mthsc206-fall-2010-notes-15.6

mthsc206-fall-2010-notes-15.6 - 15.6 Directional...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 15.6 Directional Derivatives and the Gradient Vector Let z = f ( x,y ) be the surface S . We have considered the slopes of the tangent line to the curves of intersection when we intersected S with the vertical planes y = y and x = x . These slopes were f x ( x ,y ) and f y ( x ,y ), respectively. Notice that these slopes were the instantaneous rates of change for f in the direction i (in the case of y = y ) and in the direction j (in the case x = x ) at the point ( x ,y ). Question: How would we compute the instantaneous rate of change of f in an arbitrary direction u = < a,b > at ( x ,y )? This instantaneous rate of change should measure the slope of the tangent line of the curve of intersection between S and the vertical plane determined by u through the point ( x ,y ,f ( x ,y )). Method: Let P ( x ,y ,f ( x ,y )) be the point on the surface S . Let Q ( x,y,f ( x,y )) be another point on the surface S and on the curve of intersection in the direction of u . Since Q is on the curve of intersection, the vector < x- x ,y- y > is parallel to the vector u = < a,b > . Hence < x- x ,y- y > = h < a,b > x = x + ha and y = y + hb. Thus the slope of the tangent line to the curve of intersection is given by the limit lim h f ( x,y )- f ( x ,y ) h = lim h f ( x + ha,y + hb )- f ( x ,y ) h , if it exists....
View Full Document

This note was uploaded on 10/11/2010 for the course MTHSC 206 taught by Professor Chung during the Fall '07 term at Clemson.

Page1 / 3

mthsc206-fall-2010-notes-15.6 - 15.6 Directional...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online