Finding the Derivative of f(x,y) = 0

[unscientific]

Homework Statement

Given f(x,y) = 0, find an expression for dy/dx.

Homework Equations

The Attempt at a Solution

f(x,y) = 0

df = (∂f/∂x)dx + (∂f/∂y)dy = 0

df/dx = (∂f/∂x) + (∂f/∂y)(dy/dx) = 0


dy/dx = -(∂f/∂x)/(∂f/∂y)

 

[Dick]

Homework Statement

Given f(x,y) = 0, find an expression for dy/dx.

Homework Equations

The Attempt at a Solution

f(x,y) = 0

df = (∂f/∂x)dx + (∂f/∂y)dy = 0

df/dx = (∂f/∂x) + (∂f/∂y)(dy/dx) = 0


dy/dx = -(∂f/∂x)/(∂f/∂y)

Looks ok to me. Do you have a question about that?

 

[unscientific]

Looks ok to me. Do you have a question about that?

Hmmm, so I guess I'm right?

My only question is

f(x,y) = 0,

df = 0?

 

[Dick]

Hmmm, so I guess I'm right?

My only question is

f(x,y) = 0,

df = 0?

Well, yes. Why would you think not?

 

[unscientific]

Well, yes. Why would you think not?

That makes absolute sense. If a function is zero for all x,y then the infinitesimal change df would be 0 for all x,y.

 

[Dick]

That makes absolute sense. If a function is zero for all x,y then the infinitesimal change df would be 0 for all x,y.

In this problem you shouldn't think of f as identically 0. Here's an example to think about. Suppose y=x (defining a curve). Take f(x,y)=x^2-y^2. Then f(x,y)=0 along the line y=x, but f(x,y) is not identically 0. But df=0 along the line y=x.

 

[unscientific]

In this problem you shouldn't think of f as identically 0. Here's an example to think about. Suppose y=x (defining a curve). Take f(x,y)=x^2-y^2. Then f(x,y)=0 along the line y=x, but f(x,y) is not identically 0. But df=0 along the line y=x.

So, what does f(x,y) = 0 actually mean?

 

[Dick]

So, what does f(x,y) = 0 actually mean?

You think of y as being some function of x. That's why they can ask you to find an expression for dy/dx. Then saying f(x,y)=0 means f is zero along the curve defined by y(x).

 

[unscientific]

You think of y as being some function of x. That's why they can ask you to find an expression for dy/dx. Then saying f(x,y)=0 means f is zero along the curve defined by y(x).

That makes sense, thanks! But still it doesn't explain why df = 0..

 

[haruspex]

That makes sense, thanks! But still it doesn't explain why df = 0..

You're given that f is always zero, does not change as x and y change. So df=0.

 

[epenguin]

Congratulations, you have worked out what is a fairly fundamental and often needed little formula. Which surprises many students as with a superficial look of the formula they can imagine it is going to be the other way up and the minus sign surprises.

If you draw a little picture it should become less mysterious and surprising.

Also, as mentioned, it holds for f constant - the constant doesn't have to be 0.

 

[algebrat]

This reply doesn't really stand alone, I was trying to build on what others had previously said.

If f is a nice function (for instance, a polynomial in x and y, or some other equation not doing anything crazy. Look up space filling curve.), the solution set is one-dimensional. So it is a curve in the plane.

In other words, the equation f(x,y)=0 has a set of solution points (x,y) that is a curve. Along some sections of that curve, we might be able to write the curve explicitly, as y=y(x), or g(x), whatever. Notice that along the curve, f=0.

We can use f to look at other parts of the ambient plane. f may take many values; in fact, for each value of c, we expect the equation f(x,y)=c to give various curves in the plane, a solution set for each value c. Notice that once we pick a curve as specified, f is constant, along that curve. In other words, df=0. (We could say something like, "because d(c)=0"; i.e., the differential of a constant is zero.)

The idea of an equation, or relation, determining a 1-d subset of the plane, ie curve, generalizes and is a very useful and frequently used concept in math and science, and you should totally try to absorb the idea into your brain, and keep an eye out for when it comes up.