Gradient, directional derivative and tangent plane

Last week you saw partial derivatives: $f_x$ is the rate of change only in the $x$ direction, $f_y$ only in the $y$ direction. But how fast does the function change in the $45^\circ$ direction, or in any direction? The answer to this question is the directional derivative, and the key to computing it is the gradient vector. The gradient is the most important object in multivariable calculus: it points in the direction of fastest increase, it is perpendicular to level curves, and it gives the normal of the tangent plane. A directional derivative, a “maximum rate of change” and a tangent plane question are almost guaranteed on this topic in the exam.

Golden rules:

1. In a directional derivative, always convert the direction vector into a unit vector.
2. $D_{\mathbf u}f = \nabla f\cdot\mathbf u$. The largest value is $\lvert\nabla f\rvert$, in the direction of the gradient.
3. The gradient is perpendicular to the level curve (level surface). The tangent line and tangent plane formulas follow from this sentence.
4. For $z = f(x,y)$, the tangent plane and the linearization are the same thing: $L(x,y) = f(a,b)+f_x(x-a)+f_y(y-b)$.

The gradient

Definition.

$\nabla f(x,y) = \langle f_x,\,f_y\rangle,\qquad\nabla f(x,y,z) = \langle f_x,\,f_y,\,f_z\rangle.$

It is read “nabla f” or “grad f”. When evaluated at a point, it becomes a vector.

Example 1. $f(x,y) = x^2y-3y$. $\nabla f = \langle2xy,\,x^2-3\rangle$. At $(2,1)$, $\langle4,1\rangle$.

The directional derivative

Definition. Let $\mathbf u = \langle u_1,u_2\rangle$ be a unit vector. The derivative of $f$ at the point $(a,b)$ in the direction of $\mathbf u$ is

$D_{\mathbf u}f(a,b) = \lim_{h\to0}\frac{f(a+hu_1,\,b+hu_2)-f(a,b)}{h}.$

That is, how much $f$ changes if you walk one unit from $(a,b)$ in the direction $\mathbf u$.

Theorem. If $f$ is differentiable,

$D_{\mathbf u}f = \nabla f\cdot\mathbf u = f_xu_1+f_yu_2.$

For $\mathbf u = \mathbf i$, $D_{\mathbf u}f = f_x$; the partial derivatives are special cases of the directional derivative.

Example 1 (continued). At $(2,1)$ in the direction $\langle1,1\rangle$: $\mathbf u = \frac{1}{\sqrt2}\langle1,1\rangle$, $D_{\mathbf u}f = \frac{4+1}{\sqrt2} = \frac{5}{\sqrt2}\approx3.54$.

Example 2 (three variables, direction given by two points). $f(x,y,z) = xyz^2$, from $P(1,2,1)$ toward $Q(3,3,3)$.

PQ    = <2, 1, 2>,  |PQ| = 3,  u = <2/3, 1/3, 2/3>
∇f    = <y z^2, x z^2, 2xyz>  →  at P: <2, 1, 4>
D_u f = (2·2 + 1·1 + 4·2)/3 = 13/3

PITFALL · the unit vector. If you compute with $\nabla f\cdot\mathbf v$ when $\mathbf v$ is not a unit vector, your answer will be wrong by a factor of $\lvert\mathbf v\rvert$. In Q2, if you had multiplied directly with $\langle3,4\rangle$ you would have found $27$; the correct answer is $27/5$. Divide the moment you see the phrase “in the direction of”.

Fastest increase

$D_{\mathbf u}f = \nabla f\cdot\mathbf u = \lvert\nabla f\rvert\lvert\mathbf u\rvert\cos\theta = \lvert\nabla f\rvert\cos\theta$. $\theta$ is the angle between $\mathbf u$ and the gradient. From this:

Intuition. Imagine you are standing on the slope of a mountain. The gradient is a compass pointing in the direction of steepest climb. Its length is the slope in that direction. If you walk perpendicular to the gradient, you neither climb nor descend; you walk along a contour line. If you release a ball, it rolls in the direction of $-\nabla f$.

Example 3. The temperature of a plate is $T(x,y) = 100-x^2-2y^2$. In which direction should an ant at $(1,1)$ go to cool down fastest? $\nabla T = \langle-2x,-4y\rangle = \langle-2,-4\rangle$. The fastest decrease is in the direction $-\nabla T = \langle2,4\rangle$, that is, $\frac{1}{\sqrt5}\langle1,2\rangle$. In this direction the temperature drops at a rate of $\lvert\nabla T\rvert = \sqrt{20}\approx4.47$ degrees/unit.

The gradient and level curves

Along the level curve $f(x,y) = c$, $f$ does not change, that is, $D_{\mathbf u}f = 0$ in the direction tangent to the curve. That is why $\nabla f$ is perpendicular to the level curve. The tangent line to the level curve at $(x_0,y_0)$:

$f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) = 0.$

In three dimensions it is the same idea: $\nabla F$ is perpendicular to the level surface $F(x,y,z) = c$.

Tangent plane and normal line

Level surface $F(x,y,z) = c$. The normal of the tangent plane at $P_0(x_0,y_0,z_0)$ is $\nabla F(P_0)$:

$F_x(P_0)(x-x_0)+F_y(P_0)(y-y_0)+F_z(P_0)(z-z_0) = 0.$

The normal line: $\mathbf r(t) = P_0+t\,\nabla F(P_0)$.

Graph $z = f(x,y)$. If you write $F = f(x,y)-z = 0$, then $\nabla F = \langle f_x,f_y,-1\rangle$ and the formula is

$z-z_0 = f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$

Example 4. $z = x^2+3y^2$, $(1,1,4)$: $f_x = 2$, $f_y = 6$. $z = 4+2(x-1)+6(y-1) = 2x+6y-4$.

Example 5. The ellipsoid $x^2+2y^2+3z^2 = 6$, $(1,1,1)$: $\nabla F = \langle2x,4y,6z\rangle = \langle2,4,6\rangle\parallel\langle1,2,3\rangle$. The tangent plane is $x+2y+3z = 6$. The normal line is $(1+t,\,1+2t,\,1+3t)$.

TIP · which formula? If the surface is in the form $z = \ldots$, the second formula is quick. If the surface is a mixed equation in $x$, $y$, $z$ ($xyz = 6$, $x^2+y^2-z^2 = 1$), gather everything on one side, call it $F$, and take the gradient.

Linearization and the differential

The linearization of $z = f(x,y)$ at $(a,b)$ is the equation of the tangent plane:

$L(x,y) = f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b),\qquad f(x,y)\approx L(x,y)\ \ ((x,y)\approx(a,b)).$

Example 6. $f(x,y) = xe^y$, $(1,0)$: $f = 1$, $f_x = e^y = 1$, $f_y = xe^y = 1$. $L = 1+(x-1)+y$. $f(1.1,-0.1)\approx1+0.1-0.1 = 1.0$. The true value is $0.99532$.

Total differential.

$dz = f_x\,dx+f_y\,dy.$

In measurement error questions, $dx$, $dy$ are the possible errors in the measurements; for the worst case you take all of them in the same direction (with the signs that give a positive contribution).

Common mistakes