NB This is a physics question which I feel is most suited to maths.
A hill is described h(x,y) = 300exp((x^2+y^2)).
A railway runs along the line xy=2
What's the steepest slope that the train will need to climb?
I want to do as much as I ran myself, so am principally asking for a starting hint please. Specifically: there must be a way to view a 3D surface plot like this side on as defined by the railway line. Then I could parameterise this function h = f(t) and use normal 1D diff' to solve. Is this vaguely correct reasoning? If so, I don't have a clue how to carry put the initial 'projection'
Thanks in advance.

Sm0key
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 15032016 18:30

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 15032016 18:38
(Original post by Sm0key)
NB This is a physics question which I feel is most suited to maths.
A hill is described h(x,y) = 300exp((x^2+y^2)).
A railway runs along the line xy=2
What's the steepest slope that the train will need to climb?
I want to do as much as I ran myself, so am principally asking for a starting hint please. Specifically: there must be a way to view a 3D surface plot like this side on as defined by the railway line. Then I could parameterise this function h = f(t) and use normal 1D diff' to solve. Is this vaguely correct reasoning? If so, I don't have a clue how to carry put the initial 'projection'
Thanks in advance. 
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 15032016 19:08
Parametrise x and y, and then h(x, y) = h(t), which is 1 dimensional.Last edited by EricPiphany; 15032016 at 19:13. 
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 15032016 19:18
(Original post by EricPiphany)
I would do this.
Parametrise x and y, and then h(x, y) = h(t), which is 1 dimensional. 
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 15032016 20:17
(Original post by Sm0key)
NB This is a physics question which I feel is most suited to maths.
A hill is described h(x,y) = 300exp((x^2+y^2)).
A railway runs along the line xy=2
What's the steepest slope that the train will need to climb?
a) find the gradient function for the surface in terms of
b) use that to find the directional derivative in the direction of the line, unit vector
c) then reduce the problem to one dimension via
The gradient function encapsulates information about the gradient of the surface at every point . The directional derivative extracts gradient information from the gradient function along a line defined by a vector. A trivial directional derivative is , which extracts gradient information from the function along the line defined by the xaxis, or the vector . 
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 15032016 23:02
Thanks all for the replies  much appreciated.

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 15032016 23:10
(Original post by atsruser)
Looks more like a maths question that's turned up in a physics course :)
I think that the usual way to approach this is to
a) find the gradient function for the surface in terms of
b) use that to find the directional derivative in the direction of the line, unit vector
c) then reduce the problem to one dimension via
The gradient function encapsulates information about the gradient of the surface at every point . The directional derivative extracts gradient information from the gradient function along a line defined by a vector. A trivial directional derivative is , which extracts gradient information from the function along the line defined by the xaxis, or the vector .
Please could you guide me through those steps? 
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 15032016 23:38
I'll assume that you are happy with computing that unless you say otherwise.
We need a vector in the direction of the line. Since , we have a line of gradient 1 in the xy plane, which means that we go 1 unit along the yaxis for 1 unit along the xaxis. A vector in this direction is thus
You must now
1) turn this into a unit vector and
2) compute the dot product of with the gradient that you found above.
When you've done this, you will have a function which computes the gradient of the surface in the *direction* of your line (i.e. along a vector parallel to it), but at any point in the xy plane...
... What you want, however, is a function that computes the gradient of the surface specifically on the points that lie on . But to get that function, you merely have to impose the constraint on i.e. use it to substitute for either or in , to give you, say, .
You can then find in the usual way to maximise the gradient. 
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 15032016 23:39
(Original post by Sm0key)
I wanted to do a directional derivative approach BUT didn't know how to compute the actual scalar product.
Please could you guide me through those steps? 
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 16032016 00:15
(Original post by atsruser)
We have
I'll assume that you are happy with computing that unless you say otherwise.
We need a vector in the direction of the line. Since , we have a line of gradient 1 in the xy plane, which means that we go 1 unit along the yaxis for 1 unit along the xaxis. A vector in this direction is thus
You must now
1) turn this into a unit vector and
2) compute the dot product of with the gradient that you found above.
When you've done this, you will have a function which computes the gradient of the surface in the *direction* of your line (i.e. along a vector parallel to it), but at any point in the xy plane...
... What you want, however, is a function that computes the gradient of the surface specifically on the points that lie on . But to get that function, you merely have to impose the constraint on i.e. use it to substitute for either or in , to give you, say, .
You can then find in the usual way to maximise the gradient.
Say , the unit 'direction vector' of the curve varies.
I suppose you could parametrise the 'direction vector' in terms of x too, in my head that looks very messy.Last edited by EricPiphany; 16032016 at 00:32. 
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 16032016 00:29
(Original post by EricPiphany)
I would do this.
Parametrise x and y, and then h(x, y) = h(t), which is 1 dimensional.
You would have to parametrise in terms of curve length l think.Last edited by EricPiphany; 16032016 at 00:31. 
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 16032016 10:06
(Original post by EricPiphany)
What if the curve in the xy plane wasn't a line?
Say , the unit 'direction vector' of the curve varies.
I suppose you could parametrise the 'direction vector' in terms of x too, in my head that looks very messy.
So you can follow the prescription as above with that unit vector, and finally constrain the gradient to to get a 1D problem, then differentiate. 
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 16032016 10:23
(Original post by Zacken)
Why not plug into to get and then find a way to maximise (this is probably all wrong)  otherwise, this sounds like something that can be tackled by Lagrange Multipliers, something I'm not very familiar with, in which case, I'll let somebody like atsruser or Greg jump in.
In fact, a unit increment along gives you an increment of along so this will make the calculated gradient too big (too small?). The gradient method that I showed evades this, since it allows us to differentiate naturally in the direction of the line in question, and at one point we calculate a unit tangent along that direction to make sure that we are working in terms of unit increments along that line.
[edit: in calculating the gradient along the line via Zacken's method, this happens: Suppose that the height of the function above the xy plane is . We work with an xchange of along the xaxis  however, due to the slope of , this corresponds to points on the function that are really apart along , so our zdelta turns out to be and the slope of is computed as which is too big.]Last edited by atsruser; 16032016 at 10:40. 
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 16032016 10:24
(Original post by EricPiphany)
I would do this.
Parametrise x and y, and then h(x, y) = h(t), which is 1 dimensional. 
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 16032016 11:59
(Original post by atsruser)
I don't think that this is fundamentally different from Zacken's suggestion, but it suffers from the same problem (of step size) which I described above.
(Original post by EricPiphany)
Just realized why this won't work in the general case.
You would have to parametrise in terms of curve length l think. 
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 16032016 23:23
(Original post by Zacken)
Why not plug into to get and then find a way to maximise (this is probably all wrong)  otherwise, this sounds like something that can be tackled by Lagrange Multipliers, something I'm not very familiar with, in which case, I'll let somebody like atsruser or Greg jump in.
and then solve the series of equations:
The last one is boring; it just reproduces the constraint that you started with, but the other two give you the equations in that you need to solve to find the extrema. This looks like it's equivalent to the gradient method in some sense, though [Maybe Gregorius can explain the exact connection?]Tagged: 
Gregorius
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 17032016 08:47
(Original post by atsruser)
[Maybe Gregorius can explain the exact connection?] 
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 17032016 09:46
(Original post by Gregorius)
Eeek! I was never any good at this sort of stuff. But from a geometric point of view, one needs to observe that to do the parametrization approach correctly (the idea of if xy=2 then y = x  2) the resulting curve must be normalized by arclength  or you end up out by a factor of root two as has already been observed. My guess is getting the unit vector corresponds to the arc length normalization step.
Having thought about it a bit, in this simple case, these are indeed merely the same thing, since once we have got our 3 equations by partial differentiation (one of which just gives us back the constraint) we then have to solve them (by substitution..) so Lagrange multipliers gives us nothing here.
My question was fairly stupid, in fact. 
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 17032016 09:57
(Original post by atsruser)
My question was fairly stupid, in fact.
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