane that facilitates approximations near the point of tangency.

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Subsection 11.5.3 Linearization

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

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

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In the same way, the tangent plane to the graph of a locally linear function, \(f(x,y)\) at a point \((x_0,y_0)\text{,}\) 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. 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)\tag{11.5.2} \end{equation}

is the linearization of \(f\) at \((x_0,y_0)\text{.}\) Note that \(f(x,y)\approx L(x,y)\) for points near \((x_0, y_0)\text{,}\) which is illustrated in Figure 11.5.4.

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Figure 11.5.4. A plot of a locally linear function \(f(x,y)\)

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Using a function’s linearization when you already have an algebraic expression for the function provides little value. However, in many applications the function of interest does not have an algebraic formula. For instance, if you were working for a mining company and were trying to map a pocket of some resource underground, it would be very expensive to drill out a bunch of samples to make a chart with amounts of the resource at a comprehensive grid of locations. Instead, you could be more selective in obtaining data via sampling and then use linearization to make estimates for intermediate locations. Linearization is a valuable first step in determining a way to estimate values across different input locations. In later courses, you may encounter more sophisticated ways to estimate between data points and discuss the advantages and drawbacks of these ideas.

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Example 11.5.5.

In this example, we will give the equation of the tangent plane to \(f(x,y)=6-\frac{x^2}{2}-y^2\) at the point \((1,1)\text{,}\) state the associated linearization, and use this linearization to estimate the output of \(f\) for a nearby input.

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The partial derivatives of \(f\) are \(f_x=-x\) and \(f_y=-2y\text{.}\) This gives the values \(f_x(1,1)=-1\) and \(f_y(1,1)=-2\text{.}\) Using \(f(1,1)=4.5\text{,}\) the tangent plane to \(f\) at the point \((1,1)\) is

\begin{equation*} z=4.5+(-1)(x-1)+(-2)(y-1)\text{.} \end{equation*}

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The linearization of \(f\) at \((1,1)\) is

\begin{equation*} L(x,y)=4.5+(-1)(x-1)+2(y-1)\text{.} \end{equation*}

Notice that the same information is used to find the tangent plane and the linearization. Generally speaking, we consider the tangent plane to be a geometric tool, while and the linearization is a function or algebraic tool. However, both describe \(f\) near \((1,1)\text{.}\) We can use the linearization to estimate \(f(0.9,1.2)\text{:}\)

\begin{align*} f(0.9,1.2) \amp \approx L(0.9,1.2) = 4.5+(-1)(0.9-1)+(-2)(1.2-1)\\ \amp= 4.5+(0.1)-0.4=4.8\text{.} \end{align*}

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Because we had an algebraic expression for \(f\) in the previous example, we could have evaluated \(f(0.9,1.2)\) directly. However, as the next activity shows, many applications of multivariable functions use tables or graphs where it is not possible to have the function’s output at every possible input.

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Activity 11.5.3.

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

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(a)

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 \((1,2)\text{.}\) Use the linearization to estimate the value of \(g(0.8, 2.3)\text{.}\)

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(b)

The table below 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.

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\(v \backslash T\)	\(-20\)	\(-15\)	\(-10\)	\(-5\)	\(0\)	\(5\)	\(10\)
\(10\)	\(-41\)	\(-35\)	\(-28\)	\(-22\)	\(-16\)	\(-10\)	\(-4\)
\(15\)	\(-45\)	\(-39\)	\(-32\)	\(-26\)	\(-19\)	\(-13\)	\(-7\)
\(20\)	\(-48\)	\(-42\)	\(-35\)	\(-29\)	\(-22\)	\(-15\)	\(-9\)
\(25\)	\(-51\)	\(-44\)	\(-37\)	\(-31\)	\(-24\)	\(-17\)	\(-11\)
\(30\)	\(-53\)	\(-46\)	\(-39\)	\(-33\)	\(-26\)	\(-19\)	\(-12\)
\(35\)	\(-55\)	\(-48\)	\(-41\)	\(-34\)	\(-27\)	\(-21\)	\(-14\)

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 11.3.5.

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(c)

The image below gives a contour plot of a continuously differentiable function \(f\text{.}\) 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 \(f(2.2, 0.8)\text{.}\)

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Example 11.5.6.

Suppose we have a machine that cuts 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 \(dw = \Delta w = 0.2\) cm and the height could be off by \(dh = \Delta h = 0.4\) cm.

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The area of the rectangle is

\begin{equation*} A(w,h) = wh, \end{equation*}

which means 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 rectangle could differ from the perfect rectangle. We will estimate the uncertainty in the area using (11.5.3), and find that

\begin{equation*} \Delta A \approx dA = A_w(20, 10) \enspace dw + A_h(20,10)\enspace dh \end{equation*}

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By computing partial derivatives, we find that \(A_w = h\) and \(A_h = w\text{.}\) This implies that

\begin{equation*} \Delta A \approx dA = 10 \enspace dw + 20 \enspace dh = 10 (0.2) + 20 (0.4) = 10\text{.} \end{equation*}

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

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We can apply our differential for a box of a different size as well. If we wanted to make rectangles that are 63 cm wide and 25 cm high and our machines had the same tolerances (\(dw = \Delta w = 0.2\) cm and \(dh = \Delta h = 0.4\) cm), then our target rectangle would have area 1,575 square centimeters but the uncertainty in the area would be \(30.2 cm^2\text{:}\)

\begin{align*} \Delta A \approx dA =\amp A_w(63, 25) \enspace dw + A_h(63,25) \enspace dh\\ \amp= 25 dw + 63 dh\\ \amp= 25 (0.2) + 63 (0.4) = 30.2 \end{align*}

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In the next activity, we will explore the differential in several different contexts. First, however, we highlight some of the differences in meaning and usage between the change in the output of a function, the linearization, and the differential. In particular, we will look back at Figure 11.5.7, equation (11.5.3), and equation (11.5.2).

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The linearization of a function at a point is the algebraic function equivalent to the tangent plane at that point. If a function is “nice” (meaning differentiable or continuously differentiable), then the tangent plane/linearization will be close to the actual function on a small neighborhood around the base point. This gives us both a geometric tool in the form of the tangent plane and a numerical tool in the form of the lienarization to estimate the function’s outputs in a small region around the point of interest.

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As Figure 11.5.7 shows visually, the difference between a function’s output at \((x_0,y_0)\) and at \((x,y)\) is given by \(\Delta f\text{.}\) The difference between the linearization at \((x_0,y_0)\) and \((x,y)\) is given by \(df\text{.}\) If we look at a small enough scale, the difference between the differential and \(\Delta f\) will be small. The differential is primarily an algebraic tool that will allow us to talk how the change in the output of our function varies depending on the location we consider, which changes \(f_x\) and \(f_y\text{,}\) and how much change we will allow in the input variables, represented as \(dx\) and \(dy\text{.}\)

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Activity 11.5.4.

(a)

Suppose that the elevation of a plot of land 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 east and north directions, respectively, from \((0,0)\text{.}\)

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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 \(0.2\) miles; that is, \(dx = \Delta x = 0.2\) and \(dy= \Delta y = 0.2\text{.}\) Estimate the uncertainty in your elevation using differentials.

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(b)

The pressure, volume, and temperature of an ideal gas are related by the equation

\begin{equation*} P= P(T,V) = 8.31 \frac{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.

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(c)

Use the table of values for 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 provided in part 11.5.3.b for this part. Suppose your anemometer

An instrument for measuring wind speed.

says the wind is blowing at \(25\) miles per hour and your thermometer

An instrument for measuring the temperature.

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.

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Subsection 11.5.4 Differentials

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

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Figure 11.5.7. A plot of a locally linear function \(f(x,y)\)

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Figure 11.5.7 illustrates a perspective on measuring the change in the output of \(f\text{.}\) The graph of \(f\) is shown in blue, while the graph of its tangent plane is shown in red. The change we seek to measure is identified as \(\Delta f\) in blue. 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 output of the function \(f\) has changed. We denote this change by \(\Delta f\text{,}\) where

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

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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)-L(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\text{.} \end{align*}

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For consistency, we will denote the change in the input variables as \(dx = \Delta x\) and \(dy = \Delta y\text{.}\) These quantities are equal because the change in the \(x\)- and \(y\)-coordinates are the same whether we use the function \(f\) or the linearization \(L\text{.}\) Thus,

\begin{equation} \Delta f \approx df = f_x(x_0,y_0) dx + f_y(x_0,y_0) dy\text{.}\tag{11.5.3} \end{equation}

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The equivalent expression using Leibniz notation is

\begin{equation} df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy\text{.}\tag{11.5.4} \end{equation}

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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 (11.5.3) and (11.5.4) express the relationship between these changes. Equation (11.5.4) resembles an important idea from single-variable calculus: when \(y\) depends on \(x\text{,}\) we can use the notation of differentials to write

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

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linearization approximates around a particular point but differentials show the algebraic form corresponding to the same measures of change in terms of linear combinations.

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We close this section by highlighting an idea that was used throughout this section and will be used time and again in this chapter. For a “nice” function (e.g., locally linear, differentiable, or continuously differentiable), the rate of change in the output of the function as you approach a specific point is the same as the rate of change along the linearization. In other words, we can approximate change near a point by looking at the change on the linearization. We can also evaluate the rates of change along our surface by evaluating the corresponding change on the linearization. Regardless of the direction, these rates of change will be the same at the specified point.

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Subsection 11.5.5 Differentiability and Local Linearity

In Assemblage , we gave general formula for the equation of the tangent plane at a particular point on the graph of \(z=f(x,y)\text{,}\) but we had a condition involving partial derivatives that we needed to satisfy in order for the tangent plane to make sense. We will briefly discuss some ways in which a function can fail to be locally linear; in other words, we will look at several examples of functions of two variables where the tangent plane is either not defined or is not a good approximation of the surface near a particular point.

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In single-variable calculus, one of the first functions you studied that was continuous but had a point where the graph was not locally linear was the absolute value function \(f(x) = |x|\text{.}\) When zooming in on the graph of \(f\) near the point \((0,0)\text{,}\) the graph maintains a sharp corner, and thus \(f\) is not locally linear at the point on its graph with \(x=0\text{.}\)

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Example 11.5.8.

(a)

The function \(f(x,y)=|x|+|y|\) is graphed in Figure 11.5.9. No matter how much you zoom in around the origin, the surface will not look like a plane. You can also see this if you zoom in near any point on the surface with either \(x=0\) or \(y=0\text{.}\) While this function is continuous at every point, neither partial derivative will be defined when either \(x\) or \(y\) is zero.

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Figure 11.5.9. A plot of \(f(x,y)=|x|+|y|\)

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(b)

Your experience from single-variable calculus may lead you to reasonably expect that if \(f_x(a,b)\) and \(f_y(a,b)\) both exist at a point \((a,b)\text{,}\) then \(f\) is locally linear at \((a,b)\text{.}\) This is not sufficient, however. To illustrate this, consider the function defined by \(f(x,y) = x^{1/3} y^{1/3}\text{.}\) You can see from the figure below that as you zoom in around the origin, the graph does not flatten out and look like a plane. In Exercise 11 guides you through using the limit definition of the partial derivative to show that \(f_x(0,0)\) and \(f_y(0,0)\) both exist, but that \(f\) is not locally linear at \((0,0)\text{.}\)

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Figure 11.5.10. A plot of \(f(x,y)= x^{1/3} y^{1/3}\)

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A precise discussion of differentiability of functions of more than one variable is beyond the scope of this text, but if you are interested in exploring these ideas a bit more, see Exercise 15. We will be content to define a stronger, but more easily verified, set of conditions called continuous differentiability that ensure local linearity. For our purposes, continuous differentiability is the only condition we will need to use. All of the results we encounter will apply to differentiable functions, and so also apply to continuously differentiable functions.

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Differentiability.

If \(f\) is a function of \(x\) and \(y\) for which 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 \((x_0,y_0)\text{.}\)

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If a function \(f\) of two variables is continuously differentiable at a point \((x_0,y_0)\text{,}\) then \(f\) has a tangent plane at \((x_0,y_0)\text{.}\) This means that when viewed up close, the tangent plane and the graph of the function are virtually indistinguishable. You can see this illustrated in Figure 11.5.1 if you enable the Show Tangent Plane check box.

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Important Note: As discussed before, the function \(f(x,y)=x^{1/3}y^{1/3}\text{,}\) which we saw in Figure 11.5.10, is not locally linear as we zoom in near the point \((0,0)\text{.}\) However, as Exercise 11 demonstrates, both \(f_x(0,0)\) and \(f_y(0,0)\) exist for this function. Thus, we can define the plane \(z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\text{.}\) However, this plane is not “tangent” to the graph in any reasonable sense. This is why we have specified that we require that \(f\) be continuously differentiable at a point in order to discuss the tangent plane at that point. The function \(f(x,y)=x^{1/3}y^{1/3}\) is not continuously differentiable (or differentiable) at \((0,0)\text{!}\) Differentiability for a function of two variables implies the existence of a tangent plane. However, the existence of the two first order partial derivatives of a function at a point does not imply differentiability. This is quite different from single-variable calculus, where the derivative existing at a point is equivalent to the function being locally linear at that point.

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Subsection 11.5.6 Summary

A function \(f\) of two independent variables is locally linear at a point \((x_0,y_0)\) if the graph of \(z=f(x,y)\) looks like a plane as we zoom in on the graph (around the point \((x_0,y_0)\)). 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*}

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The function \(L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\) 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 \((x_0,y_0)\text{.}\) The graph of the linearization function is the tangent plane.
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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 \((x_0,y_0)\text{.}\)
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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 the output of \(f\) that results from small changes in \(x\) and \(y\text{.}\)

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Exercises 11.5.7 Exercises

1.

Find the linearization \(L \left( x, y \right)\) of the function \(f\left( x, y \right) = \sqrt{ 196 - 4 x^{2} - 16 y^{2} }\) at \(\left( -3, -3 \right)\text{.}\)

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\(L \left( x, y \right) =\)

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Note: Your answer should be an expression in x and y; e.g. “3x - 5y + 9”

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2.

Find the equation of the tangent plane to the surface \(z = e^{1 x/17} \ln \left( 2 y \right)\) at the point \((3,1, 0.8269)\text{.}\)

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z =

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Note: Your answer should be an expression of x and y; e.g. “5x + 2y - 3”

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3.

A student was asked to find an equation for the plane that is tangent plane to the surface

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\(\displaystyle{ z = {x^{3}-y^{4}} }\)

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at the point \(\displaystyle{ (x,y) = (2,4) }\text{.}\) The student’s answer was

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\(\displaystyle{ z = {3x^{2}\mathopen{}\left(x-2\right)-248-4y^{3}\mathopen{}\left(y-4\right)}}\text{.}\)

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At a glance, how do you know this is wrong? What mistakes did the student make? Select all that apply.
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- The -248 should not be in the answer.
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- The partial derivatives were not evaluated at the point.
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- The (x - 2) and (y - 4) should be x and y.
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- The answer is not a linear function.
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- All of the above
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Find a correct equation for the tangent plane.
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\(\displaystyle{ z = }\)

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4.

(a) Check the local linearity of \(f(x,y) = e^{x}\cos\mathopen{}\left(y\right)\) near \(x=-1,\ y=1.5\) by filling in the following table of values of \(f\) for \(x=-1.1,\ -1,\ -0.9\) and \(y=1.4,\ 1.5,\ 1.6\text{.}\) Express values of \(f\) with 4 digits after the decimal point.

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\(x =\)	-1.1	-1	-0.9
\(y = 1.4\)
\(y = 1.5\)
\(y = 1.6\)

(b) Next, fill in the table for the values \(x=-1.01,\ -1,\ -0.99\) and \(y = 1.49,\ 1.5,\ 1.51,\) again showing 4 digits after the decimal point.

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\(x =\)	-1.01	-1	-0.99
\(y = 1.49\)
\(y = 1.5\)
\(y = 1.51\)

Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other).

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(c) Give the local linearization of \(f(x,y) = e^{x}\cos\mathopen{}\left(y\right)\) at \((-1,1.5)\text{:}\)

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Using the second of your tables:

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\(f(x,y) \approx\)

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Using the fact that \(f_x(x,y) = e^{x}\cos\mathopen{}\left(y\right)\) and \(f_y(x,y) = -e^{x}\sin\mathopen{}\left(y\right)\text{:}\)

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\(f(x,y) \approx\)

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5.

Suppose that \(z\) is a linear function of \(x\) and \(y\) with slope 4 in the \(x\) direction and slope 1 in the \(y\) direction.

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(a) A change of \(-0.5\) in \(x\) and \(-0.4\) in \(y\) produces what change in \(z\text{?}\)

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change in \(z =\)

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(b) If \(z=8\) when \(x=5\) and \(y=4\text{,}\) what is the value of \(z\) when \(x=5\) and \(y=4.2\text{?}\)

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\(z =\)

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6.

Find the differential of the function \(w = x^{7} \sin(y^{2} z^{1})\)

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\(dw =\)\(dx +\) \(dy +\) \(dz\)

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7.

The dimensions of a closed rectangular box are measured as 50 centimeters, 90 centimeters, and 70 centimeters, respectively, with the error in each measurement at most .2 centimeters. Use differentials to estimate the maximum error in calculating the surface area of the box.

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square centimeters

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8.

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy \(U\) (in Joules) of the ammonia is a function of the volume \(V\) (in cubic meters) of the container, and the temperature \(T\) (in degrees Kelvin) of the gas. The differential \(dU\) is given by \(dU = 840 dV + 27.32 dT\text{.}\)

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(a) How does the energy change if the volume is held constant and the temperature is decreased slightly?

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it does not change
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it decreases slightly
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it increases slightly
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(b) How does the energy change if the temperature is held constant and the volume is increased slightly?

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it does not change
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it decreases slightly
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it increases slightly
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(c) Find the approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin.

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Change in energy = . Please include in your answer.

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9.

An unevenly heated metal plate has temperature \(T(x,y)\) in degrees Celsius at a point \((x,y)\text{.}\) If \(T(2,1) = 115\text{,}\) \(T_x \, (2,1) = 20\text{,}\) and \(T_y \, (2,1) = -15\text{,}\) estimate the temperature at the point \((2.03,0.98)\text{.}\)

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\(T(2.03,0.98) \approx\) . Please include in your answer.

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10.

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

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Find the equation of the tangent plane to \(f\) at the point \((1,2)\text{.}\)
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Use the linearization to approximate the values of \(f\) at the points \((1.1, 2.05)\) and \((1.3,2.2)\text{.}\)
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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.
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11.

Let \(f\) be the function defined by \(f(x,y) = x^{1/3}y^{1/3}\text{,}\) whose graph is shown in Figure 11.5.11.

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Figure 11.5.11. The surface for \(f(x,y) = x^{1/3}y^{1/3}\text{.}\)
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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 \(f_x(0,0)\text{?}\)

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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 11.5.11. Based on this picture, do you think \(f\) is locally linear at \((0,0)\text{?}\) Why?
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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 \((0,0)\text{?}\)
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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?
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12.

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

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.
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Estimate the value of \(g(-1.8, 4.7)\text{.}\) Clearly show your work and thinking.
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Given changes of \(dx = -0.34\) and \(dy = 0.21\text{,}\) estimate the corresponding change in \(g\) that is given by its differential, \(dg\text{.}\)
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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.
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13.

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

Find the linearization \(L(x,y)\) for the function \(f\) defined by \(f(x,y) = \cos(x)(2e^{2y}+e^{-2y})\) at the point \((x_0,y_0) = (0,0)\text{.}\) Use the linearization to estimate the value of \(f(0.1, 0.2)\text{.}\) Compare your estimate to the actual value of \(f(0.1, 0.2)\text{.}\)
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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 11.5.12.

Table 11.5.12. Heat index.

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T \(\downarrow \backslash\) H \(\rightarrow\)	\(70\)	\(75\)	\(80\)	\(85\)
\(90\)	\(106\)	\(109\)	\(112\)	\(115\)
\(92\)	\(112\)	\(115\)	\(119\)	\(123\)
\(94\)	\(118\)	\(122\)	\(127\)	\(132\)
\(96\)	\(125\)	\(130\)	\(135\)	\(141\)

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.

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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 \(h(-0.1, 0.9, -1.8)\text{.}\)
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14.

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

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 \(\left(\frac{\pi}{4}-0.01, \frac{\pi}{3}+0.1\right)\text{.}\)

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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 \(R\text{.}\)

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15.

In this section we argued that if \(f = f(x,y)\) is a function of two variables and if \(f_x\) and \(f_y\) both exist and are continuous in an open disk containing the point \((x_0,y_0)\text{,}\) then \(f\) is differentiable at \((x_0,y_0)\text{.}\) This condition ensures that \(f\) is differentiable at \((x_0,y_0)\text{,}\) but it does not define what it means for \(f\) to be differentiable at \((x_0,y_0)\text{.}\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x,y)= \sqrt{|xy|}\text{.}\)

Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times [-1,1]\text{.}\) Explain why \(g\) is not locally linear at \((0,0)\text{.}\)
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Show that both \(g_x(0,0)\) and \(g_y(0,0)\) exist. If \(g\) is locally linear at \((0,0)\text{,}\) what must be the equation of the tangent plane \(L\) to \(g\) at \((0,0)\text{?}\)
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Recall that if a function \(f = f(x)\) of a single variable is differentiable at \(x=x_0\text{,}\) then

\begin{equation*} f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h} \end{equation*}

exists. We saw in single variable calculus that the existence of \(f'(x_0)\) means that the graph of \(f\) is locally linear at \(x=x_0\text{.}\) In other words, the graph of \(f\) looks like its linearization \(L(x) = f(x_0)+f'(x_0)(x-x_0)\) for \(x\) close to \(x_0\text{.}\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_0\text{.}\) We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function

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

As \(x\) approaches \(x_0\text{,}\) \(E(x)\) approaches \(f(x_0)+f'(x_0)(0) - f(x_0) = 0\text{,}\) and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_0\text{.}\) Show that, even though \(g(x,y) = \sqrt{|xy|}\) is not locally linear at \((0,0)\text{,}\) its error term

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

at \((0,0)\) has a limit of \(0\) as \((x,y)\) approaches \((0,0)\text{.}\) (Use the linearization you found in part (b).) 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 \((x_0,y_0)\text{.}\)

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As the previous part illustrates, having the error term go to \(0\) does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f = f(x)\) as a function of one variable. If we let \(x = x_0+h\text{,}\) where \(|h|\) is the distance from \(x\) to \(x_0\text{,}\) then the relative error in approximating \(f(x_0+h)\) with \(L(x_0+h)\) is

\begin{equation*} \frac{E(x_0+h)}{h}. \end{equation*}

Show that, for a function \(f = f(x)\) of a single variable, the limit of the relative error is \(0\) as \(h\) approaches \(0\text{.}\)

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Even though the error term for a function of two variables might have a limit of \(0\) at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f = f(x,y)\) at a point \((x_0,y_0)\text{,}\) then if \(x = x_0+h\) and \(y = y_0+k\text{,}\) the distance from \((x,y)\) to \((x_0,y_0)\) is \(\sqrt{h^2+k^2}\text{.}\)
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Definition 11.5.13.

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 \(k = y-y_0\text{.}\)

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A function \(f\) is differentiable if it is differentiable at every point in its domain. The function \(L\) in the definition is the linearization of \(f\) at \((x_0,y_0)\text{.}\) Verify that \(g(x,y) = \sqrt{|xy|}\) is not differentiable at \((0,0)\) by showing that the relative error at \((0,0)\) does not have a limit at \((0,0)\text{.}\) Conclude that the existence of partial derivatives at a point is not enough to ensure differentiability at that point. (Hint: Consider the limit along different paths.)
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16.

Suppose that a function \(f = f(x,y)\) is differentiable at a point \((x_0,y_0)\text{.}\) Let \(L = L(x,y) = f(x_0,y_0) + m(x-x_0) + n(y-y_0)\) as in the conditions of Definition 11.5.13. Show that \(m = f_x(x_0,y_0)\) and \(n = f_y(x_0,y_0)\text{.}\) (Hint: Calculate the limits of the relative errors when \(h = 0\) and \(k = 0\text{.}\))

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17.

We know that if a function of a single variable is differentiable at a point, then that function is also continuous at that point. In this exercise we determine that the same property holds for functions of two variables. A function \(f\) of the two variables \(x\) and \(y\) is continuous at a point \((x_0,y_0)\) in its domain if

\begin{equation*} \lim_{(x,y) \to (x_0,y_0)} f(x,y) = f(x_0,y_0) \end{equation*}

or (letting \(x=x_0+h\) and \(y = y_0 + k\text{,}\)

\begin{equation*} \lim_{(h,k) \to (0,0)} f(x_0+h,y+k) = f(x_0,y_0). \end{equation*}

Show that if \(f\) is differentiable at \((x_0,y_0)\text{,}\) then \(f\) is continuous at \((x_0,y_0)\text{.}\) (Hint: Multiply both sides of the equality that comes from differentiability by \(\lim_{(h,k) \to (0,0)} \sqrt{h^2+k^2}\text{.}\))

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