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Directional Derivatives and Gradients

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

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Directional Derivatives and Gradients

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  1. 13.6 Directional Derivatives and Gradients

  2. Directional Derivative

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

  4. Directional Derivative 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.

  5. Directional Derivative 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.

  6. Directional Derivative 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.

  7. Directional Derivative 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)) Point above P Point above Q

  8. Directional Derivative 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.

  9. Directional Derivative

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

  11. Directional Derivative Two of these are the partial derivatives fx and fy.

  12. 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) = 2xy fy(x, y) = x2 + 2y; θ = π/4 Note:

  13. 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.

  14. 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

  15. b. Graph the surface and interpret the directional derivatives.

  16. The Gradient of a Function of Two Variables

  17. 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…..

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

  19. The Gradient of a Function of Two Variables The gradient of f is a vector in the xy-plane.

  20. 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…….

  21. Example 4 Let a. Compute Note that fx = -x/5 + y2/10 and fy = (xy)/5. Therefore……

  22. 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).

  23. 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……

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

  25. 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……

  26. 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……

  27. Applications of the Gradient

  28. Applications of the Gradient

  29. 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……

  30. 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……

  31. 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……

  32. 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……

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

  34. 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?

  35. 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).

  36. Applications of the Gradient

  37. 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.

  38. 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.

  39. Functions of Three Variables

  40. Functions of Three Variables

  41. Example 9 – Finding the Gradient for a Function of Three Variables Find f(x, y, z) for the function given by f(x, y, z) = x2 + y2 – 4z and find the direction of maximum increase of f at the point (2, –1, 1). Solution: The gradient vector is given by f(x, y, z) = fx(x, y, z)i + fy(x, y, z)j + fz(x, y, z)k = 2xi + 2yj – 4k

  42. cont’d So, it follows that the direction of maximum increase at (2, –1, 1) is f(2, –1, 1) = 4i – 2j – 4k. Level surface and gradient vector at (2, –1, 1) for f(x, y, z) = x2 + y2 – 4z

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