OBJECTIVES

When you finish this section, you should be able to:

1. Find the directional derivative of a function of two variables (p. 863)
2. Find the gradient of a function of two variables (p. 866)
3. Use properties of the gradient (p. 868)
4. Find the directional derivative and gradient of a function of three variables (p. 871)

NEED TO REVIEW?

Unit vectors are discussed in Section 10.3, pp. 708-709.

DEFINITION Directional Derivative of a Function of Two Variables

Suppose $z=f(x,y)$ is a function of two variables whose domain is $D$ and $(x_{0},y_{0})$ is an interior point of $D$. If $\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}$, $0\leq \theta \leq 2\pi$, is a unit vector, the directional derivative of $f$ at $(x_{0},y_{0})$ in the direction of $\mathbf{u}$ is the number $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{D_{\mathbf{u}}f(x_{0},y_{0})=\lim\limits_{t \rightarrow 0}\dfrac{f(x_{0}+t\cos \theta ,y_{0}+t\sin \theta )-f(x_{0},y_{0})}{t}}}$$

provided the limit exists.

NEED TO REVIEW?

Conditions under which a function $z=f(x,y)$ is differentiable are discussed in Section 12.4, pp. 841-843.

THEOREM Directional Derivative of a Differentiable Function $f$

If $z=f(x,y)$ is a differentiable function, then the directional derivative of $f$ at $(x_{0},y_{0})$ in the direction of the unit vector $\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}$ is given by $$\begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{ D_{\mathbf{u}}f(x_{0},y_{0})=f_{x}(x_{0},y_{0})\cos \theta +f_{y}(x_{0},y_{0})\sin \theta }} \end{equation}$$

Proof

If $x=x_0\,+\,t\cos \theta$, $y=y_0\,+\,t\sin \theta$, and $\theta$ is fixed, we can express $z=f( x,y)$ as a function $z= f(x_{0}+t \cos \theta, y_{0} + t \sin\theta)=g(t)$. Then the derivative of $g$ at $t=0$ is $$\begin{eqnarray} g′ (0) &=& \lim\limits_{t\rightarrow 0}\dfrac{g(t)-g(0) }{t-0}=\lim_{t\rightarrow 0}\frac{f(x_{0}+t\cos \theta ,y_{0}+t\sin \theta )-f(x_{0},y_{0})}{t}\nonumber \\ &=& D_{\mathbf{u}}f(x_{0},y_{0}) \end{eqnarray}$$

Because $f$ is differentiable, we can use Chain Rule I to find $g′(t)$. $$\begin{eqnarray*} g′ (t) &=&\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{ \partial y}\frac{dy}{dt} \qquad \color{#0066A7}{\hbox{Chain Rule I.}} \\ &=&\frac{\partial f}{\partial x}\cos \theta +\frac{\partial f}{\partial y} \sin \theta \quad\! \color{#0066A7}{\dfrac{dx}{dt}=\dfrac{d}{dt} (x_{0}+t\cos \theta ) = \cos \theta ; \dfrac{dy}{dt}=\dfrac{d}{dt} (y_{0}+t\sin \theta ) =\sin \theta} \end{eqnarray*}$$

If $t=0$, then $x=x_{0}$ and $y=y_{0}$, and $$g′ (0)=f_{x}(x_{0},y_{0})\cos \theta +f_{y}(x_{0},y_{0})\sin \theta \underset{\color{#0066A7}{(2)}}{\underset{\color{#0066A7}{\uparrow} }{=}}D_\mathbf{u}f( x_{0},y_{0})$$

Solution (a) Since the partial derivatives of $f$, namely, $$f_{x}(x,y)=2xy\qquad \hbox{and}\qquad f_{y}(x,y)=x^{2}+2y$$

are continuous, the function $f$ is differentiable. So, we can use formula (1) with the unit vector $\mathbf{u}=\cos \dfrac{\pi }{4}\mathbf{i}+\sin \dfrac{\pi }{4}\mathbf{j} =\dfrac{\sqrt{2}}{2}\mathbf{i}+\dfrac{\sqrt{2}}{2 }\mathbf{j}$. Then $$\begin{eqnarray*} D_{\mathbf{u}}f(x,y)& =&f_{x}(x,y)\cos \theta +f_{y}(x,y)\sin \theta \\ &=& 2xy\cos \dfrac{\pi }{4}+(x^{2}+2y)\sin \dfrac{\pi }{4} \qquad \color{#0066A7}{f_x(x,y)=2xy;} \\ &&\hspace{11.50pc}\color{#0066A7}{f_y (x,y)=x^{2}+2y;\theta =\dfrac{\pi }{4}} \\ &=& \sqrt{2}xy+\dfrac{\sqrt{2}}{2}(x^{2}+2y) \end{eqnarray*}$$

(b) $D_{\mathbf{u}}f(1,2)=2\sqrt{2}+\dfrac{\sqrt{2}}{2}(1+4)=\dfrac{9\sqrt{2}}{2}$.

(c) When we are at the point $( 1,2)$ and moving in the direction $\mathbf{u}=\dfrac{\sqrt{2}}{2}\mathbf{i}+\dfrac{\sqrt{2}}{2} \mathbf{j}$, the function is changing at a rate of approximately 6.364 units per unit length.

NOW WORK

Problem 11.

THEOREM The Directional Derivative as a Slope

If $z=f(x,y)$ is a differentiable function, then the directional derivative $D_{\mathbf{u}}f(x_{0},y_{0})$ of $f$ at $(x_{0},y_{0})$ in the direction of $\mathbf{u}$ equals the slope of the tangent line to the curve $C$ at the point $( x_{0}, y_{0}, f(x_{0}, y_{0}))$ on the surface $z=f( x,y)$, where $C$ is the intersection of the surface with the plane perpendicular to the $xy$-plane and containing the line through $( x_{0}, y_{0}, 0)$ in the direction $\mathbf{u}$.

NOTE

The unit vector $\mathbf{u}$ in the direction of a given vector is defined in Section 10.3, p. 708.

Find the directional derivative of $f(x,y)=x\sin y$ at the point $\left( 2, \dfrac{\pi }{3}\right)$ in the direction of $\mathbf{a}=3\mathbf{i}+4 \mathbf{j}$. Interpret the result as a slope.

Solution To find a directional derivative of a function $f$ in the direction of $\mathbf{a}$, where $\mathbf{a}$ is a nonzero vector, we must first find the unit vector in the direction of $\mathbf{a}$. The unit vector $\mathbf{u}$ in the direction of $\mathbf{a}$ is $$\begin{equation*} \mathbf{u}=\frac{\mathbf{a}}{\Vert \mathbf{a}\Vert }=\dfrac{3\mathbf{i}+4 \mathbf{j}}{\sqrt{9+16}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j} \end{equation*}$$

The partial derivatives of $f$ $$f_{x}(x,y)=\sin y\qquad \hbox{and}\qquad f_{y}(x,y)=x\cos y$$

are continuous throughout the plane, so $f$ is differentiable. Using the unit vector $\mathbf{u}=\dfrac{3}{5}\mathbf{i}+\dfrac{4}{5}\mathbf{j=}\cos \theta \mathbf{\mathbf{i}+}\sin \theta \mathbf{{j}}$, we find

Figure 4 $f( x,y) =x\sin y$

At the point $\left( 2,\dfrac{\pi }{3}\right)$, $$\begin{equation*} D_{\mathbf{u}}f\!\left( 2,\frac{\pi }{3}\right) =\dfrac{3}{5}\sin \dfrac{\pi }{3}+\dfrac{4}{5}\!\left( 2\cos \dfrac{\pi }{3}\right) =\dfrac{3}{5}\!\left( \frac{\sqrt{3}}{2}\right) +\dfrac{8}{5}\!\left(\dfrac{1}{2}\right)=\frac{3\sqrt{3}+8}{10} \end{equation*}$$

The slope of the tangent line to the curve $C$ that is the intersection of the surface $f( x,y) =x\sin y$ and the plane perpendicular to the $xy$-plane that contains the line through the point $\left( 2,\dfrac{\pi }{3}, 0\right)$ in the direction $\mathbf{a}$ is approximately 1.32. See Figure 4.

NOW WORK

Problems 11 and 15.

NEED TO REVIEW?

The dot product is discussed in Section 10.4, pp. 715-716.

DEFINITION Gradient of a Function of Two Variables

Let $z=f(x,y)$ be a differentiable function. The vector ${\bf\nabla}\! f(x,y)$, read “del $f$,” is called the gradient of $f$ and is defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{ {\bf\nabla}\! f(x,y)=f_{x}(x,y)\mathbf{i}+f_{y}(x,y)\mathbf{j}}}$$

NOTE

“del $f$” is short for “delta $f$”; the symbol ${\bf\nabla}$ is an upside down Greek letter delta. In some books, ${\bf\nabla}\! f$ is written as “grad $f$.” The symbol ${\bf\nabla}$ has no meaning except when it operates on a function $z = f (x,y)$. So, ${\bf\nabla}$ is referred to as an operator, just as $\dfrac{d}{dx}$ is an operator indicating “take the derivative of a function of $x$.”

THEOREM The Directional Derivative as a Dot Product

The dot product of the vector ${\bf\nabla}\! f(x,y)$ and the unit vector $\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}$ is a scalar that is equal to the directional derivative $D_{\mathbf{u}}f(x,y)$. That is, $$\begin{equation} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{ {\bf\nabla}\! f(x,y)\,{\bf\cdot}\, \mathbf{u} =D_{\mathbf{u}}f(x,y) }}\tag{3} \end{equation}$$

Proof

$$\begin{eqnarray*} {\bf\nabla}\! f(x,y)\,{\bf\cdot}\, \mathbf{u} &=& [f_{x}(x,y) \mathbf{i}+f_{y}(x,y)\mathbf{j}] \,{\bf\cdot}\, [ \cos \theta \mathbf{i} +\sin \theta \mathbf{j}] \\ &=& f_{x}(x,y)\cos \theta +f_{y}(x,y)\sin \theta =D_{\mathbf{u}}f(x,y). \end{eqnarray*}$$

Solution (a) The gradient of $f$ at $(x,y)$ is $${\bf\nabla}\! f(x,y)=f_{x}(x,y)\mathbf{i}+f_{y}(x,y)\mathbf{j}=3x^{2}y \mathbf{i}+x^{3}\mathbf{j}$$

The gradient of $f$ at $(2,1)$ is $${\bf\nabla}\! f(2,1)=12\mathbf{i}+8\mathbf{j}$$

(b) The unit vector $\mathbf{u}$ from $(2,1)$ to $(3,5)$ is $$\begin{equation*} \mathbf{u=}\dfrac{( 3-2) \mathbf{i} +( 5-1) \mathbf{j} }{\sqrt{( 3-2) ^{2}+( 5-1) ^{2}}}=\frac{\mathbf{i}+4 \mathbf{j}}{\sqrt{17}} \end{equation*}$$

We use formula (3) for the directional derivative to find $D_{\mathbf{u} }(2,1).$ $$\begin{equation*} D_{\mathbf{u}}(2,1)={\bf\nabla\! }f(2,1)\,{\bf\cdot}\, \mathbf{u}=(12\mathbf{i}+8 \mathbf{j})\,{\bf\cdot}\, \frac{\mathbf{i}+4\mathbf{j}}{\sqrt{17}}=\frac{44}{\sqrt{17} } \end{equation*}$$

The directional derivative of $f$ at $(2,1)$ in the direction of $\mathbf{u}$ is $\dfrac{44}{\sqrt{17}}$

NOW WORK

Problem 25.

NEED TO REVIEW?

The angle between two vectors is discussed in Section 10.4, pp. 716-719.

THEOREM Properties of the Gradient

Suppose the function $z=f(x,y)$ is differentiable at $(x_{0},y_{0})$.

• If ${\bf\nabla}\! f(x_{0},y_{0})=\mathbf{0}$, then $D_{\mathbf{u} }f(x_{0},y_{0})=0$ for all directions $\mathbf{u}$.
• If ${\bf\nabla}\! f(x_{0},y_{0})\neq \mathbf{0}$, then the directional derivative $D_{\mathbf{u}}f( x_{0},y_{0})$ of $f$ at $(x_{0},y_{0})$ is a maximum when $\mathbf{u}$ is in the direction of ${\bf\nabla}\! f(x_{0},y_{0})$. The maximum value of $D_{\mathbf{u}}f(x_{0},y_{0})$ is $\left\Vert {\bf\nabla\! }f(x_{0},y_{0})\right\Vert$.
• If ${\bf\nabla}\! f(x_{0},y_{0})\neq \mathbf{0}$, then the directional derivative of $f$ at $(x_{0},y_{0})$ is a minimum when $\mathbf{u }$ is in the direction of $-{\bf\nabla}\! f(x_{0},y_{0})$. The minimum value of $D_{\mathbf{u} }f(x_{0},y_{0})$ is $-\left\Vert {\bf\nabla\! }f(x_{0},y_{0})\right\Vert$.

Solution (a) We begin by finding the gradient of $f$ at $(1,-2)$. $$\begin{array}{rcl@{\qquad}l} {\bf\nabla\! }f(x,y) &=&(2x-y)\mathbf{i}+(2y-x)\mathbf{j} & \color{#0066A7}{{\bf\nabla}\! f(x,y)=f_{x}(x,y) \mathbf{i}+f_{y}(x,y)\mathbf{j}} \\ {\bf\nabla\! }f(1,-2) &=&4\mathbf{i}-5\mathbf{j} & \color{#0066A7}{{x=1; y=-2}} \end{array}$$

The direction $\mathbf{u}$ for which $D_{\mathbf{u}}f(1,-2)$ is maximum occurs when ${\bf\nabla}\! f$ and the unit vector $\mathbf{u}$ have the same direction. That is, the directional derivative is a maximum when the direction is $\mathbf{u}=\dfrac{4\sqrt{41}}{41}\mathbf{i}-\dfrac{5\sqrt{41}}{ 41}\mathbf{j}$.

(b) The maximum value of the directional derivative at $(1,-2)$ equals the magnitude of the gradient, namely $$\Vert {\bf\nabla}\! f(1,-2)\Vert =\sqrt{16+25}=\sqrt{41}$$

NOW WORK

Problem 35.

THEOREM

Suppose $z=f(x,y)$ is differentiable at $(x_{0},y_{0})$ and ${\bf\nabla}\! f (x_{0},y_{0}) \neq \mathbf{0}$.

• The value of $z=f(x,y)$ at $(x_{0},y_{0})$ increases most rapidly in the direction of ${\bf\nabla}\! f(x_{0},y_{0})$ and decreases most rapidly in the direction of $-{\bf\nabla}\! f(x_{0},y_{0})$.
• The value of $z= f( x,y)$ remains the same for directions orthogonal to ${\bf\nabla}\! f(x_{0},y_{0})$.

A metal plate is placed on the $xy$-plane in such a way that the temperature $T$ in degrees Celsius at any point $P=(x,y)$ is inversely proportional to the distance of $P$ from $(0,0)$. Suppose the temperature of the plate at the point $(-3,4)$ equals ${50^{\circ}{\rm C}}$.

Solution (a) The temperature $T$ at any point $(x,y)$ is inversely proportional to the distance of $( x,y)$ from the origin $(0,0)$. We can model $T=T( x,y)$ as $$T(x,y)=\dfrac{k}{\sqrt{x^{2}+y^{2}}}$$

where $k$ is the constant of proportionality. Since $T=50^{\circ}{\rm C}$ when $(x,y)=(-3,4)$, then $$k=T\sqrt{x^{2}+y^{2}}=50\sqrt{( -3) ^{2}+4^{2}}=50(5)=250$$

So, $T=T( x,y) =\dfrac{250}{\sqrt{x^{2}+y^{2}}}$.

Figure 5 $T=T( x,y) =\dfrac{250}{\sqrt{x^{2}+y^{2}}}$

(b) The gradient of $T=T( x,y) =\dfrac{250}{\sqrt{ x^{2}+y^{2}}}$ is $${\bf\nabla }T(x,y)=T_{x}(x,y)\mathbf{i}+T_{y}(x,y)\mathbf{j}=-\frac{250x }{(x^{2}+y^{2})^{3/2}}\mathbf{i}-\frac{250y}{(x^{2}+y^{2})^{3/2}}\mathbf{ j}$$

At $( -3,4)$, $${\bf\nabla }T(-3,4)=-\frac{250 ( -3) }{ [ ( -3) ^{2}+4^{2}] ^{3/2}}\mathbf{i}-\frac{250 ( 4) }{ [ ( -3) ^{2}+4^{2}] ^{3/2}}\mathbf{j}=\frac{750}{125} \mathbf{i}-\frac{1000}{125}\mathbf{j}=6\mathbf{i}-8\mathbf{j}$$

(c) The temperature increases most rapidly in the direction ${\bf\nabla} T(-3,4)=6\mathbf{i}-8\mathbf{j}$.

(d) The temperature decreases most rapidly in the direction $-{\bf\nabla }T(-3,4)=-6\mathbf{i}+8\mathbf{j}$.

(e) The rate of change in $T$ at $(-3,4)$ equals $0$ for directions orthogonal to ${\bf\nabla} T(-3,4)=6\mathbf{i}-8\mathbf{j}$. That is, the rate of change in $T$ is $0$ in either of the two directions orthogonal to $6\mathbf{i}-8\mathbf{j}$, namely $\pm ( 8\mathbf{i}+6 \mathbf{j})$.

NEED TO REVIEW?

Level curves are discussed in Section 12.1, pp. 812-814.

NOW WORK

Problem 47.

NOTE

The grade of a mountain is a measure of its steepness. The name gradient given to ${\bf\nabla}\! f$ means the direction of steepest ascent.

THEOREM The Gradient Is Normal to the Level Curve

Suppose the function $z=f(x,y)$ is differentiable at a point $P_{0}=(x_{0},y_{0})$. If ${\bf\nabla}\! f(x_{0},y_{0})\neq \mathbf{0}$, then ${\bf\nabla}\! f(x_{0},y_{0})$ is normal to the level curve of $f$ at $P_{0}$.

RECALL

A normal vector to a curve at a point $P$ is defined as a vector with initial point at $P$ and orthogonal (perpendicular) to the tangent line to the curve at $P$.

Proof

Let $f(x,y)=k$ be the level curve through $P_{0}$. Suppose this level curve is represented parametrically by $x=x(t)$ and $y=y(t)$, with $x_{0}=x(t_{0})$ and $y_{0}=y(t_{0})$. Then $$\begin{equation*} f( x(t),y(t)) =k \end{equation*}$$

Differentiating $f( x(t),y(t))$ with respect to $t$, we get $$\begin{equation*} f_{x}( x(t),y(t))\, x′ (t)+f_{y}( x(t),y(t))\, y′ (t)=0 \qquad \color{#0066A7}{\hbox{Use Chain Rule I.}} \end{equation*}$$

or, equivalently, $$\begin{equation*} {\bf\nabla\! }f(x,y)\,{\bf\cdot}\, \lbrack x′ (t)\mathbf{i}+y′ (t) \mathbf{j}]=0 \end{equation*}$$

Since $x′ (t) \mathbf{i}+y′ (t) \mathbf{j}$ is tangent to the level curve, the vector ${\bf\nabla\! }f( x,y)$ is normal to the level curve. In particular, when $t=t_{0}$, ${\bf\nabla}\! f(x_{0},y_{0})$ is normal to the level curve of $f$ at $(x_{0},y_{0})$.

For the function $f(x,y)=\sqrt{x^{2}+y^{2}}$, graph the level curve containing the point $(3,4)$ and graph the gradient ${\bf\nabla\! }f(x,y)$ at this point.

Solution See Figure 8(a). The graph of the equation $z=\sqrt{x^{2}+y^{2}}$ is the upper half of a circular cone whose traces are circles. So, the level curves are concentric circles centered at $( 0,0)$. Because $f(3,4)=\sqrt{9+16}=5$, the level curve through $(3,4)$ is the circle $x^{2}+y^{2}=25$. Since $$\begin{equation*} {\bf\nabla}\! f(x,y)=\frac{x}{\sqrt{x^{2}+y^{2}}}\mathbf{i}+\frac{y}{\sqrt{ x^{2}+y^{2}}}\mathbf{j} \end{equation*}$$

the gradient at $(3,4)$ is $$\begin{equation*} {\bf\nabla}\! f(3,4)=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j} \end{equation*}$$

This vector is orthogonal to the level curve $x^{2}+y^{2}=25$, as shown in Figure 8(b).

Figure 8

NOW WORK

Problem 45.

DEFINITION Directional Derivative of a Function of Three Variables

The directional derivative $D_{\mathbf{u} }f(x_{0},y_{0},z_{0})$ of a function $w=f(x,y,z)$ at $(x_{0},y_{0},z_{0})$ in the direction of a unit vector $\mathbf{u=}$ $\cos \alpha \mathbf{i} +\cos \beta \mathbf{j}+\cos \gamma \mathbf{k}$ is given by $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{ D_{\mathbf{u}}f(x_{0},y_{0},z_{0})=\lim\limits_{t\rightarrow 0} \dfrac{f(x_{0}+t\cos \alpha ,y_{0}+t\cos \beta ,z_{0}+t\cos \gamma )-f(x_{0},y_{0},z_{0})}{t} }}$$

provided the limit exists.

DEFINITION Gradient of a Function of Three Variables

Let $w = f(x,y,z)$ be a differentiable function. Then the gradient of $f$ at $(x_{0},y_{0},z_{0})$ is the vector $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{ {\bf\nabla}\! f(x_{0},y_{0},z_{0})=f_{x}(x_{0},y_{0},z_{0})\mathbf{i} +f_{y}(x_{0},y_{0},z_{0})\mathbf{j}+f_{z}(x_{0},y_{0},z_{0})\mathbf{k} }}$$

THEOREM The Gradient Is Normal to the Level Surface

Suppose the function $w = f(x, y, z)$ is differentiable at a point $P_{0}=(x_{0}, y_{0}, z_{0})$. If ${\bf\nabla\! }f(x_{0}, y_{0}, z_{0})\neq \mathbf{0}$, the gradient ${\bf\nabla\! }f(x_{0}, y_{0}, z_{0})$ is normal to the level surface of $f$ through $P_{0}$.