Question

Give a formula for a vector field in the x,y plane with the property that F points in the direction of the origin and has a magnitude inversely proportional to the to the square of the distance to the origin.

Answer 1

|F| = (a)(1/x^2+y^2) ???

Answer 2

both terms under the 1 ^^^

Answer 3

F must be some f(x,y) i + g(x,y)j

Answer 4

it is circular yes; a<cos(t),sin(t)> would produce a circle i believe

Answer 5

lets say:

r = a<cos(t),sin(t)> where a is any constant greater than 0 for good luck

r' = tangent vector

r'' = normal vector

Answer 6

the normal vector is what i believe you are after, and modify it so it points inward

Answer 7

i think they just want the equation of a vector field with the property that each vector points towards the origin

Answer 8

and the magnitude satisfies the requirement

Answer 9

its only in the x,y plane so how could there be a normal vector?

Answer 10

a normal vector is just a vector that points out of the curve; which is useful for defining planes

Answer 11

ohh i thought a normal vector was orthogonal to a plane?

Answer 12

orthogonal to the curve, regardless of R^n

Answer 13

hmm, acceleration here might be inward ... i tend to get those backwards i think

Answer 14

well lets see if we have a point at say (1,0) and we want the vector to point at the origin we would need F = -i

Answer 15

nd if we have a point at say (0,1) we would want F= -J

Answer 16

at (1,1) we would need -xi -yj

Answer 17

would F = -x i - yj satisfy the origin requirement?

Answer 18

yes, as long as x and y are functions of t; x(t) and y(t)

Answer 19

the trick is to define rules for x(t) and y(t) so that they are not so generic

Answer 20

and x(t) = a cos(t), y(t) = a sin(t) should fit since those define a circle to begin with

Answer 21

why do we need x as a function of t? it seems as though all we need is an f(x,y) and g(x,y) for this: F = f(x,y)i + g(x,y) j

Answer 22

because in order to get "the property that F points in the direction of the origin" we have to define a circle, and there are no functions the that define a circle without going to trig ...

Answer 23

and trig defines a circle in relation to x and y being functions of time

Answer 24

unless of course im reading it wrong, which is highly probable today :)

Answer 25

thats the solution from the sol. manual

Answer 26

ahh, ok. and its using n for the normal ...

let me go over it to see if im on the right track at least :)

Answer 27

ok so at some point in the x,y plane, the distance to the origin in \[\sqrt{x^2 +y^2}\]

Answer 28

had similar ideas, but i was overthinking it

Answer 29

and the inverse is (x^2 + y^2)^-1/2

Answer 30

hmm i dunno i still gont get how they got that final answer

Answer 31

lets go at it from this manner:

r^2 = x^2 + y^2 is the standard eqaution a circle right? such that

r = \(\sqrt{x^2+y^2}\)

Answer 32

yes..

Answer 33

r^2 = magnitude but |F| is inversely porportional

|F| = 1/r^2

|F| = k/(x^2+y^2) is what it reads to me so far

Answer 34

ohhh! ok

Answer 35

k being some constant correct?

Answer 36

yes, a constant of variation if i recall it right

Answer 37

the normal to function is what they appear to define next, at least the unit normal vector

Answer 38

so why do they use -x/sqrtx^2+y^2 i -y/sqrtx^2+y^2j to represent the direction instead of just -xi-yj

Answer 39

i believe it is so that they have a unit vector to work with. I vector that has a magnitude of "1"