Need some maths help (scalar fields)

SheepBeast · 7 May 2012 at 23:52

HAELPZ MEH PLOX

Got a uni exam in 4 days so I'm cramming for it. I've been doing practice problems but I don't have any answer sheets so if I'm stuck there's nothing to help me. Hence I though I'd tap the vast conduit of knowledge that is OcUK GD

Here's the question:

I've done part (i) and I think I can do part (iii) so it's just (ii) which is the problem.

I understand that I need to find Grad[h], the gradient function of the scalar field, so I've done that. I get Grad[h] = -2x[unit vector in x] -4y[unit vector in y] but I'm not quite sure where to go from there. I can see that if we call an arbitrary point on the hiker's path r (a vector) then the direction of the steepest slope at that point is Grad[h[r]]. But I can't get my head around a way to express the path as a function y[x], which is what I know I need to do.

Can anyone help? I know a dedicated maths/physics forum would be a better place for this but I'm not a member of any. If no-one here can help then I might take my business elsewhere

Scotty123 · 8 May 2012 at 00:18

You know dx/dh and dy/dh so can use the chain rule to get it in terms of just x and y.

You have a differential equation for the path in x-y space, just integrate it and use the initial condition.

y = x^2 + 5 .. I think. Been a few years since vector calculus

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SheepBeast · 9 May 2012 at 00:11

@Scotty123 thanks very much that actually helped a lot

. Didn't think of it that way.

Alex74 · 9 May 2012 at 00:14

Yea, tbh the hint should somewhat give it away

SheepBeast · 9 May 2012 at 01:23

Alex74 said:
Yea, tbh the hint should somewhat give it away

yeah lol..... i'm not feeling to good about this upcoming exam

Morbius · 9 May 2012 at 11:03

Scotty123 said:
y = x^2 + 5 .. I think. Been a few years since vector calculus .

Close but I think you've made a small mistake with your constant of integration.

You get Integral(dy/2y) = Integral(dx/x) which leaves you with

(1/2) log y = log x + c

where c is the constant of integration. Without loss of generality, let c = log k

(1/2) log y = log kx

Manipulate the logs and let A=k^2

then you get y = Ax^2

When x=sqrt(2), y=7 which implies that A=7/2 and the path is

y = (7/2)x^2