Question

find unit vectors that give steepest ascent and descent at point, find vector that points in direction of no change.

Answer 1

\[f(x,y)=x^4-x^2y+3y^2+7\]
point (-1,1)

Answer 2

\[gradient = <4x^3-2xy , x^2+6y>\]

Answer 3

no wrong

Answer 4

<4x^3-2xy,-x^2+6y>

Answer 5

alright, typo, next step?

Answer 6

next plug in x=-1 and y=1

Answer 7

you get <-2,5>

Answer 8

yep

Answer 9

then find the unit vector by just diving by modulus and you get???

Answer 10

so for steepest ascent it goes in the same direction as the gradient?

Answer 11

\[<-2/\sqrt{29},5/\sqrt{29}>\]

Answer 12

right and for the steepest descent just multiply the ascent by -1 and you get??

Answer 13

steepest decent i just times each component by -1?

Answer 14

ya

Answer 15

now for 0 change?

Answer 16

and last part you know? you need to find a vector perpendicular to the gradient

Answer 17

so, dot product = 0?

Answer 18

Let the vector by <a,b> then a(-2/sqrt{29})+b(5/sqrt{29})=0 this gives -2a+5b=0 gives a=(5/2)b

Answer 19

in particular take b=2 then a=5 therefore (5,2) is such a vector now you may divide by the modulus and get <5/sqrt(29),2/sqrt(29)> the required vector

Answer 20

@TylerD

Answer 21

hmm, for some reason the answer was -5,-2

Answer 22

yeah it is also a choice..

Answer 23

ya i think i c

Answer 24

a=(5/2)b if you put b=-2 then a becomes -5

Answer 25

you didnt give me options

Answer 26

actually there exists infinite number of solutions

Answer 27

since a direction is given by a line and a line is composed of infinite number of points so the solution is not unique . the solution is given by the equation a=(5/2)b where you will get value of a if you plug in b

Answer 28

@TylerD