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Section 15.6: Directional and

Goals : 1. To calculate the directional of a multivariable 2. To calculate the of a multivariable function 3. To solve applications involving a gradient

Recall : Partial Derivatives allow us to calculate the rate of change of a two-variable function f in the x and y directions. In this section, we will learn how to compute the rate of change of f in any direction.

The question of the day is how do we go about measuring the instantaneous rate of change of a two- variable function f in any direction on the domain of f? Well, as you've seen, vectors are very good at pointing in a direction of our choosing. So let u be a unit vector pointing in some pre-specified direction (in the xy -plane). Then a line through some point (x0 , y0 ) in the direction of u = u1,u2 is + = + given in vector form by r = x0 , y0 t u1,u2 (the parametric of the line are x x0 tu 1 and = + y y0 tu 2 ). Because u has magnitude 1, the t gives the distance from (x0 , y0 ) to the point (x, y) . Now, if we restrict the domain of f to the points on this line, then z = f (x, y ) = + + f (x0 tu 1, y0 tu 2 ) is a traced out on the corresponding to f. Note that, with this restriction, f is ultimately a function of one variable: t. Thus, we can define the using a limit definition in much the same way we did in single variable .

Definition : Assume the domain of the function f( x , y ) is restricted as above and u = u1,u2 is a

unit vector. Then the directional derivative of f at (x0 , y 0 ) in the direction of u is

fxy(,)− fx (, y ) = 0 0 Dfu ( x0 , y 0 ) lim , t→0 t

if this limit exists.

Notes: 1. Since z = f (x, y ) is a one variable function of t on the restricted domain, we can actually find the directional derivative of f as a function of x and y by applying the , assuming f is differentiable. In this case, the directional derivative of f in

the direction of a unit vector u = u1,u2 is

dz∂ z dx ∂ z dy D f( x , y ) = = + = f (x, y)u + f (x, y)u . u dt∂ x dt ∂ y dt x 1 y 2

2. We can rewrite the last formula in terms of a dot product as follows: = • Du f (x, y) ( f x (x, y )i + f y (x, y) j) u, where u is a unit vector. The first factor of this dot product is called the gradient .

Definition : Suppose z = f (x, y ) and first partials exist. Then the gradient of f, denoted ∇f (x, y ) , is the vector ∇ = f (x, y) f x (x, y) i + f y (x, y) j

Notes : 1. ∇f is read "del f." The symbol del is considered an operator and has no value in and of itself. Another notation for the gradient is grad f (x, y) . 2. The gradient is a vector that lies in the xy -plane, not in space. 3. With the new notation, the directional derivative can be written as = ∇ • Du f (x, y) f (x, y ) u (or, using the alternative form of the dot product = ∇ φ φ ∇ Du f (x, y) || f (x, y ||) cos where is the angle between f (x, y ) and u).

= ∇ φ 4. Using the form Du f (x, y) || f (x, y ||) cos of the directional derivative, we arrive at the properties of the gradient listed below.

PROPERTIES OF GRADIENT JUSTIFICTION OF PROPERTY

The maximum rate of change of f is ||∇f (x, y ) || OR cos φ = 1 when φ = 0 f increases most rapidly in the direction of the gradient ∇f (x, y)

The rate of change of f is 0 when u is orthogonal to ∇ f (x, y) . π OR cos φ = 0 when φ = To stay on a level curve of f, go in an orthogonal 2 direction to ∇f (x, y ) .

The minimum rate of change of f is − ∇f (x, y) which occurs when u is in the opposite direction of GRAD z . cos φ = −1when φ = π OR f decreases most rapidly in the opposite direction of ∇f (x, y)

Note : Directional Derivatives and Gradients generalize to functions of three variables. See page 983.

The Plane (Method 2)

Another way to find the of a plane is to use a point on the plane and a vector (see section 13.5). Suppose a surface is represented by an implicit function of z, say F(, x y , z ) = c. To find a normal vector to the tangent plane, we use the fact that, for a three-variable function of the form = ∇ = w Fxyz(, , ) , the gradient, Fx(0 , y 0 , z 0 ) , is orthogonal to the level surface F(, x y , z ) c at

(x0 , y0 , z0 ) and, therefore, normal to the corresponding tangent plane at (x0 , y0 , z0 ) . In other words, we treat an implied function defined by the equation F(, x y , z ) = c as a level surface of a three-variable ∇ function. Thus, Fx(0 , y 0 , z 0 ) is a normal vector to the tangent plane at (x0 , y0 , z0 ) . So, the equation of the tangent plane to a surface at (x0 , y0 , z0 ) is given by

−+ −+ −= Fxyzxxx(,,)(000 0 ) Fxyz y (,,)( 000 yy 0 ) Fxyzzz z (,,)( 000 0 )0

Note : This last result only holds if f is differentiable at (x0 , y0 , z0 ) .