Directional derivatives and gradient vectors (Sect. 14.5) Directional derivative of functions of two variables.

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1 Directional derivatives and gradient vectors (Sect. 14.5) Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector. Directional derivative of functions of two variables. Remark: The directional derivative generalizes the partial derivatives to an direction. Definition The directional derivative of the function f : D R 2 R at the point P 0 = ( 0, 0 ) D in the direction of a unit vector u = u, u is given b the limit ( Du f ) P0 = lim t 0 1[ f (0 + u t, 0 + u t) f ( 0, 0 ) ]. t Remarks: The line b r 0 = 0, 0 tangent to u is r(t) = r 0 + tu. (a) Equivalentl, ( D u f ) P0 = lim t 0 1[ ] f (r(t)) f (r(0)). t (b) If ˆf (t) = f (r(t)), then holds ( D u f ) P0 = ˆf (0).

2 Directional derivatives generalize partial derivatives The partial derivatives f and f are particular cases of directional derivatives ( D u f ) P0 = lim t 0 1 t [ f (0 + u t, 0 + u t) f ( 0, 0 ) ] : u = 1, 0 = i, then ( D i f ) P0 = f ( 0, 0 ). u = 0, 1 = j, then ( D j f ) P0 = f ( 0, 0 ). z f (,) f (, ) 0 f (, ) 0 0 z f(,) f(t) f (, ) 0 0 (D uf ) P 0 f (,) 0 i (, ) 0 0 j P = (, ) u u = 1 Directional derivatives generalize partial derivatives Find the derivative of f (, ) = at P 0 = (1, 0) in the direction of θ = π/6 counterclockwise from the -ais. Solution: A unit vector in the direction of θ is u = cos(θ), sin(θ). For θ = π/6 we get u = 1/2, 3/2. The line containing the vector r 0 = 1, 0 and tangent to u is r(t) = 1, 0 + t 2 1, 3 (t) = 1 + t 2, (t) = 3 t 2. Hence ˆf (t) = f ((t), (t)) is given b ( ˆf (t) = 1 + t ) 2 3t ˆf (t) = 1 + t + t 2. Since (D u f ) P0 = ˆf (0), and ˆf (t) = 1 + 2t, then (D u f ) P0 = 1.

3 Directional derivative of functions of two variables Remark: The condition u = 1 in the line r(t) = 0, 0 + u t implies that the parameter t is the distance between the points ((t), (t)) = ( 0 + u t, 0 + u t) and ( 0, 0 ). In other words: The arc length function of the line is l = t. Proof: d = 0, 0 = u t, u t = t u, d = t. Equivalentl, r (t) = u l = t 0 r (τ) dτ = t 0 dτ l = t. Remark: The directional derivative ( D u f ) is the pointwise rate P0 of change of f with respect to the distance along the line parallel to u passing through P 0. Directional derivatives and gradient vectors (Sect. 14.5) Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector.

4 Directional derivative and partial derivatives Remark: The directional derivative ( D u f ) is the P0 derivative of f along the line r(t) = 0, 0 + u t. z f(,) f(t) (D f ) u P 0 P = (, ) u u = 1 Theorem If the function f : D R 2 R is differentiable at P 0 = ( 0, 0 ) and u = u, u is a unit vector, then ( Du f ) P0 = f ( 0, 0 ) u + f ( 0, 0 ) u. Directional derivative and partial derivatives Proof: The line r(t) = 0, 0 + u, u t has parametric equations: (t) = 0 + u t and (t) = 0 + u t; Denote f evaluated along the line as ˆf (t) = f ((t), (t)). Now, on the one hand, ˆf (0) = ( D u f ) P0, since ˆf 1 ] (0) = lim [ˆf (t) ˆf (0) t 0 t ˆf 1[ (0) = lim f (0 + u t, 0 + u t) f ( 0, 0 ) ] = D u f ( 0, 0 ). t 0 t On the other hand, the chain rule implies: ˆf (0) = f ( 0, 0 ) (0) + f ( 0, 0 ) (0). Therefore, ( D u f ) P0 = f ( 0, 0 ) u + f ( 0, 0 ) u.

5 Directional derivative and partial derivatives Compute the directional derivative of f (, ) = sin( + 3) at the point P 0 = (4, 3) in the direction of vector v = 1, 2. Solution: We need to find a unit vector in the direction of v. Such vector is u = v u = 1 1, 2. v 5 We now use the formula ( D u f ) P0 = f ( 0, 0 ) u + f ( 0, 0 ) u. That is, ( D u f ) P0 = cos( )(1/ 5) + 3 cos( )(2/ 5). Equivalentl, ( D u f ) P0 = (7/ 5) cos( ). Then, ( D u f ) P0 = (7/ 5) cos(13). Directional derivatives and gradient vectors (Sect. 14.5) Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector.

6 Directional derivative of functions of three variables Definition The directional derivative of the function f : D R 3 R at the point P 0 = ( 0, 0, z 0 ) D in the direction of a unit vector u = u, u, u z is given b the limit ( Du f ) P0 = lim t 0 1[ f (0 + u t, 0 + u t, z 0 + u z t) f ( 0, 0, z 0 ) ]. t Theorem If the function f : D R 3 R is differentiable at P 0 = ( 0, 0, z 0 ) and u = u, u, u z is a unit vector, then ( Du f ) P0 = f ( 0, 0, z 0 ) u + f ( 0, 0, z 0 ) u + f z ( 0, 0, z 0 ) u z. Directional derivative of functions of three variables Find ( D u f ) for f (,, z) = z 2 at the point P0 P 0 = (3, 2, 1) along the direction given b v = 2, 1, 1. Solution: We first find a unit vector along v, u = v v u = 1 6 2, 1, 1. Then, ( D u f ) is given b ( D u f ) = (2)u + (4)u + (6z)u z. We conclude, ( D u f ) P0 = (6) (8) (6) 1 6, that is, ( D u f ) P0 = 26 6.

7 Directional derivatives and gradient vectors (Sect. 14.5) Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector. The gradient vector and directional derivatives Remark: The directional derivative of a function can be written in terms of a dot product. (a) In the case of 2 variable functions: D u f = f u + f u D u f = ( f ) u, with f = f, f. (b) In the case of 3 variable functions: D u f = f u + f u + f z u z, D u f = ( f ) u, with f = f, f, f z.

8 The gradient vector and directional derivatives Definition The gradient vector of a differentiable function f : D R 2 R at an point (, ) D is the vector f = f, f. The gradient vector of a differentiable function f : D R 3 R at an point (,, z) D is the vector f = f, f, f z. Notation: For two variable functions: f = f i + f j. For two variable functions: f = f i + f j + f z k. Theorem If f : D R n R, with n = 2, 3, is a differentiable function and u is a unit vector, then, D u f = ( f ) u. The gradient vector and directional derivatives Find the gradient vector at an point in the domain of the function f (, ) = Solution: The gradient is f = f, f, that is, f = 2, 2. z 2 f(,) = + 2 Remark: f = 2r, with r =,. f D f u

9 Directional derivatives and gradient vectors (Sect. 14.5) Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector. Properties of the the gradient vector Remark: If θ is the angle between f and u, then holds D u f = f u D u f = f cos(θ). The formula above implies: The function f increases the most rapidl when u is in the direction of f, that is, θ = 0. The maimum increase rate of f is f. The function f decreases the most rapidl when u is in the direction of f, that is, θ = π. The maimum decrease rate of f is f. Since the function f does not change along level curve or surfaces, that is, D u f = 0, then f is perpendicular to the level curves or level surfaces.

10 Properties of the the gradient vector Find the direction of maimum increase of the function f (, ) = 2 /4 + 2 /9 at an arbitrar point (, ), and also at the points (1, 0) and (0, 1). Solution: The direction of maimum increase of f is the direction of its gradient vector: f = 2, 2 9 At the points (1, 0) and (0, 1) we obtain, respectivel, f =. 1 2, 0. f = 0, 2. 9 Properties of the the gradient vector. Given the function f (, ) = 2 /4 + 2 /9, find the equation of a line tangent to a level curve f (, ) = 1 at the point P 0 = (1, 3 3/2). Solution: We first verif that P 0 belongs to the level curve f (, ) = 1. This is the case, since (9)(3) 4 The equation of the line we look for is 1 9 = 1. r(t) = 1, t v, v, where v = v, v is tangent to the level curve f (, ) = 1 at P 0.

11 Properties of the the gradient vector Given the function f (, ) = 2 /4 + 2 /9, find the equation of a line tangent to a level curve f (, ) = 1 at the point P 0 = (1, 3 3/2). Solution: Therefore, v f at P 0. Since, f = 2, 2 9 ( f ) P0 = 1 2, = 2 2, 1. 3 Therefore, 0 = v ( f ) P0 1 2 v = 1 3 v v = 2, 3. The line is r(t) = 1, t 2, 3. Properties of the the gradient vector f f 1 2 f f = 1 2, 0, f = 0, 2, r(t) = 1, t 2,

12 Further properties of the the gradient vector Theorem If f, g are differentiable scalar valued vector functions, g 0, and k R an constant, then holds, (a) (kf ) = k ( f ); (b) (f ± g) = f ± g; (c) (fg) = ( f ) g + f ( g); ( f ) (d) = g ( f ) g f ( g) g 2.