Section 4.4 - Differentials, Differentiability, Linearizations, and Tangent Planes

Section Objectives

1. Determine if a function of two variables is differentiable.
2. Compute the total differential of a function and use it to approximate change.
3. Find an equation of the plane tangent to a given surface at a point.
4. Find parametric equations for the line normal to a given surface at a point.
5. Use tangent planes (i.e., linearizations) to approximate function values.

Differentials

Let $f$$f$ be a two-variable function whose 1st-order partial derivatives exist at $(x,y)$$(x,y)$. Also let $\mathrm{\Delta }x$$\Delta x$ and $\mathrm{\Delta }y$$\Delta y$ be increments of $x$$x$ and $y$$y$, respectively. The differentials of $x$$x$ and $y$$y$ are defined by

and the total differential of $z=f(x,y)$$z=f(x,y)$ is defined by

Examples

• Let $z=3x^2-7xy+8y^2-5$$z=3x^2-7xy+8y^2-5$. Determine the differential $dz$$dz$.

• Let $w=z^2+e^y\cos x$$w=z^2+ e^y\cos x$. Determine the differential $dw$$dw$. (There are three independent variables! We must generalize from the definition.)

Roughly speaking, a total differential tells us how a infinitesimal change in the dependent variable is related to small changes in the independent variables. In this way, the total differential $dz$$dz$ approximates the increment $\mathrm{\Delta }z$$\Delta z$:

where $\mathrm{\Delta }z=f(x+\mathrm{\Delta }x,y+\mathrm{\Delta }y)-f(x,y)$$\Delta z = f(x+\Delta x, y+\Delta y) - f(x,y)$. To "approximate by differentials" is to use the approximation $\mathrm{\Delta }z\approx dz$$\Delta z \approx dz$.

Examples

• Use differentials to approximate the change in $z=\sqrt{4-x^2-y^2}$$z=\sqrt{4-x^2-y^2}$ as $(x,y)$$(x,y)$ changes from $(1,1)$$(1,1)$ to $(1.01,0.97)$$(1.01,0.97)$. Compare the approximate change to the exact change.

• The period of a pendulum of length $L$$L$ is given by

where $g$$g$ is the acceleration due to gravity. A pendulum is moved from a location where $g=32.09$$g=32.09$ ft/s${}^2$$^2$ to a location where $g=32.23$$g=32.23$ ft\s${}^2$$^2$. There was also a temperature change that resulted in a change in the length of the pendulum from $2.5$$2.5$ ft to $2.48$$2.48$ ft. Use differentials to approximate the corresponding change in the pendulum's period.

Differentiability

For functions of a single variable, "having a derivative" and "being differentiable" mean the same thing. For functions of several variables, however, the partial derivative is a weaker concept than the corresponding Calculus I derivative. For example, for functions of a single variable, wherever the derivative exists, the function is automatically continuous. For multi-variable functions, the existence of partial derivatives does not necessarily imply continuity.

Our goal now is to generalize Calculus I differentiability to multi-variable functions so that the new generalized idea retains the "strength" it initially had.

Definition

Let $z=f(x,y)$$z=f(x,y)$. The function $f$$f$ is differentiable at $(x,y)$$(x,y)$ if $\mathrm{\Delta }z=f(x+\mathrm{\Delta }x,y+\mathrm{\Delta }y)-f(x,y)$$\Delta z= f(x+\Delta x,y + \Delta y)-f(x,y)$ can be written in the form

where ${ϵ}_1$$\epsilon_1$ and ${ϵ}_2$$\epsilon_2$ are functions of $x$$x$, $y$$y$, $\mathrm{\Delta }x$$\Delta x$, and $\mathrm{\Delta }y$$\Delta y$ having the the property that $({ϵ}_1,{ϵ}_2)\to (0,0)$$(\epsilon_1,\epsilon_2) \to (0,0)$ as $(\mathrm{\Delta }x,\mathrm{\Delta }y)\to (0,0)$$(\Delta x, \Delta y) \to (0,0)$.

According to this definition, to be differentiable means that a function can be well-approximated by differentials and that the approximation gets better and better as $\mathrm{\Delta }x$$\Delta x$ and $\mathrm{\Delta }y$$\Delta y$ get small. This is completely analogous to the definition of differentiability in Calculus I, but it wasn't presented there in quite the same way. With this definition of differentiability, we can continue to say that differentiability implies continuity: If a function of any number of variables is differentiable at a point, then it is continuous at that point.

Examples

• Use the definition to show that $f(x,y)=x^2+xy-5y$$f(x,y)=x^2+xy-5y$ is differentiable everywhere in ${\mathbb{R}}^2$$\mathbb{R}^2$.

You have noticed by now (after only one example!) that the definition of differentiability is rather difficult to work with. Fortunately, we have the following theorem.

Theorem

If $f(x,y)$$f(x,y)$ is a function such that $f$$f$, $f_x$$f_x$, and $f_y$$f_y$ all exist and are continuous in a neighborhood of $(a,b)$$(a,b)$, then $f$$f$ is differentiable at $(a,b)$$(a,b)$.

Linearizations

Let's return now to the idea of approximation by differentials:

Solving the approximation for $f(x+\mathrm{\Delta }x,y+\mathrm{\Delta }y)$$f(x+\Delta x,y+ \Delta y)$ gives

If we substitute $x_0$$x_0$ for $x$$x$ and $y_0$$y_0$ for $y$$y$, the last expression takes the form

According to this approximation, at an arbitrary point $(x,y)$$(x,y)$ that is close to $(x_0,y_0)$$(x_0,y_0)$, we should expect

The right-hand side of this expression is a linear function of the variables $x$$x$ and $y$$y$, and this motivates the following definition.

Definition

Suppose the function $f$$f$ has contiuous first partial derivatives in a neighborhood of the point $(x_0,y_0)$$(x_0,y_0)$. The linearization of $f$$f$ at $(x_0,y_0)$$(x_0,y_0)$ is given by

The approximation $f(x,y)\approx L(x,y)$$f(x,y) \approx L(x,y)$ is called the standard linear approximation at $(x_0,y_0)$$(x_0,y_0)$.

If $f$$f$ is a differentiable function, we should expect this approximation to be pretty good near the point $(x_0,y_0)$$(x_0,y_0)$. Of course, the farther we get from $(x_0,y_0)$$(x_0,y_0)$, the less we should expect from the approximation.

Examples

• Find the linearization of $f(x,y)=\sqrt{41-4x^2-y^2}$$f(x,y)=\sqrt{41-4x^2-y^2}$ at $(2,3)$$(2,3)$. Then use the linearization to approximate $f(2.1,2.9)$$f(2.1,2.9)$.

• Find the linearization of $g(x,y,z)=x^3-2xy+x{z}^2-6xyz$$g(x,y,z)=x^3-2xy+xz^2-6xyz$ at $(2,1,-1)$$(2,1,-1)$. Then use the linearization to approximate $g(1.94,1.1,-0.98)$$g(1.94,1.1,-0.98)$.

• Use a linearization (or differentials) to estimate $\sqrt{123}\sqrt[4]{17}$$\sqrt{123} \, \sqrt[4]{17}$.

Tangent planes

You may have not noticed, but the graph of the linearization $z=L(x,y)$$z=L(x,y)$ is a plane. Take a closer look:

This is a linear equation in the three variables $x$$x$, $y$$y$, and $z$$z$ (everything else is a number). Therefore, its graph is a plane in 3-dimensional space. In fact, that plane is the tangent plane. Analogous to a tangent line, the plane is tangent to the graph $f$$f$ at the point $(x_0,y_0)$$(x_0,y_0)$. We will come back to this idea in section 4.6, but for now, let's do a few examples.

Examples

• Let $f(x,y)=2x^2-3xy+8y^2+2x-4y+4$$f(x,y)=2x^2-3xy+8y^2+2x-4y+4$. Find an equation of the plane tangent to the graph of $f$$f$ at the point where $(x,y)=(2,-1)$$(x,y)=(2,-1)$. Then use technology to sketch their graphs.

x
var("x,y")
f=2*x^2-3*x*y+8*y^2+2*x-4*y+4
g=34+13*(x-2)-26*(y+1)
A=plot3d(f,(x,0,4),(y,-3,1),color="lightgray", opacity=0.8)
B=plot3d(g,(x,0,4),(y,-3,1),color="yellow", opacity=0.5)
C=point([(2,-1,34)],size=50, color="black")
show(A+B+C,aspect_ratio=[2,2,0.05])

• Let $f(x,y)=e^{4x^2+6y^2}$$f(x,y)=e^{4x^2+6y^2}$. Find an equation of the plane tangent to the graph of $f$$f$ at the point where $(x,y)=(0,0)$$(x,y)=(0,0)$.

• For a general function $f$$f$, show that the vector $⟨f_x(x_0,y_0),f_y(x_0,y_0),-1⟩$$\langle f_x(x_0,y_0), f_y(x_0,y_0), -1 \rangle$ is normal to the tangent plane at $(x_0,y_0)$$(x_0,y_0)$.

• Use the result of the example above to find a set of parametric equations for the line normal to the graph of $z=5x^2-2y^2$$z=5x^2-2y^2$ at the point $(2,1,18)$$(2,1,18)$.