{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Section14_5

Section14_5 - Section 14.5 Directional derivatives and...

This preview shows pages 1–3. Sign up to view the full content.

(3/23/08) Section 14.5 Directional derivatives and gradient vectors Overview: The partial derivatives f x ( x 0 , y 0 ) and f y ( x 0 , y 0 ) are the rates of change of z = f ( x, y ) at ( x 0 , y 0 ) in the positive x - and y -directions. Rates of change in other directions are given by directional derivatives . We open this section by defining directional derivatives and then use the Chain Rule from the last section to derive a formula for their values in terms of x - and y -derivatives. Then we study gradient vectors and show how they are used to determine how directional derivatives at a point change as the direction changes, and, in particular, how they can be used to find the maximum and minimum directional derivatives at a point. Topics: Directional derivatives Using angles of inclination Estimating directional derivatives from level curves The gradient vector Gradient vectors and level curves Estimating gradient vectors from level curves Directional derivatives To find the derivative of z = f ( x, y ) at ( x 0 , y 0 ) in the direction of the unit vector u = ( u 1 , u 2 ) in the xy -plane, we introduce an s -axis, as in Figure 1, with its origin at ( x 0 , y 0 ), with its positive direction in the direction of u , and with the scale used on the x - and y -axes. Then the point at s on the s -axis has xy -coordinates x = x 0 + su 1 , y = y 0 + su 2 , and the value of z = f ( x, y ) at the point s on the s -axis is F ( s ) = f ( x 0 + su 1 , y 0 + su 2 ) . (1) We call z = F ( s ) the cross section through ( x 0 , y 0 ) of z = f ( x, y ) in the direction of u . braceleftbigg x = x 0 + su 1 y = y 0 + su 2 Tangent line of slope F prime (0) = D u f ( x 0 , y 0 ) FIGURE 1 FIGURE 2 327

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
p. 328 (3/23/08) Section 14.5, Directional derivatives and gradient vectors If ( x 0 , y 0 ) negationslash = (0 , 0), we introduce a second vertical z -axis with its origin at the point ( x 0 , y 0 , 0) (the origin on the s -axis) as in Figure 2. Then the graph of z = F ( s ) the intersection of the surface z = f ( x, y ) with the sz -plane. The directional derivative of z = f ( x, y ) is the slope of the tangent line to this curve in the positive s -direction at s = 0, which is at the point ( x 0 , y 0 , f ( x 0 , y 0 )). The directional derivative is denoted D u f ( x 0 , y 0 ), as in the following definition. Definition 1 The directional derivative of z = f ( x, y ) at ( x 0 , y 0 ) in the direction of the unit vector u = ( u 1 , u 2 ) is the derivative of the cross section function (1) at s = 0 : D u f ( x 0 , y 0 ) = bracketleftbigg d ds f ( x 0 + su 1 , y 0 + su 2 ) bracketrightbigg s =0 . (2) The Chain Rule for functions of the form z = f ( x ( t ) , y ( t )) (Theorem 1 of Section 14.4) enables us to find directional derivatives from partial derivatives. Theorem 1 For any unit vector u = ( u 1 , u 2 ) , the (directional) derivative of z = f ( x, y ) at ( x 0 , y 0 ) in the direction of u is D u f ( x 0 , y 0 ) = f x ( x 0 , y 0 ) u 1 + f y ( x 0 , y 0 ) u 2 . (3) Remember formula (3) as the following statement: the directional derivative of z = f ( x, y ) in the direction of u equals the x -derivative of f multiplied by the x -component of u , plus the y -derivative of f multiplied by the y -component of u .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern