11.6 Directional Derivative & The Gradient Vector. Working Definitions

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1 1 Ma Kidogchi Kenneth The directional derivative o at 0 0 ) in the direction o the nit vector is the scalar nction deined b: where q is angle between the two vectors placed tail-to-tail and 0 < q < p. q cos ) ) ) D Working Deinitions j i ) ) )) ) grad I is a well behaved nction o two variables and then: the gradient o is the vector nction deined b: is oten call the "Del" operator.

2 Sample Comptation Given ) = e ind: a) the rate o change o at the point 0 0 ) = 0) in direction o the vector v b) = grad) and c) the magnitde o grad) =. 1 Ma 016 Kidogchi Kenneth

3 Sample Comptation 1 Ma Kidogchi Kenneth

4 Sample Comptation Given ) = e ind: a) the rate o change o at the point 0 0 ) = 0) in direction o the vector v b) = grad) and c) the magnitde o grad) =. Soltion to a): nit vector in the direction o v ) e e Gradient o D ) ) Directional derivative o in the direction o v 1 Ma Kidogchi Kenneth

5 Sample Comptation Cont.) D ) Soltion to b): ind e 0) e 3 e.1 e = grad) Directional derivative o in the direction o v Directional derivative o in the direction o at ) = 0) v ) e e Gradient o Soltion to c): 0) 1 0) 5 ind the magnitde o grad) =..4 Gradient o at ) = 0) Magnitde o the gradient o at ) = 0) 1 Ma Kidogchi Kenneth

6 Directional Derivative o z = ) Interpretation Given z = ) and nit vector 1 0 ) 10 This directional derivative is the rate o change i.e. the slope o the tangent line) at a point on the grid crve that traverses the srace in the direction o <1 0 > a nit vector parallel to the -ais. 1 Ma Kidogchi Kenneth

7 Directional Derivative o z = ) Interpretation Given z = ) and nit vector 0 1 ) 01 This directional derivative is the rate o change i.e. the slope o the tangent line) at a point on the grid crve that traverses the srace in the direction o <0 1> a nit vector parallel to the -ais. 1 Ma Kidogchi Kenneth

8 Directional Derivative o z = ) Interpretation Given z = ) and nit vector ) û p p cos sin 4 4 This directional derivative is the rate o change at a point on C the space crve that traverses the srace in the direction o the vector <1 1> which makes and angle p/4 with respect to <1 0> at a point on C. Eample 1 1 Ma Kidogchi Kenneth

9 Directional Derivative o z = ) Eample Sppose o are climbing a hill whose srace is described b: z = 1000 /00 /100) where and z are measred in metres and o are standing at the point with coordinates 0 0 z 0 ) = ). The positive -ais points east and the positive -ais points north. a) I o walk de soth will o start to ascend or descend? At what rate? b) I o walk northwest will o start to ascend or descend? At what rate? c) In which direction is the slope the largest? What is the rate o ascent in that direction? At what angle above the horizon does the path begin in that direction? 1 Ma Kidogchi Kenneth

10 Directional Derivative o z = ) Eample z = 1000 /00 /100) and 0 0 z 0 ) = ). a) I o walk de soth will o start to ascend or descend? At what rate? b) I o walk northwest will o start to ascend or descend? At what rate? c) In which direction is the slope the largest? What is the rate o ascent in that direction? At what angle above the horizon does the path begin in that direction? m54_c11.6_e.mw 1 Ma Kidogchi Kenneth

11 Directional Derivatives & Gradient Sample Comptation m54_c11.6_e3.mw: Given the srace: ) = a) Find the directional derivative ) i û is the nit vector given b the angle q p/6 b) evalate 1) and c) ind the maimm vale o the directional derivative and eplain its signiicance. Soltion: a) ) b) 1) c) D ) ma 1 Ma Kidogchi Kenneth

12 1 Ma Kidogchi Kenneth Directional Derivatives & Gradient General Case I 1 n ) a nction o n variables and its derivatives are dierentiable on some interval o interest and is an n-component nit vector then: n n n e e e m54_c11.6_e4.mw: Electric potential de to a point charge is: where k is a constant. Let Fz) = V/k then: a) ind gradf) b) plot the eqipotential srace Fz) = 1 and Fz) = and draw gradient vectors at enogh points in space to evalate their properties. ) z k z V

13 Tangent Planes to Level Sraces Given Fz) = k where k is a constant then Fz) represents a level srace. I the coordinates are nctions o some parameter t: z) t)t)zt)) and a crve C on the srace can be described b the parametric position vector then: Ft)t)zt)) r t) t) t) z t) F d F d F dz 0 dt dt z dt dr F 0 dt The tangent plane to the level srace Fz) = k at the point 0 0 z 0 ) is: F 0 0 z 0 ) 0 ) + F 0 0 z o ) 0 ) + F z 0 0 z 0 ) z z 0 ) =0 with normal vector: F ) 0 0 z0 1 Ma Kidogchi Kenneth

14 Given z = ) 11.6 Directional Derivative & The Gradient Vector Signiicance o the Gradient Vector ) represents the rate o change o z in the direction o a nit vector =< > at the point ). ) ) ) o o o ) o The maimm rate o change o ) occrs when is parallel to grad ) which is the magnitde o the gradient vector. D ) ) ma For the contor crve ) = k ) is perpendiclar to the crve at ). Contor Crve 1 Ma Kidogchi Kenneth

15 The Gradient and Tangent Plane to a Level Srace Find the tangent plane to the srace: z 9 = 3 at the point z)= 1 3). m54_c11.6_e5.mw. 1 Ma Kidogchi Kenneth

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