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 your work in single variable calculus, the tangent line has served 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 visualized the value of the derivative as the slope of the line tangent to the graph of our function. 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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Additionally, we saw the correspondence between a function’s graph being locally linear and the function being differentiable. If we zoom in on the graph of a differentiable function around a point, then the graph will look like the tangent line to the function at that point. Figure 11.5.1 shows the graph of a curve with sliders that allow you to to zoom in around the labeled point on the curve or change the location for the labeled point on the curve. The displayed function’s graph is differentiable because at every point on the curve, you can zoom-in around the labeled point and the graph of the function will look linear (for a small region around the point). You should visually verify at several points that this function is differentiable at the labeled point by verifying the graph is locally linear.

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Figure 11.5.1. A plot with the ability to zoom in on a point of \(y=f(x)\text{,}\) where \(f\) is a nonlinear, differentiable function

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When viewing our differentiable function on a very small scale, the graph will be very close to the graph of the tangent line. Thus we can use the derivative as the slope of our tangent line in order to measure how tilted our graph is on a small scale. Because the graph of the function and the tangent line are so similar for a small region around the given point, the tangent line will allow us to approximate the output value of the function or the change in the output value for inputs near the given point. Tangent lines were used in many different applications and extensions of derivatives; For example, horizontal tangent lines gave critical points for optimization problems and tangent lines were used to approximate zeros or roots of functions when developing the Newton-Raphson Method.

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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, would be locally linear and approximated well by a tangent line if the function was 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, i.e. a line. When we generalized to higher dimensions, we saw how lines and planes would be considered flat graphs but lines would be considered a 1-dimensional flat graph (because there is only direction to move along the graph) and planes would be considered a 2-dimensional flat graph.

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 see what happens when we look at 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 11.5.2.

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We will examine the behavior of \(f\) near the input \((x_0,y_0) = (1,1)\) which has output of \(f(1,1)=4.5\text{.}\) In particular, we wish to view the graph of \(z=f(x,y)\) on smaller and smaller scales around the point \((1,1,4.5)\text{.}\) In Figure 11.5.2 you can use the slider labeled “Zoom” to change the scale on the plot. With a zoom of 0, you can see how the surface is curved very easily, but as you get past zoom level 3, the surface looks like a tilted plane. On plots like Figure 11.5.2, you should pay attention to the scale on the edge of the plot to see how small of a scale you are viewing around your point.

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Figure 11.5.2. A plot of \(f(x,y)=6-\frac{x^2}{2}-y^2\) with the point \((1,1,\frac{9}{2})\) 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, we see that the graph of this particular two-variable function looks like a plane, as seen in Figure 11.5.2. In the following preview activity, we will explore how to state an equation for this plane in terms of values related to our function \(f(x,y)\text{.}\) We will 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 11.5.3.

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 11.5.3) 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 \(b\) will be related to \(f(x,y)\text{.}\)

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(a)

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

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(b)

We want our plane to match the height of our surface at \((x_0,y_0)\) and the value \(z_0\) from Key Idea 11.5.3 will need to be the \(z\)-coordinate value where our plane will intersect the surface. What is the \(z\)-coordinate of the point where the tangent plane and the surface should intersect?

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(c)

If we want our plane to have the same behavior as the surface \(z=f(x,y)\) near our input \((x_0,y_0)\text{,}\) then the plane will need to match the behavior of the traces \(x=1\) and \(y=1\) near our point of interest.

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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 in Figure 11.5.4. Draw the tangent lines to the each of the traces when the input (either \(x\) or \(y\) as appropriate) is 1.

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Figure 11.5.4. The traces of \(f(x,y)\) with \(y=y_0=1\) and \(x=x_0=1\text{.}\)
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(d)

Give the slopes of the tangent lines to the traces that you drew in the previous part and write a few 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 \(f_y\text{.}\) You will likely want to talk about how each of these slopes/partial derivatives relates to the traces (of the surface and the plane) in Figure 11.5.5.

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Figure 11.5.5. A plot of \(f(x,y)=6-\frac{x^2}{2}-y^2\) with tangent plane at the point \((1,1,\frac{9}{2})\)

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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)= \underline{\hspace{1cm}} + \underline{\hspace{1cm}}(x-\underline{\hspace{1cm}}) +\underline{\hspace{1cm}}(y-\underline{\hspace{1cm}}) \end{equation*}

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Subsection 11.5.2 Differentiability and Local Linearity

Before we state a general formula for the equation of the tangent plane at a particular point on the graph of \(z=f(x,y)\text{,}\) we will discuss a technical condition and look at a couple of examples of functions that are NOT locally linear.

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As we have noted, when we look at the graph of a “nice” 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.

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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 \((x_0,y_0)\text{.}\) How can we tell when a function of two variables is locally linear at a point?

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Example 11.5.6.

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If you looked at a graph of the function \(f(x,y)=|x|+|y|\text{,}\) you would see that no matter how much you zoom in around the origin (or any point on the surface with either \(x\) or \(y\) zero), the surface will never look like a plane. This function is continuous at every point, but neither partial derivative will be defined when either \(x\) or \(y\) are zero.

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Figure 11.5.7. A plot of \(f(x,y)=|x|+|y|\)

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(b)

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. Let us consider, the function \(f\) defined by \(f(x,y) = x^{1/3} y^{1/3}\text{.}\) In Exercise 11.5.7.11 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 \((0,0)\text{.}\) You can see from the figure below that as you zoom in around the origin, the graph will not flatten out and look planar.

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Figure 11.5.8. A plot of \(f(x,y)= x^{1/3} y^{1/3}\)

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It would take us too far afield to provide a rigorous discussion of differentiability of functions of more than one variable (but Exercise 11.5.7.15 covers some of these details). We will be content to define a stronger, but more easily verified, set of conditions that ensure local linearity.

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Differentiability.

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 \((x_0,y_0)\text{.}\)

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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 \((x_0,y_0)\text{.}\) This means that when viewed up close, the tangent plane and the graph of the function are virtually indistinguishable. We won’t formally define differentiability of multivariable functions in this text. For our purposes, continuous differentiability is the only condition we will 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.

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Subsection 11.5.3 The Tangent Plane

Now that we have a good set of conditions to check if it makes sense to give a tangent plane, we return to the result of Preview Activity 11.5.1 and have the following general formula for the tangent plane.

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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).\tag{11.5.1} \end{equation}

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Important Note: As can be seen in Exercise 11.5.7.11, it is possible that \(f_x(x_0,y_0)\) and \(f_y(x_0,y_0)\) exist for a function \(f\) (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 \((x_0,y_0)\text{.}\) The graph of \(f\) (Figure 11.5.8) does not look linear when we zoom in around the point \((x_0,y_0)\text{,}\) so the plane given 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 differentiability. This is quite different than what happens in single variable calculus, where the derivative existing is equivalent to the function being locally linear.

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Having the equation for a tangent plane in the form given by \(z = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)\) will allow us to quickly state some important information about the function \(f\) at the point \((x_0,y_0)\text{.}\) For example, if we have a particular tangent plane given by \(z = 7 - 2(x-3) + 4(y+1)\text{,}\) then 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 \((3,-1)\text{.}\)

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Activity 11.5.2.

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Find the equation of the tangent plane to function given by \(f(x,y) = 2 + 4x - 3y\) at the point \((1,2)\text{.}\) You should simplify your equation for the tangent plane as much as possible and explain why this result should make sense in terms of the function \(f\text{.}\)

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(b)

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

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Subsection 11.5.4 Linearization

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 11.5.9. If you use the zoom slider, you can look at small region around the highlighted point to see how the tangent line is a good approximation for a small region around the highlighted point. In fact, the smaller the scale around the point, the better the approximation is.

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Figure 11.5.9. A plot of a locally linear function \(f\)

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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 \(f\text{.}\)

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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. 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}

Finally, note that \(f(x,y)\approx L(x,y)\) for points near \((x_0, y_0)\text{,}\) which is illustrated in Figure 11.5.10.

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Figure 11.5.10. A plot of a locally linear function \(f(x,y)\)

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Using the linearization may seem like a needless complication to your life, but in many applications the actual function you are interested in is unknown. 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 grid of locations. You would want to get the most information possible and be able to estimate between the locations where you took data. This is where the linearization is a valuable first step in determining a way to estimate between different input locations. In later courses (like differential equations or numerical methods), 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 11.5.11.

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. Our first partial derivatives of \(f\) will be \(f_x=-x\) and \(f_y=-2y\) which take values of \(f_x(1,1)=-1\) and \(f_y(1,1)=-2\) for our point of interest. Using \(f(1,1)=4.5\text{,}\) the tangent plane to \(f\) at the point \((1,1)\) will be given by \(z=4.5+(-1)(x-1)+(-2)(y-1)\text{.}\)

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The linearization of \(f\) at (1,1) will be \(L(x,y)=4.5+(-1)(x-1)+2(y-1)\text{.}\) Notice that the same information is used to find the tangent plane and the linearization; the tangent plane is a geometric tool and the linearization is a function/algebraic tool, where both describe \(f\) near (1,1). We can use the Linearization to estimate what \(f(0.9,1.2)\) would be.

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

Because we have the algebraic form of our function would could have evaluated \(f(0.9,1.2)\) but 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 point.

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

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(b)

Table 11.5.12 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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Table 11.5.12. Wind chill as a function of wind speed and temperature

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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)

Figure 11.5.13 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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Figure 11.5.13. A contour plot of \(f(x,y)\)
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Subsection 11.5.5 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 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 \((x,y)\text{.}\)

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Figure 11.5.14. A plot of a locally linear function \(f(x,y)\)

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Figure 11.5.14 illustrates the change in the output of \(f\) that we are looking for as \(\Delta f\) in blue. Specifically, we want to look at the change near the input \((x_0,y_0)\) and we know the value \(f(x_0,y_0)\text{.}\) 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 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) \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 \end{align*}

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For consistency, we will denote the change in the independent variables as \(dx = \Delta x\) and \(dy = \Delta y\) 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\tag{11.5.3} \end{equation}

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Expressed equivalently in Leibniz notation, we have

\begin{equation} df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.\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 \(x\text{,}\) it follows in the notation of differentials that

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

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We will illustrate the use of differentials with an example.

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Example 11.5.15.

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*}

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 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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Since \(A_w = h\) and \(A_h = w\text{,}\) we have

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

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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 easily apply our differential for a different sized box as well. If we wanted to make boxes with width of 63 cm and height of 25 cm and our machines had the same tolerances (\(dw = \Delta w = 0.2\) cm and \(dh = \Delta h = 0.4\) cm), then our target box 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. We will take a moment here to point out 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 11.5.14, (11.5.3), and (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 a geometric (tangent plane) and a numerical (linearization) tool to estimate our function’s outputs in a small region around the point of interest.

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As Figure 11.5.14 shows visually, the difference between our functions output at \((x_0,y_0)\) and \((x,y)\) is given by \(\Delta f\) whereas 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 a primarily algebraic tool that will allow us to talk how the change in the output of our function will vary depending on the location we want to look at (\(f_x\) and \(f_y\)) and how much change we will allow in the input variables (\(dx\) and \(dy\)).

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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 some base point \((0,0)\text{.}\)

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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 \(0.2\) units; 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)

Table 11.5.12 is a 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. 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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We will take a moment here to point out an idea that was used throughout this section and will be used time and again in this chapter. For a “nice” function (locally linear/differentiable/continuously differentiable), the rate of change in the output of the function as you approach a point will be 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 changes along our surface by evaluating the corresponding change on the linearization and regardless of the direction, these rates of change will be the same at the base 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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\(\displaystyle{ z = {3x^{2}\!\left(x-2\right)-248-4y^{3}\!\left(y-4\right)}}\text{.}\)

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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 \(f(x,y) = e^{x}\cos\!\left(y\right)\) 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 \(f(x,y) = e^{x}\cos\!\left(y\right)\) 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 \(f_x(x,y) = e^{x}\cos\!\left(y\right)\) and \(f_y(x,y) = -e^{x}\sin\!\left(y\right)\text{:}\)

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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 units 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 units 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 11.5.16.

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Figure 11.5.16. 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 11.5.16. 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 11.5.17.

Table 11.5.17. 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 11.5.18.

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 11.5.18. 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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