Calculus 3 Section 14.5 Reading Assignment: Directional Derivatives and Gradient Vectors

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Section 14.5 Reading Assignment: Directional Derivatives and Gradient Vectors Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. The reading begins with a limit definition ofthe directional derivative. Butjust like the limit definitions of derivatives we've considered so far, there will be a better way of computing this derivative. The limit definition can help understand what the directional derivative is conceptually though. Exercise 1. Read the Subsection "Interpretation of the Directional Derivative" (p. 846 84?). Thinking of z = f(x, y) as a surface, describe what the directional derivative means geometrically. Be careful to include the role ofthe direction in your answer. The subsection \"Calculations and Gradients" (p. 847 849} is probably the most computationally important here. We introduce the gradient vector. The triangle used for the gradient is called \"nabla" or "del." This is the only case where we won't write a vector arrow despite denoting a vector. The directional derivative is computed as a dot product. Read Example 2 (p. 848). Notice in this example that the direction must be "normalized\" into a unit vector by dividing by its magnitude before taking the dot product. This is easily forgotten, so it's worth actively keeping this in mind. 846 Chapter 14 Partial Derivatives Solution Applying the definition in Equation (1), we obtain ds a. Po = lim Eq. (1) +5 2 + - f(1, 2) = lim Substitute. - (1- + 1 -2) = lim 1 + 25 + + 2 + 35 - 3 V7 = lim 5.5 VZ lim = lim + 5-+0 The rate of change of f(x, y) = x + xy at A(1, 2) in the direction u is 5/ V2.Surface S: Tangent line FIGURE 14.28 The slope of the trace curve Cat A is lim slope (PO): this is the directional derivative as - Defles Interpretation of the Directional Derivative The equation z = f(x, y) represents a surface S in space. If zo = f(1, y), then the point P(Xo. ), 2) lies on S. The vertical plane that passes through P and P(x, ),) parallel to u intersects S in a curve C (Figure 14.28). The rate of change of f in the direction of u is the slope of the tangent to C at P in the right-handed system formed by the vectors u and k.14.5 Directional Derivatives and Gradient Vectors 847 When u = i, the directional derivative at P is afax evaluated at (x, y). When u = j, the directional derivative at P, is of /by evaluated at (30, y). The directional deriva- tive generalizes the two partial derivatives. We can now ask for the rate of change of f in any direction u, not just the directions i and j. For a physical interpretation of the directional derivative, suppose that I = f(x, y) is the temperature at each point (x, y) over a region in the plane. Then /(xo, y) is the tem- perature at the point Pi(xo. )) and Duf|p, is the instantaneous rate of change of the tem- perature at A stepping off in the direction u. Calculation and Gradients We now develop an efficient formula to calculate the directional derivative for a different tiable function f. We begin with the line x= No + 541. (2) through Po(xo. "), parametrizationre length parameter s increasing in the direction of the unit vector u = a i + waj. Then by the Chain Rule we find dix dy d's ax ay ds Chain Rule for differentiable f = From Eqs. (2). dr/ds - at Fa ay Fa and dy/ds = us = i + (3) Gradient of f at Direction u Equation (3) says that the derivative of a differentiable function f in the direction of u at P is the dot product of u with a special vector, which we now define.DEFINITION The gradient vector (or gradient) of f(x. y) is the vector of Of = A The value of the gradient vector obtained by evaluating the partial derivatives at a point A(x y) is written VFF. or The notation Vf is read "grad f" as well as "gradient of f" and "del f." The symbol V by itself is read "del." Another notation for the gradient is grad f. Using the gradient notation, we restate Equation (3) as a theorem. THEOREM 9-The Directional Derivative Is a Dot Product If fox, y) is differentiable in an open region containing P(xo. "), then ds = VSp . U. (4) the dot product of the gradient Vf at A, with the vector u. In brief. Dof = Vf . u.848 Chapter 14 Partial Derivatives EXAMPLE 2 Find the derivative of f(x, y) = xe + cos (xy) at the point (2, 0) in the direction of v = 3i - 4j- Solution Recall that the direction of a vector v is the unit vector obtained by dividing v by its length: n The partial derivatives of f are everywhere continuous and at (2, 0) are given by V-i+ 2j f.(2, 0) = (e' - y sin (xy)) =0-0 =1 f,(2, 0) = (xel - x sin (xy)) = 20" - 2.0 = 2. 12.0) The gradient of f at (2. 0) is P (2, 0) 3 flem = f42, 0ji + f (2, 0)j = i + 2j (Figure 14.29). The derivative of f at (2, 0) in the direction of v is therefore Eq. (4) with the D_fly, notation FIGURE 14.29 Picture Vf as a vector = (i + 2j). - 1 . in the domain of f. The figure shows a number of level curves of f. The rate at which f changes at (2, 0) in the direction Evaluating the dot product in the brief version of Equation (4) gives u is Vf . u = -1, which is the component Dif = Vf -u = \\Vf|lu| cos 0 = [Vf| cos 6. of V/ in the direction of unit vector u (Example 2). where o is the angle between the vectors u and Vf, and reveals the following properties.Properties of the Directional Derivative Dif = VS .u = [V/| cos d 1. The function f increases most rapidly when cos # = 1, which means that 0 = 0 and u is the direction of Vf. That is, at each point P in its domain, f increases most rapidly in the direction of the gradient vector Vf at P. The derivative in this direction is Dof = \\Vf| cos (0) = |Vfl- 2. Similarly, f decreases most rapidly in the direction of -Vf. The derivative in this direction is Def = [Vf| cos (#) = -|Vfl- 3. Any direction u orthogonal to a gradient Vf * 0 is a direction of zero change in f because o then equals w /2 and Duf = \\Vf| cos (w/2) = [Vf] . 0 = 0. As we discuss later, these properties hold in three dimensions as well as two. EXAMPLE 3 Find the directions in which f(x. )) = (x/2) + (13/2) (a) increases most rapidly at the point (1, 1), and (b) decreases most rapidly at (1, 1). (c) What are the directions of zero change in f at (1, 1)? Solution (a) The function increases most rapidly in the direction of Vf at (1, 1). The gradient there is Vflan = (i + x)) =i+j.14.5 Directional Derivatives and Gradient Vectors 849 Its direction is NG i+i i + j li + il VOO' HOY (b) The function decreases most rapidly in the direction of -Vf at (1, 1), which is (1. 1, 1) -1 =. 1, Di (c) The directions of zero change at (1, 1) are the directions orthogonal to Vf: Zero change Most rapid in f decrease in f and -1 = V2 Most rapid V/=i+j increase in f See Figure 14.30. FIGURE 14.30 The direction in which f(x, y) increases most rapidly at (1, 1) is the direction of Vflan = i + j- It corre- sponds to the direction of steepest ascent Gradients and Tangents to Level Curves on the surface at (1, 1, 1) (Example 3). If a differentiable function f(x, y) has a constant value c along a smooth curve r - gogi + hinj (making the curve part of a level curve of f). then f(e(), h() - c. Dif- ferentiating both sides of this equation with respect to / leads to the equationsof de of dh ax dr + ay di Chain Rule (5) i+ i+ ch = 0. The level curve f(x. y) = f(x0. Val dr Equation (5) says that Vf is normal to the tangent vector dr / dr, so it is normal to the curve. This is seen in Figure 14.31, where Vf is a nonzero vector (it is possible for Vf to be the zero vector). FIGURE 14.31 The gradient of a dif- ferentiable function of two variables at a At every point (xo, )) in the domain of a differentiable function f(x, y), the gradi- point is always normal to the function's ent of f is normal to the level curve through (xo, ") (Figure 14.31). level curve through that point. Equation (5) validates our observation that streams flow perpendicular to the contours in topographical maps (see Figure 14.26). Since the downflowing stream will reach its destination in the fastest way, it must flow in the direction of the negative gradient vectors from Property 2 for the directional derivative. Equation (5) tells us these directions are perpendicular to the level curves. This observation also enables us to find equations for tangent lines to level curves. They are the lines normal to the gradients. The line through a point P(x, Jo) normal to a nonzero vector N = Ai + Bj has the equation A(x - x) + B(y - >) = 0 (Exercise 39). If N is the gradient Vflex, = f.(xo vali + f, (xo- wj. and this gradient is not the zero vector, then this equation gives the following formula