Multi-variable systems?

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Boris Lagutin

Re: Multi-variable systems?

Post by Boris Lagutin »

Thank you, Joe. First of all, It would be interesting to know:

1) How to calculate a surface area of the parabaloid region colored with red color on the drawing?

2) Whether we use a double integral to find the area?

Thanks a lot.
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joe
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Re: Multi-variable systems?

Post by joe »

F . dA can be set equal to three different expressions:
i) F dA cos theta or
ii) dA times the component of F parallel to dA or
iii) F times the component of dA parallel to F

You don't need to calculate that area, because you are not going to use (i) or (ii).
In this case you use (iii), so the only integration you need (if F is constant) is the integration to get A1
Then the flux is F * A1
Boris Lagutin

Re: Multi-variable systems?

Post by Boris Lagutin »

Could someone clarify: May a gradient of G = sqrt(x^2 + y^2) = R where R is a radius of a contour curve be equal to:

gradient of G = (x/R)i + (y/R)j ?

P.S. As I see a gradient of G so it should be 2x + 2y resulting from (x^2)i + (y^2)j = R^2

Thank you.
a.s.h.
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Re: Multi-variable systems?

Post by a.s.h. »

Boris Lagutin wrote:Could someone clarify: May a gradient of G = sqrt(x^2 + y^2) = R where R is a radius of a contour curve be equal to:

gradient of G = (x/R)i + (y/R)j ?

P.S. As I see a gradient of G so it should be 2x + 2y resulting from (x^2)i + (y^2)j = R^2

Thank you.
Your first expression for grad(G) is correct. We have

∂/∂x (sqrt(x^2 + y^2)) = 2x/(2sqrt(x^2 +y^2)) (using chain rule)

= x/sqrt(x^2 + y^2)

= x/R, where R = R(x,y) = sqrt(x^2 + y^2).

Similarly, ∂/∂y (sqrt(x^2 + y^2)) = y/R.

So grad(G) = (x/R)i + (y/r)j.
Boris Lagutin

Re: Multi-variable systems?

Post by Boris Lagutin »

Oh! Thank you, a.s.h.

Definitely I need some relax. Study even in Christmas holidays. So I have just quickly imagined (even didn't calculate) its partial derivatives and decided that those partial derivatives were produced in a way unknown for me. Oh, My God.

Happy Holidays, a.s.h.
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joe
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Re: Multi-variable systems?

Post by joe »

There doesn't appear to be a drawing.
If your surface is a parabola of rotation, then you need only a single integration:
the element of area will look something like 2*pi*r*ds
and you'll need to integrate along s.
Boris Lagutin

Re: Multi-variable systems?

Post by Boris Lagutin »

Thank you, Joe.

Happy Holidays!
Boris Lagutin

Re: Multi-variable systems?

Post by Boris Lagutin »

Given z = f(x,y) and a point (x_1,y_1,z_2). Could somebody explain how to find the shortest distance between the point (x_1,y_1,z_1) and a surface formed by z = f(x,y) by using the distance formula: d = sqrt[ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 ] ??? Some kind of optimization problem.

Thanks a lot.
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joe
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Re: Multi-variable systems?

Post by joe »

I can only see the brute force method:
Write down the distance D between the given point and an arbitrary point (X,Y,Z) on the curve (using the d= equation you have). So you have
D = D(X,Y,Z)
At the shortest distance, ∂D/∂X, ∂D/∂Y and ∂D/∂Z are all zero. That gives you three equations to solve simultaneously, and three is just the number you need to solve for X,Y and Z.

But there may be a sneaky method that is more elegant.

Joe
a.s.h.
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Re: Multi-variable systems?

Post by a.s.h. »

Boris Lagutin wrote:Given z = f(x,y) and a point (x_1,y_1,z_2). Could somebody explain how to find the shortest distance between the point (x_1,y_1,z_1) and a surface formed by z = f(x,y) by using the distance formula: d = sqrt[ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 ] ??? Some kind of optimization problem.

Thanks a lot.
In fact, we only need two partial derivatives here, because a point on the surface has only two independent variables x and y; the value of z is then given by f(x,y). Also, a typical but handy trick for these problems is to work with the square of the distance, S := (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2, rather than the distance itself. This is because it is much more pleasant to differentiate things without the square root, and optimising the distance is equivalent to optimising the squared distance (i.e. the solution(s) you get from optimising S is (are) the exact same as those for optimising the distance D).

cont.
a.s.h.
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Re: Multi-variable systems?

Post by a.s.h. »

On the surface, we have z = f(x,y), so the squared distance of a point on the surface to our given (fixed) point (x_1, y_1, z_1) is S = (x-x_1)^2 + (y-y_1)^2 + [f(x,y)-z_1]^2 .

So to optimise S, we just take the partials of S with respect to x and y, and set these to 0 as Joe described. I.e.

∂S/∂x = 2(x - x_1) + 2[f(x,y)-z_1].∂f/∂x --> set equal to 0

∂S/∂y = 2(y - x_1) + 2[f(x,y)-z_1].∂f/∂y --> set equal to 0.

This gives us two equations in two unknowns. The optimum solution must satisfy these equations. (We have assumed throughout that f is a differentiable function, which is a safe assumption for most applications.)

If we then find the x,y values satisfying these equations and can verify which one(s) correspond(s) to the desired optimum, the z-value of the point is found by plugging these x,y values into the function f(x,y).
Last edited by a.s.h. on Sat Jan 09, 2016 10:10 pm, edited 1 time in total.
a.s.h.
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Re: Multi-variable systems?

Post by a.s.h. »

More generally, if our surface is given implicitly, by an equation like g(x,y,z) = k (k a constant), we can use the method of Lagrange multipliers. In this case, we want to optimise the function S = S(x,y,z) = (x – x_1)^2 + (y – y_1)^2 + (z – z_1)^2 subject to the constraint g(x,y,z) = k (which means that our point must lie on the surface). To do this, we compute the gradients of our objective function and constraint functions, i.e. ∇S and ∇g, and we set ∇S = λ∇g. This, along with the equation g(x,y,z) = k (the constraint equation) gives us a system of four equations in four unknowns (x,y,z and λ), which our optimal solution must satisfy. Generally these equations will be non-linear though and hard to solve. The Lagrange multipliers method is taught in one of the MIT Multivariable Calculus videos you mentioned before I think, so you can watch that if you want to review it.
Boris Lagutin

Re: Multi-variable systems?

Post by Boris Lagutin »

Thanks a lot, a.s.h.! Let me consider the first situation described by you. Ok. We got partial derivatives with respect to x,y of the squared distance formula (S) but how to find values of x,y which make the partial derivatives equal to zero?! So if we got x,y of the first power in the partials we could simply use a system of equations: 2 partial derivatives with respect to x and y appropriately, each of which equals zero.

However, what if we got partial derivatives with respect to x,y as polynomials of high order powers like a*x^5+b*x^3+c*x^2+d where a, b, c, d are just constants? I suppose it could happen if z=f(x,y) were a very complicated function.

Thank you.
a.s.h.
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Re: Multi-variable systems?

Post by a.s.h. »

In general, there is no guaranteed method of solving non-linear simultaneous equations. Sometimes they cannot be solved analytically. In the case of your polynomial example, for a general polynomial of degree 5 or higher, there is in fact no general formula for the roots of the polynomial (there are formulas for degree 2, 3 and 4).

So if your simultaneous equations are too nasty to deal with by hand, you would need a computer to solve them numerically.
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