Directional Derivatives and Gradients

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Directional Derivative

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Directional Derivative

You are standing on the hillside pictured and want to determine the hill’s incline toward the z-axis.

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If the hill were represented by z = f (x, y), you would already

know how to determine the slopes in two different

directions—the slope in the y-direction would be given by

the partial derivative fy(x, y), and the slope in the x-direction

would be given by the partial derivative fx(x, y).

In this section, you will see that these two partial

derivatives can be used to find the slope in any direction.

Directional Derivative

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To determine the slope at a point on a surface, you will define a new type of derivative called a directional derivative.

Begin by letting z = f (x, y) be a surface and P(x0, y0) be a point in the domain of f, as shown.

The “direction” of the directional

derivative is given by a unit vector

u = cos i + sin j

where is the angle the vector

makes with the positive x-axis.

Directional Derivative

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To find the desired slope, reduce the problem to two dimensions by intersecting the surface with a vertical plane passing through the point P and parallel to u, as shown.

This vertical plane intersects the

surface to form a curve C.

The slope of the surface at

(x0, y0, f (x0, y0)) in the direction

of u is defined as the slope of

the curve C at that point.

Directional Derivative

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You can write the slope of the curve C as a limit that looks much like those used in single-variable calculus.

The vertical plane used to form C intersects the xy-plane in a line L, represented by the parametric equations

x = x0 + t cos and y = y0 + t sin so that for any value of t, the point Q(x, y) lies on the line L.

For each of the points P and Q, there is a corresponding point on the surface.

(x0, y0, f (x0, y0))

(x, y, f (x, y))

Directional Derivative

Point above P

Point above Q

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Moreover, because the distance between P and Q is

you can write the slope of the secant line through (x0, y0, f (x0, y0)) and (x, y, f (x, y)) as

Finally, by letting t approach 0, you arrive at the following definition.

Directional Derivative

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Directional Derivative

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There are infinitely many directional derivatives of a surface at a given point—one for each direction specified by u, as shown.

Directional Derivative

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Two of these are the partial derivatives fx and fy.

Directional Derivative

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Example 1

Find the directional derivative Duf(x, y) of f(x, y) = x2y + y2 in the direction of

Since the partial derivatives of f, namely fx(x, y) = 2xy and fy(x, y) = x2 + 2y are continuous, the function f is differentiable.

fx(x, y) = 2xyfy(x, y) = x2 + 2y; θ = π/4

Note:

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What is Duf(1, 2)?

When we are at the point (1, 2) and moving in the direction of u, the function is changing at a rate of approximately 6.364 units per length.

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Example 2

Consider the paraboloid z = f(x, y) = ¼(x2 + 2y2) + 2. Let P0 be the point (3, 2) and consider the unit vectors u = <1/√2, 1/√2> and v = <1/2, -√3/2>.

a. Find the directional derivative of f at P0 in the direction of u and v.

We see that fx = x/2 and fy = y; evaluated at (3, 2), we have fx(3, 2) = 3/2 and fy(3, 2) = 2. The directional derivatives in the directions u and v are

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b. Graph the surface and interpret the directional derivatives.

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The Gradient of a Function of Two Variables

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The directional derivative of a differentiable function z = f(x, y) at a point (x0, y0) in the direction of the unit vector u = cos θi + sin θj is given by……

We can express Duf(x0, y0) as the dot product of two vectors as follows…..

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The vector fx(x0, y0)i + fy(x0, y0)j is called the gradient o f f at (x0, y0).

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The Gradient of a Function of Two Variables

The gradient of f is a vector in the xy-plane.

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Example 3

Find and for f(x, y) = x2 + 2xy – y3.

Computing fx = 2x + 2y and fy = 2x – 3y2, we have……

Substituting x = 3 and y = 2 gives…….

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Example 4Let

a. Compute

Note that fx = -x/5 + y2/10 and fy = (xy)/5. Therefore……

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b. Compute Duf(3, -1), where

Before computing the directional derivative, it is important to verify that u is a unit vector (in this case, it is). The required directional derivative is……

The figure on the next slide shows the tangent to the intersection curve in the plane corresponding to u whose slope is Duf(3, -1).

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c. Compute the directional derivative of f at (3, -1) in the direction of the vector <3, 4>.

In this case, the direction is given in terms of a nonunit vector. The vector <3, 4> has length 5, so the unit vector in the direction of <3, 4> is u = <3/5, 4/5>. The directional derivative at (3, -1) in the direction of u is……

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which gives the slope of the surface in the direction of u at (3, -1).

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Example 5

Find the gradient of f(x, y) = x3y at the point (2, 1).

The gradient of f at (x, y) is……

The gradient of f at (2, 1) is……

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Use the gradient to find the directional derivative of f at (2, 1) in the direction from (2, 1) to (3, 5).

The unit vector u from (2, 1) to (3, 5) is……

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Applications of the Gradient

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Applications of the Gradient

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Example 6

Consider the bowl-shaped paraboloid z = f(x, y) = 4 + x2 + 3y2. If you are located on the paraboloid at the point (2, -1/2, 35/4), in which direction should you move in order to ascend on the surface at the maximum rate? What is the rate of change?

At the point (2, -1/2), the value of the gradient is……

Therefore, the direction of steepest ascent in the xy-plane is in the direction of the gradient vector <4, -3> (or u = 1/5<4, -3>, as a unit vector). The rate of change is……

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If you are located at the point (2, -1/2, 35/4), in which direction should you walk in order to descend on the surface at the maximum rate? What is the rate of change?

The direction of steepest descent is the direction of…..

(or u = 1/5<4, -3>, as a unit vector). The rate of change is……

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At the point (3, 1, 16), in what direction(s) is there no change in the function values?

At the point (3, 1), the value of the gradient is……

The function has zero change if we move in either of the two directions orthogonal to <6, 6>; these two directions are parallel to <6, -6>. In terms of unit vectors, the directions of no change are……

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Example 7

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 50o C.

Find T = T(x, y).

where k is the constant of proportionality. Since T = 50o C when (x, y) = (-3, 4), then……

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Find the gradient of T at the point (-3, 4).

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The temperature increases most rapidly in the direction

The temperature decreases most rapidly in the direction

In what directions is the rate of change of T at (-3, 4) equal to zero?

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The rate of change in T at (-3, 4) equals zero for directions orthogonal to That is, the rate of change in T is zero in either of the two directions orthogonal to 6i – 8j, namely ±(8i + 6j).

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Applications of the Gradient

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Example 8

For the function graph the level curve containing the point (3, 4) and graph the gradient at this point.

The graph of z = f(x, y) is the upper half of a circular cone whose traces are circles.

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So, the level curves are concentric circles centered at (0, 0). Since f(3, 4) = √(9 + 16) = 5, the level curve through (3, 4) is the circle x2 + y2 = 25. Since……

the gradient at (3, 4) is……

This vector is orthogonal to the level curve x2 + y2 = 25.

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Functions of Three Variables

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Functions of Three Variables

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Example 9 – Finding the Gradient for a Function of Three Variables

Consider the function f(x, y, z) = x2 + 2y2 + 4z2 – 1 and its level surface f(x, y, z) = 3.

a. Find and interpret the gradient at the points P(2, 0, 0), Q(0, √2, 0), R(0, 0, 1), and S(1, 1, ½) on the level surface.

The gradient is……

Evaluating the gradient at the four points we find that…..

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The level surface f(x, y, z) = 3 is an ellipsoid.

In each case, the gradient points in the direction that f has its maximum rate of increase. Of particular importance is the fact – the gradient is orthogonal to the level surface.

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b. What are the actual rates of change of f in the directions of the gradients in part (a)?

The actual rate of increase of f at (a, b, c) in the direction of the gradient is…..