Please answer A-E but use this equation down below: Homework: Chapter 12 Section

Question

Please answer A-E but use this equation down below:
Homework: Chapter 12 Section 6 Score: 0.44 of 1 pt sof 15(15 complete) Hw Score: 89.63%, 13.44 of 15 pts 12.6.35 Question Help * A function f and a point P are given. Let correspond to the direction of the directional derivative. Complete parts a. throughe a. Find the gradent and evaluate it at P The grad ent at P-1 1 Type exact answers, using radicais as needed.) b. Find the angles 0 (with respect to the positive x-axis) associated with the directions of maximum increase, maximum decrease, and zero change What angles are associated with the drection of maximum increase? needed ) and 2s. Type an exact answer, using s as needed. Use a comma to separase answers as Type any angles in radians between O What angles are associated with the dinection of maximum decreas 0 and 2x. Type an exact answer, using s as needed Use a comma to separate answers as needed ) What angles are associaed with the dinection of zero change? any angjes in radans between o and 2x Type an exact answer, using s as needed. Use a comma to separate answers as needed) Question is complete. Tap on the red indicators to see incorrect answers All parts showing 2 3 4 5 6

Explanation / Answer

given f(x,y)= sqrt(6+3x^2+3y^2) at the point (-1,1)

a)

gradient,f= (1/2(6+3x2+3y2))*(0+6x+0),(1/2(6+3x2+3y2))*(0+0+6y)

gradient,f= 3x/(6+3x2+3y2),3y/(6+3x2+3y2)

at point P(-1,1)

gradient,f= 3(-1)/(6+3(-1)2+3*12),3*1/(6+3(-1)2+3*12)

gradient,f= -3/12,3/12

gradient,f= -(12)/4,(12)/4

gradient,f=  -(3)/2,(3)/2

direction of maximum increase

=tan-1(((3)/2)/(-(3)/2))

= -tan-1(1)

= -(/4)

=3/4  

direction of maximum decrease

=tan-1(((-3)/2)/((3)/2))

=2 -tan-1(1)

=2 -(/4)

=7/4  

direction of zero change = (3/4)-(/2), (3/4)+(/2)

direction of zero change = (/4), (5/4)

c)

g()= ((-3)/2)cos +((3)/2)sin

d)

g'()= ((3)/2)sin +((3)/2)cos

g'()=0

((3)/2)sin +((3)/2)cos=0

=>sin =-cos

=>tan =-1

=> =3/4 ,7/4  

g(3/4)= ((-3)/2)cos(3/4) +((3)/2)sin(3/4)

g(3/4)= (3/2)

g(3/4)= (6)/2

=3/4  

maximum value:g()= (6)/2

e)

from part a

f=  -(3)/2,(3)/2 ,=3/4  

maximum increase =|f|

maximum increase =[(-(3)/2)2+((3)/2)2]

maximum increase =(3/2)

maximum increase =(6)/2

yes

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