Question

Dealing with a conceptual/word problem with Gradients/Vector Fields, I don't understand why the answer is what it is. One moment.

Answer 1

"Give a formula \[F=M(x,y)i+N(x,y)j\]
for the vector field in the plane that has the property that F points toward the origin with magnitude inversely proportional to the square of the distance from (x,y) to the origin. (The field is not defined at (0,0)."

Answer 2

Tried drawing it out?

Answer 3

My confusion is that the answer is \[-a \bigg[x i +yj\bigg](x^2+y^2)^{-3/2}\]

Answer 4

And I don't understand why it's to the -3/2ths power. I can reasonably imagine you having something like the square of the distance formula in the denominator due to the stated inverse square relationship, but I don't understand why it's to the 3/2ths power.

A is just any positive, nonzero arbitrary constant/

Answer 5

the description of this vector field matches exactly with the gravitational field

Answer 6

That's dandy, but I don't understand anything more from that, lol.

Answer 7

but gravity decreases as square of distance in 3D so they are not quite the same

Answer 8

Try to derive M and N yourself and see if you get the same result when you simplify it.

Answer 9

I'm really not sure how I should do this "backwards", I just imagined something approximating the formula the way it described it, but I'll take another shot.

Answer 10

Distance formula in 2D plane:\[\sqrt{x^2+y^2}\]
The square of the distance formula is thus\[x^2+y^2\]Having an inverse relationship, you could express the field as\[\frac{1}{x^2+y^2}\]

Answer 11

(I could elaborate, but it's very clear that there's already something wrong with that, so I'm going to stop there, assuming people immediately have corrections to my thought process.)

Answer 12

Oh, and let both of those be negative, yeah.

Answer 13

Lets start with preliminary functions \(M_0, N_0\) which just give a unit vector pointing in the right direction.

\(M_0(x, y) = \frac{-x}{|(x, y)|} = \frac{-x}{(x^2 + y^2)^{1/2}} \)
\(N_0(x, y) = \frac{-y}{|(x, y)|} = \frac{-y}{(x^2 + y^2)^{1/2}} \)

Then we need to add in the factor consisting of its magnitude. Which should be
$$\frac{a}{x^2 + y^2}$$
So we have
\(M = \frac{-x}{(x^2 + y^2)^{1/2}}\cdot\frac{a}{x^2 + y^2}\)
\(N = \frac{-y}{(x^2 + y^2)^{1/2}}\cdot\frac{a}{x^2 + y^2}\)

Answer 14

Taking a look, one second.

Answer 15

Alright, yeah, this makes sense. Thanks.

Answer 16

It always helps to try and derive it yourself. You will be glad you did since you will understand exactly how the formula works in the end.