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.

spanDEFINITIONspan 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 \[ g′ (0)=f_{x}(x_{0},y_{0})\cos \theta +f_{y}(x_{0},y_{0})\sin \theta \underset{\color{#0066A7}{(2)}}{\underset{\color{#0066A7}{\uparrow} }{=}}D_\mathbf{u}f( x_{0},y_{0}) \]

Finding the Directional Derivative of a Function

1. Find the directional derivative \(D_{\mathbf{u}}f(x,y)\) of \(f(x,y)=x^{2}y+y^{2}\) in the direction of \(\mathbf{u}=\cos \dfrac{\pi }{4} \mathbf{i}+\sin \dfrac{\pi }{4}\mathbf{j}\).
2. What is \(D_{\mathbf{u}}f(1,2)\)?
3. Interpret \(D_{\mathbf{u}}f(1,2)\).

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.

Interpreting the Directional Derivative as a Slope

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.

spanDEFINITIONspan 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

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

Finding the Gradient of a Function of Two Variables

1. Find the gradient of \(f(x,y)=x^{3}y\) at the point \((2,1)\).
2. Use the gradient to find the directional derivative of \(f\) at \((2,1)\) in the direction from \((2,1)\) to \((3,5)\).

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 \(-\left\Vert {\bf\nabla\! }f(x_{0},y_{0})\right\Vert\).

Using Properties of the Gradient

1. Find the direction for which the directional derivative of \(f(x,y)=x^{2}-xy+y^{2}\) at \((1,-2)\) is a maximum.
2. Find the maximum value of the directional derivative.

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 \({\bf\nabla}\! f(x_{0},y_{0})\).

Modeling Temperature Change

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 \({50^{\circ}{\rm C}}\).

1. Find \(T=T( x,y)\).
2. Find the gradient of \(T\) at the point \(( -3,4)\).
3. In what directions does the temperature increase most rapidly?
4. In what directions does the temperature decrease most rapidly?
5. In what directions is the rate of change of \(T\) at \((-3,4)\) equal to \(0\)?

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 \((x_{0},y_{0})\).

Using Properties of the Gradient

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.

spanDEFINITIONspan 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.

spanDEFINITIONspan 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 \(P_{0}\).