Suppose f:R^2->R is a differentiable function whose gradient is nowhere 0 and that satisfies ∂f/∂x = 2(∂f/∂y).

Find, with proof, the level curves of f.

Show that there is a differentiable function F:R->R s.t. f(x,y) = F(2x+y)

So I integrated both sides of the differential equation

∫(∂f/∂x)dx = 2∫(∂f/∂y)dy

f(x) = 2f(y)

Somehow, I assume this gets me f(x,y) = 2x + y, since that's used in the second part of the problem. How do I make this conclusion (ie, how do I go from f(x) = 2f(y) to f(x,y) = 2x + y)? It seems like this should be a basic calculus thing, but I keep getting hung up on it... (maybe this is the universe's way of telling me that I shouldn't do math late at night, haha)

If it helps, the topic of this section is supposed to be the gradient and using the gradient to define tangent planes/surfaces (specifically, ∇f(ā) · (x̄-ā) = 0 )

Update:

I also have a sketch of the proof:

Suppose g(t) = f(t, -2t)

g'(t) = ∂f/∂x - 2(∂f/∂y) = 0

So g is constant; ie, g has the same value at any point t.

But then I get stuck again--how does this get me to 2x y?

Update 2:

The last thing was supposed to say "how does this get me to 2x y?"

Relevance

Essentially the same calculation that you did, tells you that if you first fix an arbitrary constant c and let

g(t) = f(t, -2t + c),

then g'(t) = 0 for all t. In particular, for any t we have

f(t, -2t + c) = g(t) = g(0) = f(0, c).

Fixing any (x,y) in R^2, putting c = 2x + y and t = x in this, we learn

f(x,y) = f(x, -2x + 2x + y) = f(0, 2x + y).

Since x and y were arbitrary, this shows that the function F(t) = f(0,t) of one variable satisfies f(x,y) = F(x + 2y) for all x and y.

It might help to express the idea used here in geometric language. The given information implies that f is constant along lines in the xy plane that have slope -2. This is the family of curves y = -2x + c, where c varies --- or equivalently, the family of curves 2x + y = c. So for a given (a,b), the value of f(a,b) doesn't "really" depend on (a,b), as much as it simply depends on 2a + b (a single number that, once you know it, tells you which line of slope -2 you are on. So there must be is some function F with f(x,y) = F(2x+y) (although it doesn't immediately tell us what F is). To specifically identify F, we can simply note that if we take x = 0 in f(x,y) = F(2x+y) we learn that f(0,y) = F(y) for all y, and this completely determines F: it must be that F(t) = f(0,t) for all t. Alternatively, we can derive a formula for F by thinking geometrically. If I give you a point (a,b), what line of slope -2 does it lie on? It lies on the line y = -2x + c when b = -2a + c, or when c = b - 2a. Since the point (0,c) = (0, b-2a) also lies on this line, and f is constant on lines of slope -2. we have f(a,b) = f(0,b-2a).

How to come up with the lines of slope -2 from the given information? Well, suppose we fix some value k and think of the level curve f(x,y) = k as implicitly defining y as a function of x. Taking the derivative with respect to x we learn df/dx(x,y) + df/dy(x,y) dy/dx = 0. Using the given information that df/dx(x,y) = 2 df/dy(x,y) we deduce that 2 df/dy(x,y) + df/dy(x,y) dy/dx = 0. Since the gradient of f is nowhere 0, df/dy is nowhere zero, so we can divide both sides of this last equation by df/dy(x,y) and learn that 2 + dy/dx = 0 or that dy/dx = -2, implying that y = -2x + c for some c, so that the level curve has to be a straight line of slope -2. The calculation that I began this answer with then confirms that the converse is also true (that is: not only does the given information imply that any level curve has to be a straight line of slope -2; it implies that any straight line of slope -2 is indeed a level curve of f). I hope this helped.

• Login to reply the answers