Question

ok, gonna try this again. i was in tears after being ridiculed this morning.

Match the functions f with the plots of their gradient vector fields labeled A-D.

f(x,y)=sqrt(x^2+y^2)
2. f(x,y)=x^2−y^2
3. f(x,y)=x^2+y^2
4. f(x,y)=xy

Answer 1

1.* f(x,y)=sqrt(x^2+y^2)
2. f(x,y)=x^2−y^2
3. f(x,y)=x^2+y^2
4. f(x,y)=xy

Answer 2

that didn't actually answer anything. just restated what i already said. and you're supposed to be blocked. leave me alone.

Answer 3

@ganeshie8 , can you help me here? pretty please?

Answer 4

the easiest function is
3. f(x,y)=x^2+y^2

Answer 5

gradient = <2x, 2y>

which clearly points in the same direction as position vector, right ?

Answer 6

at each point (x, y) in the plane, the gradiant points in the same direction as position vector. the magnitude would be twice larger

Answer 7

oh, so it would be D. ok. not that clear on drawing vectors.

Answer 8

C looks good

Answer 9

use this to draw if u want
http://kevinmehall.net/p/equationexplorer/vectorfield.html

Answer 10

for x^2+y^2 it's C? if it's the same magnitude, wouldn't it be D?

Answer 11

yeah, i tried using that this morning and it gave me fields that weren't any of my choices so i was confused.

Answer 12

<2x, 2y> the size of vectors must increase as you go away from center

Answer 13

i don't see them increasing in option D, so..

Answer 14

oh. didn't know they had to decrease as they moved from the origin. is that cause they double in size each time?

Answer 15

exactly!

Answer 16

as the name says, `gradient vector` is just a vector defined at `each point`

Answer 17

so it's C for x^2+y^2

Answer 18

Yep!

Answer 19

ok, but then x^2-y^2 is a similar vector <2x, -2y> which would suggest to me that it goes toward the origin, but i don't see any like that.

Answer 20

<2x, -2y>

Notice that when y is positive, -2y is negative
that means the vector shoots DOWN in I, II quadrants

Answer 21

b is the only one that does that, but the vectors there are curved. why?

Answer 22

because each vector has "x" component also

Answer 23

the y component is down
but the x component is positive in quadrant I, so it faces right+down in quadrant I

Answer 24

ok, so x^2-y^2 is B. is xy D cause it doesn't increase or decrease? stays constant? just <y, x>

Answer 25

its number 3

Answer 26

<y, x> is NOT constant
it changes with position

Answer 27

i mean the magnitude is not increasing like when it was doubling for x^2+y^2. sorry

Answer 28

an easy way to identify this is :
on `y axis`, x=0

so the gradient would be : <y, 0>

Answer 29

that means, on y axis, the x component of gradient is increases as u move up