Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
74 LECTURE 11. GRADIENTS AND THE DIRECTIONAL DERIVATIVE Definition 11.2 A function f(x, y) : R 2 ↦→ R is said to be locally linear at the point (a, b) if there exist numbers h and k such that where f(a + h, b + k) = f(a, b) + hf x (a, b) + kf y (a, b) + hɛ(h, k) + kδ(h, k) (11.2) lim ɛ(h) = 0 h→0 lim δ(k) = 0 k→0 Definition 11.3 A function is said to be differentiable at P if it is locally linear at P. Definition 11.4 A function is said to be differentiable on an open set R if it is differentiable at every point in R. If we define the vectors Then equation 11.2 becomes P = (a, b) h = (h, k) e(h) = (ɛ, δ) f(P + h) = f(P) + h · (f x (P), f y (P)) + h · e Rearranging terms, dividing by h = ‖h‖, and taking the limit as ‖h‖ → 0 f(P + h) − f(P) h · (f x (P), f y (P)) h · e lim = lim + lim ‖h‖→0 ‖h‖ ‖h‖→0 ‖h‖ ‖h‖→0 ‖h‖ Defining the unit vector ĥ = h/‖h‖, f(P + h) − f(P) lim = lim ĥ · (f x (P), f y (P)) + lim ĥ · e ‖h‖→0 ‖h‖ ‖h‖→0 ‖h‖→0 The first limit on the right does not depend on the length ‖h‖ because h only appears as a unit vector, which has length 1, so that f(P + h) − f(P) lim = ‖h‖→0 ‖h‖ ĥ · (f x(P), f y (P)) + lim ĥ · e ‖h‖→0 The second limit on the right depends on h through the vector e; but the components of the vector e approach 0 as the components of h approach 0. Since the dot product of a vector of length 1 with a vector length 0 is zero, f(P + h) − f(P) lim = ‖h‖→0 ‖h‖ ĥ · (f x(P), f y (P)) (11.3) This gives us a generalized definition of the derivative in the direction of any vector h. We first define the gradient vector of functions of two and three variables, which we will use heavily in the remainder of this course. Revised December 6, 2006. Math 250, Fall 2006
LECTURE 11. GRADIENTS AND THE DIRECTIONAL DERIVATIVE 75 Definition 11.5 The gradient of a function f(x, y) : R 2 ↦→ R of two variables is given by ∇f(x, y) = gradf(x, y) = i ∂f ∂x + j∂f (11.4) ∂y The gradient of a function f(x, y, z) : R 3 ↦→ R of three variables is ∇f(x, y) = gradf(x, y) = i ∂f ∂x + j∂f ∂y + k∂f ∂z (11.5) Note that if we apply the definition in three-dimensions to a function of two variables, we obtain the same result as the first definition, because the partial derivative of f(x, y) with respect to z is zero (z does not appear in the equation). Example 11.1 Find the gradient of f(x, y) = sin 3 (x 2 y) Solution. ∇f = i ∂f ∂x + j∂f ∂y = i ∂ ∂x sin3 (x 2 y) + j ∂ ∂y sin3 (x 2 y) [ = i 3 sin 2 (x 2 y)) ∂ ] [ ∂x sin(x2 y) + j 3 sin 2 (x 2 y)) ∂ ] ∂y sin(x2 y) = i [ 3 sin 2 (x 2 y)) cos(x 2 y)(2xy) ] + j [ 3 sin 2 (x 2 y)) cos(x 2 y)(x 2 ) ] = 3x sin 2 ( x 2 y ) cos ( x 2 y ) (2yi + xj) Example 11.2 Find the gradient of f(x, y) = x 2 y + y 2 z + z 2 x Solution. ∇f = i ∂f ∂x + j∂f ∂y + k∂f ∂z = i ∂ ( x 2 y + y 2 z + z 2 x ) + j ∂ ( x 2 y + y 2 z + z 2 x ) + ∂x ∂y k ∂ ( x 2 y + y 2 z + z 2 x ) ∂z = i(2xy + z 2 ) + j(x 2 + 2yz) + k(y 2 + 2zx) Theorem 11.1 Properties of the Gradient VectorSuppose that f and g are functions and c is a constant. Then the following are true: ∇[f + g] = ∇f + ∇g (11.6) ∇(cf) = c∇f (11.7) ∇(fg) = f∇g + g∇f (11.8) Math 250, Fall 2006 Revised December 6, 2006.
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74 LECTURE 11. GRADIENTS AND THE DIRECTIONAL DERIVATIVE<br />
Definition 11.2 A function f(x, y) : R 2 ↦→ R is said to be locally linear at the<br />
point (a, b) if there exist numbers h and k such that<br />
where<br />
f(a + h, b + k) = f(a, b) + hf x (a, b) + kf y (a, b) + hɛ(h, k) + kδ(h, k) (11.2)<br />
lim ɛ(h) = 0<br />
h→0<br />
lim δ(k) = 0<br />
k→0<br />
Definition 11.3 A function is said to be differentiable at P if it is locally linear<br />
at P.<br />
Definition 11.4 A function is said to be differentiable on an open set R if it<br />
is differentiable at every point in R.<br />
If we define the vectors<br />
Then equation 11.2 becomes<br />
P = (a, b)<br />
h = (h, k)<br />
e(h) = (ɛ, δ)<br />
f(P + h) = f(P) + h · (f x (P), f y (P)) + h · e<br />
Rearranging terms, dividing by h = ‖h‖, and taking the limit as ‖h‖ → 0<br />
f(P + h) − f(P) h · (f x (P), f y (P)) h · e<br />
lim<br />
= lim<br />
+ lim<br />
‖h‖→0 ‖h‖<br />
‖h‖→0 ‖h‖<br />
‖h‖→0 ‖h‖<br />
Defining the unit vector ĥ = h/‖h‖,<br />
f(P + h) − f(P)<br />
lim<br />
= lim ĥ · (f x (P), f y (P)) + lim ĥ · e<br />
‖h‖→0 ‖h‖<br />
‖h‖→0<br />
‖h‖→0<br />
The first limit on the right does not depend on the length ‖h‖ because h only<br />
appears as a unit vector, which has length 1, so that<br />
f(P + h) − f(P)<br />
lim<br />
=<br />
‖h‖→0 ‖h‖ ĥ · (f x(P), f y (P)) + lim ĥ · e<br />
‖h‖→0<br />
The second limit on the right depends on h through the vector e; but the components<br />
of the vector e approach 0 as the components of h approach 0. Since the dot product<br />
of a vector of length 1 with a vector length 0 is zero,<br />
f(P + h) − f(P)<br />
lim<br />
=<br />
‖h‖→0 ‖h‖ ĥ · (f x(P), f y (P)) (11.3)<br />
This gives us a generalized definition of the derivative in the direction of any vector<br />
h. We first define the gradient vector of functions of two and three variables, which<br />
we will use heavily in the remainder of this course.<br />
Revised December 6, 2006. Math 250, Fall 2006
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