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Linearization: Tangent Planes and Differentials
This section relies on the ideas of planes from and first partial derivatives as measuring the tilt of a surface plot at a particular location from .
Linearization: Tangent Planes and Differentials
This section relies on the ideas of planes from and first partial derivatives as measuring the tilt of a surface plot at a particular location from . Some instructors will only cover a subset of the linearization tools covered here (tangent plane, total differential, linearization for approximation, differentiability, locally linear) but these tools are introduced and defined in such a way that each conceptually flows from the idea of locally linear functions. To cover all the topics in this section will likely take a couple of class meetings.
We have moved examples of functions that are not differentiable to the later subsection, . This subsection does not have activities because we feel that the deeper ideas involved in showing a function is not differentiable are not usually suitable for students at this level. There is an extension of the example function to to show that a function that has partial derivatives existing is NOT the same as the function being differentiable. There are several exercises that work through a more complete definition of differentiability including .
Introduction
Throughout 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 as the slope of the line tangent to the function's graph. This allowed us to see the derivative's value in terms of an important behavior of the function's graph.
In , we saw that a curve in space, given 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 the vector-valued function of one variable that describes a curve in space will be differentiable if the curve looks like a line on a small scale.
In this section, we will explore the various ways in which we can extend to functions of two variables the related ideas of differentiability and local linearity. Geometrically, this means we will use a tangent plane as a flat analog for surfaces of the form \(z=f(x,y)\).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 algebraically and numerically describe the change in the output of \(f(x,y)\).
To start our investigation, let's consider the graph of a nice two-variable function on a small scale. We will look at \[\begin{aligned}\end{aligned}\] whose graph is shown in Figure.
You should examine the behavior of \(f\) near the input \((x_0,y_0) = (1,1)\), which has output of \(f(1,1)=4.5\). You can change how closely Figure shows the graph of \(z=f(x,y)\) on smaller and smaller scales around the point \((1,1,4.5)\). 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.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
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 , 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.
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 \((x_0,y_0)\), 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 \[\begin{aligned}\end{aligned}\].
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 \((x_0,y_0)\). For example, if a function \(f\) has a tangent plane given by \(z = 7 - 2(x-3) + 4(y+1)\), then we can immediately read the following information from the given form of the tangent plane equation:
\(f(3,-1) = 7\),
\(f_x(3,-1)=-2\) is the slope of the trace to both \(f\) and the tangent plane in the \(x\)-direction at \((3,-1)\), and
\(f_y(3,-1) = 4\) is the slope of the trace of both \(f\) and the tangent plane in the \(y\)-direction at \((3,-1)\).
Activity
Find the equation of the tangent plane to \(f(x,y) = x^2y\) at the point \((1,2)\).
Suppose that the tangent plane to the graph of a continuously differentiable function \(z=g(x,y)\) is given in the form \[\begin{aligned}\end{aligned}\]. 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 \(g_y\). Be sure to identify at what point(s) you have found the values of the partial derivatives.
In single-variable calculus, an important use of the tangent line is to approximate a differentiable function. Near the point \(x_0\), the tangent line to the graph of \(f\) at \(x_0\) is close to the graph of \(f\) for input values close to \(x_0\), as shown in Figure. 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.
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Linearization
In single-variable calculus, we let \(L\) denote the function whose graph is the tangent line, and thus \[\begin{aligned}\end{aligned}\]. Furthermore, \(f(x) \approx L(x)\) near \(x_0\). We call \(L\) the linearization of \(f\).
Similarly, the tangent plane to the graph of \(f(x,y)\), a locally linear function of two variables, at a point \((x_0,y_0)\) provides a good approximation near \((x_0, y_0)\). We define the linearization, \(L\), to be the two-variable function whose graph is the tangent plane. Thus, \[\begin{aligned}\end{aligned}\] is the linearization of \(f\) at \((x_0,y_0)\). Note that \(f(x,y)\approx L(x,y)\) for points near \((x_0, y_0)\), as illustrated in Figure.
Using a function's linearization when you already have an algebraic expression for the function provides little added value. In many applications the function of interest does not have a known 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 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 other sampling techniques and then use linearization to make estimates for intermediate locations. Linearization is a valuable first step in estimating values between different input locations. In later math courses, you may encounter more sophisticated ways to estimate between data points and discuss the advantages and drawbacks of these ideas.
Example
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 \((1,1)\), state the associated linearization, and use this linearization to estimate the output of \(f\) for a nearby input.
The partial derivatives of \(f\) are \(f_x=-x\) and \(f_y=-2y\). This gives the values \(f_x(1,1)=-1\) and \(f_y(1,1)=-2\). Using \(f(1,1)=4.5\), the tangent plane to \(f\) at the point \((1,1)\) is \[\begin{aligned}\end{aligned}\].
The linearization of \(f\) at \((1,1)\) is \[\begin{aligned}\end{aligned}\]. 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 the linearization is a function or algebraic tool. However, both describe \(f\) near \((1,1)\). We can use the linearization to estimate \(f(0.9,1.2)\): \[\begin{aligned}f(0.9,1.2) \amp \approx L(0.9,1.2) = 4.5+(-1)(0.9-1)+(-2)(1.2-1) \\ \amp= 4.5+(0.1)-0.4=4.2\end{aligned}\].
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Differentiability and Local Linearity
In our equation for a tangent plane, we gave the general formula for the equation of the tangent plane at a particular point on the graph of \(z=f(x,y)\), 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.
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 \(f(x) = |x|\). When zooming in on the graph of \(f\) near the point \((0,0)\), the graph maintains a sharp corner, and thus \(f\) is not locally linear at the point on its graph with \(x=0\).
Remember 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 \((x_0,y_0)\). Determining when a function of two variables is locally linear at a point involves more nuance than in the single-variable setting, as the next example illustrates.
Example
The function \(f(x,y)=|x|+|y|\) is graphed in . 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 \(y=0\). While this function is continuous at every point, neither partial derivative will be defined when either \(x\) or \(y\) is zero.
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 \((a,b)\), then \(f\) is locally linear at \((a,b)\). This is not sufficient for multivariable functions, however. To illustrate this, consider the function defined by \(f(x,y) = x^{1/3} y^{1/3}\). 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. 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 \((0,0)\).
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 \((x_0,y_0)\), then \(f\) is continuously differentiable at \((x_0,y_0)\).
Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.
Practice (2)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Suppose that a function \(f = f(x,y)\) is differentiable at a point \((x_0,y_0)\). Let \(L = L(x,y) = f(x_0,y_0) + m(x-x_0) + n(y-y_0)\) as in the conditions of Definition. Show that \(m = f_x(x_0,y_0)\) and \(n = f_y(x_0,y_0)\). (Hint: Calculate the limits of the relative errors when \(h = 0\) and \(k = 0\).)
Жавобни кўрсатиш
Assume that \(f=f(x,y)\) is differentiable at a point \((x_0, y_0)\). Then by definition there exists a linear function \[\begin{aligned}\end{aligned}\] such that the relative error \[\begin{aligned}\end{aligned}\] goes to \(0\) as \((h, k)\) goes to \((0,0)\), where \(E(x, y) = f(x, y) - L(x, y), h = x - x_0, k = y-y_0\). Since we know that this limit exists and is equal to \(0\) we know that the limit must be \(0\) along any path to \((0, 0)\).
Consider the relative error. Along the path \(h = 0\) we have \[\begin{aligned}\frac{E(x_0 + h, y_0 + k)}{\sqrt{h^2 + k^2}} \amp = \frac{E(x_0, y_0 + k)}{\sqrt{k^2}} \\ \amp = \frac{f(x_0, y_0 + k) - L(x_0, y_0 + k)}{\sqrt{k^2}} \\ \amp = \frac{f(x_0, y_0 + k) - \left ( f(x_0, y_0) + m(x_0 - x_0) + n(y_0 + k - y_0) \right )}{\sqrt{k^2}} \\ \amp = \frac{\left ( f(x_0, y_0 + k) - f(x_0, y_0) \right ) - n(k)}{\vert k \vert }\end{aligned}\]. Recall that \[\begin{aligned}\end{aligned}\]. Since the limit of the relative error is \(0\) we must have \[\begin{aligned}0 \amp = \lim_{k \to 0} \frac{E(x_0 + h, y_0 + k)}{\sqrt{h^2 + k^2}} \\ \amp = \lim_{k \to 0} \frac{\left ( f(x_0, y_0 + k) - f(x_0, y_0) \right ) - n(k)}{\vert k \vert }\end{aligned}\]. From here we can show that \[\begin{aligned}0 \amp = \lim_{k \to 0} \frac{\left ( f(x_0, y_0 + k) - f(x_0, y_0) \right ) - n(k)}{k} \\ \amp = \lim_{k \to 0} \left ( \frac{f(x_0, y_0 + k) - f(x_0, y_0) }{k} - n \right )\end{aligned}\] which implies \[\begin{aligned}\end{aligned}\].
Similiarly, by taking the limit along the path \(k=0\) we can show that \[\begin{aligned}\end{aligned}\].
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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 \[\begin{aligned}\end{aligned}\] or (letting \(x=x_0+h\) and \(y = y_0 + k\), \[\begin{aligned}\end{aligned}\] Show that if \(f\) is differentiable at \((x_0,y_0)\), then \(f\) is continuous at \((x_0,y_0)\). (Hint: Multiply both sides of the equality that comes from differentiability by \(\lim_{(h,k) \to (0,0)} \sqrt{h^2+k^2}\).)
Жавобни кўрсатиш
Assume \(f\) is differentiable at \((x_0, y_0)\). Then by definition \[\begin{aligned}\end{aligned}\], so we have \[\begin{aligned}0 \amp = \lim_{(h, k) \to (0,0)} \frac{f(x_0 + h, y_0 + k) - L(x_0 + h, y_0 + k)}{\sqrt{h^2 + k^2}} \\ \amp = \lim_{(h, k) \to (0,0)} \frac{f(x_0 + h, y_0 + k) - (f(x_0, y_0) + m(x - x_0) + n(y - y_0) )}{\sqrt{h^2 + k^2}} \\ \amp = \lim_{(h, k) \to (0,0)} \frac{f(x_0 + h, y_0 + k) - f(x_0, y_0) + mh + nk}{\sqrt{h^2 + k^2}}\end{aligned}\]. Multiply both sides by \(\lim_{(h, k) \to (0,0)} \sqrt{h^2 + k^2}\) to get \[\begin{aligned}0 \amp = \left (\lim_{(h, k) \to (0,0)} \sqrt{h^2 + k^2} \right ) \left ( \frac{f(x_0 + h, y_0 + k) - f(x_0, y_0) + mh + nk}{\sqrt{h^2 + k^2}} \right) \\ \amp = \lim_{(h, k) \to (0,0) } \left(f(x_0 + h, y_0 + k) - f(x_0, y_0) + mh + nk \right )\end{aligned}\]. We know \(\lim_{(h, k) \to (0,0)} (mh + nk) = 0\), so we have \[\begin{aligned}\lim_{(h, k) \to (0,0)} \left(f(x_0 + h, y_0 + k) - f(x_0, y_0) \right )\amp = \lim_{(h, k) \to (0,0)} \left((x_0 + h, y_0 + k) - f(x_0, y_0) + (mh + nk) - (mh + nk) \right ) \\ \amp = \lim_{(h, k) \to (0,0)}\left(f(x_0 + h, y_0 + k) - f(x_0, y_0) + mh + nk \right ) - \lim_{(h, k) \to (0,0)} (mh + nk) \\ \amp = 0 - 0 \\ \amp = 0\end{aligned}\]. Thus \[\begin{aligned}\end{aligned}\] and so \(f\) is continuous at \((x_0, y_0)\).
Symbols used here
The value f(x) approaches as x approaches a.
The non-negative number whose square (n-th power) is x.
Equal to the precision shown, not exactly.
Derivative with respect to x, holding the other variables fixed.
Vector of partial derivatives; points uphill.
Antiderivative (indefinite) or signed area from a to b (definite).
Integral over a region of the plane; integral around a closed curve.
A quantity with magnitude and direction; a column of numbers.
Σ u_i v_i; the length of v, √(v·v).
How to: Linearization: Tangent Planes and Differentials
- What does it mean for a function of two variables to be locally linear at a point?
- How do we find the equation of the plane tangent to a locally linear function at a point?
- How can a linearization be used to approximate the output of a multivariable function?
Questions people ask
What is a partial derivative?
The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.
What does the gradient point at?
Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.
Ўзингизни синаб кўринг
Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.
Кўпроқ Multivariable Calculus
Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems