MATH 19520/51 Class 8
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1 MATH 19520/51 Class 8 Minh-Tam Trinh University of Chicago
2 1 Directional derivatives. 2 Gradient vectors. 3 Review level sets. 4 Tangent planes to level surfaces of functions of three variables.
3 Directional Derivatives A displacement vector v can always be rewritten ( ) v (1) v = v. v The scalar v tells us its length and the unit vector v/ v tells us its direction.
4 Directional Derivatives A displacement vector v can always be rewritten ( ) v (1) v = v. v The scalar v tells us its length and the unit vector v/ v tells us its direction. Examples of unit vectors in R 2 : (2) (1, 0), ( 3 5, 4 5 ), (0, 1), ( 1, 0), ( 1 2, 1 2 ).
5 If f is a function of x and y, then f x, resp. f y, is the change in f per unit length in the direction of (1, 0), resp. (0, 1).
6 If f is a function of x and y, then f x, resp. f y, is the change in f per unit length in the direction of (1, 0), resp. (0, 1). But what s the change per unit length in the direction of ( 3 5, 4 5 )?
7 If f is a function of x and y, then f x, resp. f y, is the change in f per unit length in the direction of (1, 0), resp. (0, 1). But what s the change per unit length in the direction of ( 3 5, 4 5 )? By proportionality, we want (3) 3/5ths of the change per unit length in the (1, 0)-direction + 4/5ths of the change per unit length in the (0, 1)-direction. That is, f x (3/5) + f y (4/5).
8 If u = (u 1, u 2 ) R 2 is a unit vector, then the derivative of f (x, y) in the direction of u is (4) D u f = f x u 1 + f y u 2. Formally, (5) D u f (x, y) = lim ɛ 0 f (x + u 1 ɛ, y + u 2 ɛ) f (x, y) ɛ assuming the limit exists.
9 Example (Stewart, 14.6, Example 2) Let f (x, y) = x 3 3xy + 4y 2, and let u R 2 be the unit vector at an angle of π/6 counterclockwise from (1, 0). Find u and D u f (1, 2).
10 Example (Stewart, 14.6, Example 2) Let f (x, y) = x 3 3xy + 4y 2, and let u R 2 be the unit vector at an angle of π/6 counterclockwise from (1, 0). Find u and D u f (1, 2). If u = (a, b), then a = cos(π/6) = 3/2 and b = sin(π/6) = 1/2.
11 Example (Stewart, 14.6, Example 2) Let f (x, y) = x 3 3xy + 4y 2, and let u R 2 be the unit vector at an angle of π/6 counterclockwise from (1, 0). Find u and D u f (1, 2). If u = (a, b), then a = cos(π/6) = 3/2 and b = sin(π/6) = 1/2. Since f x = 3x 2 3y and f y = 3x + 8y, we compute f x (1, 2) = 3 and f y (1, 2) = 13. Then (6) D u f (1, 2) = =
12 The graph of z = x 3 3xy + 4y 2.
13 In general, if f is a function of n variables x 1,..., x n, then the derivative of f in the direction of u = (u 1,..., u n ) is (7) D u f = n i=1 f x i u i. Do you see how to write this sum as a dot product?
14 Gradient Vectors If f is a function of x 1,..., x n, then the gradient of f is ( f f =,..., f ) (8). x 1 x n (Pronounced grad f or del f or nabla f. ) This is a vector function, i.e., a function whose output is a vector not a number.
15 Gradient Vectors If f is a function of x 1,..., x n, then the gradient of f is ( f f =,..., f ) (8). x 1 x n (Pronounced grad f or del f or nabla f. ) This is a vector function, i.e., a function whose output is a vector not a number. We have (9) D u f = f u for any unit vector u.
16 Theorem The gradient f points in the direction of the unit vector u that maximizes the directional derivative D u f.
17 Theorem The gradient f points in the direction of the unit vector u that maximizes the directional derivative D u f. Proof. By the angle formula for dot product, (10) D u f = f u cos θ = f cos θ, where θ is the angle between f and u. The right-hand side is maximized when θ = 0.
18 Example Let f (x, y) = xe y. Find the maximum directional derivative of f at (2, 0), and (the unit vector in) the direction where it occurs.
19 Example Let f (x, y) = xe y. Find the maximum directional derivative of f at (2, 0), and (the unit vector in) the direction where it occurs. We compute f (x, y) = (e y, xe y ), from which f (2, 0) = (1, 2). This has the same direction as the unit vector (1,2) (1,2) = ( 1, 5 2 ). 5
20 Example Let f (x, y) = xe y. Find the maximum directional derivative of f at (2, 0), and (the unit vector in) the direction where it occurs. We compute f (x, y) = (e y, xe y ), from which f (2, 0) = (1, 2). This has the same direction as the unit vector (1,2) (1,2) = ( 1, 5 2 ). 5 This is the direction of the maximum directional derivative, so we compute (11) D (1/ 5, 2/ 5) f = (1, 2) ( 1 5, 2 5 ) = 5.
21 Geometry of the Gradient Moving along the gradient gives maximum rate of change. Moving along/within a level set gives zero rate of change.
22 Geometry of the Gradient Moving along the gradient gives maximum rate of change. Moving along/within a level set gives zero rate of change. Within the domain of f, gradients are locally orthogonal to level sets.
23 Review of Level Sets Remember... 1 A function of two variables x, y has level curves in the (x, y)-plane. 2 A function of three variables x, y, z has level surfaces in (x, y, z)-space.
24 Try to imagine/draw the function f (x, y) = x 3 + y 2.
25 Try to imagine/draw the function f (x, y) = x 3 + y 2. Some of its level curves in the (x, y)-plane:
26 Now imagine the function F (x, y, z) = xy z 2.
27 Now imagine the function F (x, y, z) = xy z 2. Here are the upper halves of some of its level surfaces in (x, y, z)-space:
28 Tangent Planes to Level Surfaces of 3-Variable Functions Earlier, we learned how to find tangent planes to graphs in (x, y, z)-space of functions in two variables x and y. Now, we will learn how to find tangent planes to level surfaces in (x, y, z)-space of functions in three variables x, y, z. Warning! Do not confuse these procedures!
29 Let S = {(x, y, z) R 3 : F (x, y, z) = α} be a level surface, and let (a, b, c) be a point on this surface, i.e., g(a, b, c) = α.
30 Let S = {(x, y, z) R 3 : F (x, y, z) = α} be a level surface, and let (a, b, c) be a point on this surface, i.e., g(a, b, c) = α. The coordinate equation of the tangent plane to S at (a, b, c) is: (12) F x (a, b, c)(x a) + F y (a, b, c)(y b) + F z (a, b, c)(z c) = 0.
31 Let S = {(x, y, z) R 3 : F (x, y, z) = α} be a level surface, and let (a, b, c) be a point on this surface, i.e., g(a, b, c) = α. The coordinate equation of the tangent plane to S at (a, b, c) is: (12) F x (a, b, c)(x a) + F y (a, b, c)(y b) + F z (a, b, c)(z c) = 0. Taking u = (x, y, z), the corresponding vector equation is (13) F (a, b, c) ( u (a, b, c)) = 0.
32 Example (Stewart, 14.6, Example 8) Find a coordinate equation for the tangent plane to the ellipsoid x y2 + z2 9 = 3 at ( 2, 1, 3).
33 Example (Stewart, 14.6, Example 8) Find a coordinate equation for the tangent plane to the ellipsoid x y2 + z2 9 = 3 at ( 2, 1, 3). Set F (x, y, z) = x2 4 + y2 + z2 9. Then (14) F (x, y, z) = ( ) x 2z, 2y,, 2 9 so F ( 2, 1, 3) = ( 1, 2, 2 3 ). We get the coordinate equation (15) 1(x + 2) + 2(y 1) 2 (z + 3) = 0. 3 It simplifies to x + 2y 2 3 z = 6.
34 The lower half of the ellipsoid x2 4 + y2 + z2 9 = 3.
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