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A function is defined by $f(x,y)=x^4 - 6x^2y^2 + y^4 -2x^2 + 2y^2$.

If we let $z=f(x,y)$ and let a particle travel in the direction which $f$ decreases most rapidly, how do I show that $xy(x^2 - y^2 -1) = \text{ constant}$?

I'm unclear of how to obtain the constant. I solved for the directional derivative and tried integration, but to no avail. I see some form of Lagrange multipliers here (hence the constant) but I don't know how to approach it. Please help, thank you!

user107224
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  • I have tried to use directional derivatives to help me, but they seem to integrate very weirdly. Could someone help me on this? – user107224 Mar 07 '16 at 23:07

1 Answers1

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Some theory

If the path is parametrized by $r(t) = (x(t),y(t))$, then by assumption $r'(t)= (x'(t),y'(t))$ is parallel to $-\nabla f (x(t),y(t))$, or $r'(t) = a(t) (-f_x(r(t)),-f_y(r(t)))$ for some positive constant $a$ depending on $t$.

This implies $ (-f_y(r(t)),f_x(r(t)))\cdot r'(t)=0$, and if we can show that for some $g$, $(-f_y,f_x)=(g_x,g_y)$, then the chain rule applied to $g(r(t))$, gives

$$\frac{d}{dt} g(r(t)) = (g_x(r(t)),g_y(r(t))) \cdot r'(t)=(-f_y(r(t)),f_x(r(t)))\cdot r'(t)=0,$$

$\Rightarrow g$ is constant along the path.

Solution

Let's then try to find $g$ such that $(g_x,g_y) = (-f_y,f_x)$. First observe that

$$f_x = 4x^3 - 12xy^2 -4x, ~f_y = 4y^3 - 12x^2y +4y.$$

or

$$f_x = 4x(x^2-3y^2-1),~f_y = 4y(y^2-3x^2+1).$$

If $g_y =f_x$, integrating $f_x$ with respect to $y$ then gives:

$$g = 4x (yx^2 -y^3-y)+c(x)=4xy(x^2-y^2-1)+c(x).$$

Use this and the formula for $f_y$ to calculate $g_x$:

$$g_x = 4y(x^2-y^2-1)+4xy 2x+ c'(x) =-4y^3 +12x^2 y - 4y +c'(x).$$

This is equal to $-f_y$ if and only if $c'(x)=0$, that is, $c$ is a constant function.

In other words, the conditions can be only satisfied by a function of the form $g(x,y) = 4xy(x^2-y^2-1)+c$. But clearly this function satisfies the conditions. We're done.

Fnacool
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  • Hi, thank you for the answer! I have two queries. Firstly, why can we make the assumption that the gradient is parallel to r'(t)? Also, I can see mathematically why g is a level surface, but I can't see it intuitively. Could you explain why? – user107224 Mar 10 '16 at 07:50
  • The gradient gives direction and size of maximal increase. Minus the gradient gives the direction and size of maximal decrease (see this article for example http://math.stackexchange.com/questions/223252/why-is-gradient-the-direction-of-steepest-ascent) 2.
    • – Fnacool Mar 10 '16 at 13:28
  • I'm not sure what you meant by writing "$g$ is a level surface". However, this seems like a continuation of the first question. Some explanation: the rate of change at given point in a given direction is the dot product of the gradient at the point and the direction, and so if the path always points in a direction orthogonal to the gradient (dot product = 0), the rate of change is always zero, so that the value is constant along the path.
    • – Fnacool Mar 10 '16 at 13:59
  • Regarding 1), I meant it as why we use parallel to r' rather than r? Since the particle moves in the direction of greatest descent, shouldn't the path itself be parallel to the negative of the gradient, rather than its rate of change? – user107224 Mar 11 '16 at 19:11
  • $r$ represents the position (where we're standing) and $r'$ represents where the path goes (where we're facing or going to), and which is what we want to control: we need to go in the same direction the function decreases most rapidly. – Fnacool Mar 11 '16 at 19:27