Help with #59 please Gradients in three dimensions Consider the following functi

Question

Help with #59 please

Gradients in three dimensions Consider the following functions f, points P, and unit vectors u. a. Compute the gradient of f and evaluate it at P. b. Find the unit vector in the direction of maximum increase of f at P. c. Find the rate of change of the function in the direction of maximum increase at P. d. Find the directional derivative at P in the direction of the given vector. 55. f(x,y,z) = x^2 + 2y^2 + 4z^2 + 10; P(1,0,4); 1/2,0,1/2 56. f(x,y,z)= 4 - x^2 + 3y^2 + z^2/2; P(0,2,-1); 0.1/2,-1/2 57. f(x,y,z) = 1 + 4xyz; P(1,-1,-1); 1/3,1/3,-1/3 58. f(x,y,z) = xy + yz + xz + 4; P(2, -2, 1); 0, -1/2,-1/2 59. f(x,y,z) = 1 + sin (x + 2y - z); P(pi/6,pi/6,-pi/6); 1/3,2/3,2/3

Explanation / Answer

59)f =1+sin(x+2y -z)

a)gradient  f =<cos(x+2y -z) ,2cos(x+2y -z), -cos(x+2y -z)>

at (pi/6,pi/6,pi/6)

f =<cos((pi/6)+2(pi/6) -(pi/6)) ,2cos((pi/6)+2(pi/6) -(pi/6)), -cos((pi/6)+2(pi/6) -(pi/6))>

f =<cos((pi/3)) ,2cos((pi/3)), -cos((pi/3))>

f =<1/2 ,2/2 ,-1/2>

f =<1/2 ,1 ,-1/2>

b)|f|=[(1/2)2+12+(-1/2)2]

|f|=(9/4)

|f|=3/2

unit vector in direction of maximum increase =f/|f|

unit vector in direction of maximum increase =<1/2 ,1 ,-1/2>/(3/2)

unit vector in direction of maximum increase =<1/3 ,2/3 ,-1/3>

c) rate of change of function in direction of maximum increase =|f|=3/2

d)given vector u =<1/3,2/3,2/3>

|u|=[(1/3)2+(2/3)2+(2/3)2]=(9/9)=1

directional derivative =f .u /|u|

directional derivative =<1/2 ,1 ,-1/2>.<1/3,2/3,2/3>/1

directional derivative =(1/2 *1/3) +(1 *2/3) +(-1/2 *2/3)

directional derivative =(1/6) +(2/3) +(-1/3)

directional derivative =(1/6) +(1/3)

directional derivative =(1/6) +(2/6)

directional derivative =3/6

directional derivative =1/2

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