14.5 Directional Derivatives and the Gradient Vector

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1 14.5 Directional Derivatives and the Gradient Vector 1. Directional Derivatives. Recall z = f (x, y) and the partial derivatives f x and f y are defined as f (x 0 + h, y 0 ) f (x 0, y 0 ) f x (x 0, y 0 ) = lim h 0 h f (x 0, y 0 + h) f (x 0, y 0 ) f y (x 0, y 0 ) = lim h 0 h and represent the rates of change of in the x- and y-directions, that is, in the directions of the unit vectors i and j. Suppose that we now wish to ind the rate of change of z at (x 0, y 0 ) in the direction of an arbitrary unit vector u =< a, b >. To do this we consider the surface S with the equation z = f (x, y) (the graph of f) and we let z 0 = f (x 0, y 0 ). Then the point P(x 0, y 0, z 0 ) lies on S. The vertical plane that passes through P in the direction of u intersects S in a curve C. The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u. The directional derivative if f at (x 0, y 0 ) in the direction of a unit vector u =< a,b > is D u f (x 0, y 0 ) = lim h 0 f (x 0 + ha, y 0 + hb) f (x 0, y 0 ) h (1) if the limit exists. 2. Theorem. If f is differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u =< a,b > and D u f (x, y) = f x (x, y)a + f y (x, y)b (2) 1

2 3. If u makes an angle θ with the positive x-axis, then we can write u =< cosθ,sinθ > and D u f (x, y) = f x (x, y)cosθ + f y (x, y)sinθ (3) Exercise 1. Find D u f (x, y) if f (x, y) = x 3 3x y + 4y 2 and u is the unit vector given by an angle θ = π 6. What is D u f (1,2) 4. Gradient Vector. Notice that D u f (x, y) = f x (x, y)a + f y (x, y)b (4) =< f x (x, y), f y (x, y) > < a,b > (5) =< f x (x, y), f y (x, y) > u (6) If f is a function of two variables x and y, then the gradient of f is the vector f defined by Also, Exercise 2. Find f when f (x, y) = sin x + e x y f (x, y) =< f x (x, y), f y (x, y) >= f x i + f y j (7) D u f (x, y) = f (x, y) u (8) Exercise 3. Find the directional derivative of the function f (x, y) = x 2 y 3 4y at the point (2,-1) in the direction of the vector v = 2i + 5j 5. Function of 3 Variables. The directional derivative if f at (x 0, y 0, z 0 ) in the direction of a unit vector u =< a,b,c > is D u f (x 0, y 0, z 0 ) = lim h 0 f (x 0 + ha, y 0 + hb, z 0 + hc) f (x 0, y 0, z 0 ) h (9) if the limit exists. D u f (x, y, z) = f x (x, y, z)a + f y (x, y, z) + f z (x, y, z)c = f (x, y, z) u where f (x, y, z) =< f x, f y, f z > Exercise 4. If f (x, y, z) = x sin(y z) (a) Find the gradient of f (b) Find the directional derivative of f at (1,3,0) at the direction of v =< 1,2, 1 > 2

3 6. Maximizing Directional Derivatives. Suppose is a differentiable function of two or three variables. The maximum value of the directional derivative D u f (x) is f (x) and it occurs when u has the same direction as the gradient vector f (x). Exercise 5. (a) If f (x, y) = xe y, fin the rate of change of f at the point P(2,0) in the direction from P to Q(0.5,2). (b) In what direction does f have the maximum rate of change? What is the maximum rate of change? 7. Tangent Planes to Level Surfaces. The Tangent Planes to the level surface F (x, y, z) = k at P(x 0, y 0, z 0 ) is the plane passes through P and with normal vector F (x 0, y 0, z 0 ). The equation is F x (x 0, y 0, z 0 )(x x 0 ) + F y (x 0, y 0, z 0 )(y y 0 ) + F z (x 0, y 0, z 0 )(z z 0 ) = 0 (10) The normal line to S at P is the line passing through P and perpendicular to the tangent plane. The direction of the normal line is therefore given by the gradient vector F (x 0, y 0, z 0 ) and so its symmetric equations are x x 0 F x (x 0, y 0, z 0 ) = y y 0 F y (x 0, y 0, z 0 ) = z z 0 F z (x 0, y 0, z 0 ) Exercise 6. Find the equations of the tangent plane and normal line at the point (-2,1,-3) to the ellipsoid x y 2 + z2 9 = 3 3

4 14.6 Problem Set 1. Find the directional derivative of f at the given point in the direction indicated by the angle θ. (a) f (x, y) = x 3 y 4 + x 4 y 3, (1,1), θ = π 6 (b) f (x, y) = ye x, (0,4), θ = 2π 3 2. For the following questions, find the gradient of f, the gradient at the point P, and find the rate of change of f at P in the direction of u (a) f (x, y) = sin(2x + 3y), P(-6,4), u =< 3 2, 1 2 > (b) f (x, y, z) = x 2 y z x y z 3, P(2,-1,1), u =< 0, 4 5, 3 5 > 3. Find the directional derivative of the function at the point P, and find the rate of change of f at P in the direction of u (a) f (p, q) = p 4 p 2 q 3, P(2,1), u =< 1,3 > (b) f (x, y, z) = xe y + ye z + ze x, P(0,0,0), u =< 5,1, 2 > (c) f (x, y, z) = ln(3x + 6y + 9z), P(1,1,1), u =< 4,12,6 > 4. Find the directional derivative of f (x, y) = x y at P(2,8) in the direction of Q(5,4). 5. Find the directional derivative of f (x, y, z) = x y + y z + zx at P(1,-1,3) in the direction of Q(2,4,5). 6. Find the maximum rate of change of f at the given point and the direction in which it occurs. (a) f (x, y) = 4y x, (4,1) (b) f (x, y) = sin(x, y), (1,0) (c) f (x, y, z) = x+y z, (1,1,-1) 7. Find the directions in which the directional derivative of f (x, y) = ye x y at the point (0,2) has the value Find all points at which the direction of fastest change of the function f (x, y) = x 2 + y 2 2x 4y is i + j 9. Find the equations of the tangent plane and the normal line to the given surface at the given point (a) 2(x 2) 2 + (y 1) 2 + (z 3) 2 = 10, (3,3,5) (b) y = x 2 z 2, (4,7,3) (c) x y z 2 = 6, (3,2,1) 4

5 14.7 Max and Min 1. (Local Max/Min. ) A function of two variables has a local maximum at (a,b) if f (x, y) f (a,b) when (x,y) is near (a,b). [This means that f (x, y) f (a,b) for all points (x,y) in some disk with center (a,b).] The number f(a,b) is called a local maximum value. If f (x, y) f (a,b) when (x,y) is near (a,b), then f has a local minimum at (a,b) and f(a,b) is a local minimum value 2. If the inequalities hold for all points in the domain of f, then f has an absolute maximum (or absolute minimum) at (a,b). 3. Theorem. If f has a local maximum or minimum at (a,b) and the first-order partial derivatives of f exist there, then f x (a,b) = 0 and f y (a,b) = 0 4. If the graph of f has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal. 5. Stationary Point. A point (a,b) is called a critical point (or stationary point) of f if f x (a,b) = 0 and f y (a,b) = 0, or if one of these partial derivatives does not exist 6. If f has local max/min at (a,b), then (a,b) is a critical point of f. However, not all critical points give rise to max/min. At a critical point, a function could have local max/min or neither. Exercise 7. Let f (x, y) = x 2 + y 2 2x 6y Find local max/min. Exercise 8. Find the extreme values of f (x, y) = y 2 x Second Derivative Test. Suppose the second partial derivatives of f are continuous on a disk with center (a,b), and suppose that f x (a,b) = 0 and f y (a,b) = 0 [that is, (a,b) is a critical point of f]. Let D = D(a,b) = f xx (a,b)f y y (a,b) [f x y (a,b)] 2 (11) 5

6 (a) If D>0 and f x x(a,b) > 0, then f(a,b) is a local min. (b) If D>0 and f x x(a,b) < 0, then f(a,b) is a local max. (c) If D<0, then f(a,b) is not a local max or min. In this case, (a,b) is a saddle point of f and the graph of f crosses its tangent plane at (a,b). (d) If D=0, inconclusive. That is, f(a,b) may be local max, or local min, or (a,b) may be a saddle point of f. Exercise 9. Find the local max and min and saddle points of f (x, y) = x 4 + y 4 4x y + 1 Exercise 10. Find and classify the critical points of Also find the highest point on the graph of f. f (x, y) = 10x 2 y = 5x 2 4y 2 x 4 2y 4 Exercise 11. Find the shortest distance from the point (1,0,-2) to the plane x + 2y + z = 4. Exercise 12. A rectangular box without a lid is to be made from 12 m 2 of cardboard. Find the maximum volume of such a box. 8. Extreme Value Theorem for Functions of Two Variables. If f is continuous on a closed, bounded set D in R 2, then f attains an absolute maximum value f (x 1, y 1 ) and an absolute minimum value f (x 2, y 2 ) at some points (x 1, y 1 ) and (x 2, y 2 ) in D. 9. To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D: 1. Find the values of f at the critical points of f in D 2. Find the extreme values of f on the boundary of D. 3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Exercise 13. Find the absolute maximum and minimum values of the function f (x, y) = x 2 2x y + 2y on the rectangle D = {(x, y) 0 x 3,0 y 2} 6

7 14.7 Problem Set 1. Find the local max/min values and saddle points of the function. (a) f (x, y) = x 2 + x y + y 2 + y (b) f (x, y) = x y 2x 2y x 2 y 2 (c) f (x, y) = (x y)(1 x y) (d) f (x, y) = x y + 1 x + 1 y (e) f (x, y) = e x cos y 2. Show f (x, y) = x 2 + 4y 2 4x y + 2 has an infinite number of critical points and that D=0 at each one. Then show f has a local and absolute minimum at each critical point. 3. Find absolute max/min in the region D. (a) f (x, y) = x 2 + y 2 2x, D is the closed triangular region with vertices (2,0), (0,2) and (0,-2) (b) f (x, y) = x + y x y, D is the closed triangular region with vertices (0,0), (0,2) and (4,0) (c) f (x, y) = x 2 + y 2 + x 2 y + 4, D = {(x, y) x 1, y 1} (d) f (x, y) = 4x + 6y x 2 y 2, D = {(x, y) 0 x 4,0 y 5} 4. Find the shortest distance from the point (2,0,-3) to the plane x + y + z = 1 5. Find the point on the plane x 2y + 3z = 6 that is closest to the point (0,1,1). 6. Find three positive numbers whose sum is 100 and whose product is a maximum 7. Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible. 8. Find the dimensions of the rectangular box with largest volume if the total surface area is given as 64 cm 2. 7

8 LAGRANGE MULTIPLIER Lagrange Multiplier 1. Find the extreme values of f (x, y, z) subject to the constraint g (x, y, z) = k. Step 1: Find all x, y, z and λ such that f (x, y, z) = λ g (x, y, z) (12) and g (x, y, z) = k (13) Step 2: Evaluate f at all points (x,y,z) from step 1. The largest of these values is the max of f. The smallest value is the min. Alternative of Step 1: Let F (x, y, z) = f (x, y, z) λ(g (x, y, z) k), find solution of F x = 0 F y = 0 F z = 0 F λ = 0 Exercise 14. Find the extreme value of f (x, y) = x 2 y ln x subject to 8x + 3y = 0. Exercise 15. Find the extreme value of f (x, y) = x y subject to 3x 2 + y 2 = 6. Exercise 16. A rectangular box without a lid is to be made from 12 m 2 of cardboard. Find the maximum volume of such a box. Exercise 17. Find the extreme values of the function f (x, y) = x 2 +2y 2 on the circle x 2 + y 2 = 1. Exercise 18. Find the points on the sphere x 2 + y 2 + z 2 = 4 that are closest to and farthest from the point (3,1,-1) 2. Two constraints. Step 1: Find all x, y, z and λ such that and f (x, y, z) = λ g (x, y, z) (14) f (x, y, z) = µ h(x, y, z) (15) g (x, y, z) = k (16) h(x, y, z) = c (17) 8

9 LAGRANGE MULTIPLIER Step 2: Evaluate f at all points (x,y,z) from step 1. The largest of these values is the max of f. The smallest value is the min. Alternative of Step 1: Let F (x, y, z) = f (x, y, z) λ(g (x, y, z) k) µ(h(x, y, z) c), find solution of F x = 0 F y = 0 F z = 0 F λ = 0 F µ = 0 Exercise 19. Find the maximum value of the function f (x, y, z) = x + 2y + 3z on the curve of intersection of the plane x y + z = 1 and the cylinder x 2 + y 2 = 1 9

10 PROBLEM SET Problem Set 1. Find the max/min values of the function subject to the given constant (a) f (x, y) = x 2 + y 2, x y = 1 (b) f (x, y) = 3x + y, x 2 + y 2 = 1 (c) f (x, y) = y 2 x 2, 1 4 x2 + y 2 = 1 (d) f (x, y, z) = 2x + 2y + z, x 2 + y 2 + z 2 = 1 (e) f (x, y, z) = x 2 + y 2 + z 2, x + y + z = 12 (f) f (x, y, z) = x y z, x 2 + 2y 2 + 3z 2 = 1 2. Find the extreme values of f subject to both constraints (a) f (x, y, z) = x + 2y, x + y + 1 = 1, y 2 + z 2 = 4 (b) f (x, y, z) = 3x y 3z, x + y z = 0, x 2 + 2z 2 = 1 (c) f (x, y, z) = y z + x y, x y = 1, y 2 + z 2 = 1 3. Find extreme value of f on the region described by the inequality (a) f (x, y) = x 2 + y 2 + 4x 4y, x 2 + y 2 9 (b) f (x, y) = 2x 2 + 3y 2 4x 5, x 2 + y

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