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- 6 Jun 2011
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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
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
Last edited:
. Didn't think of it that way.