Implicit functions in R2 vs. functions of two variables in R3

1. Sep 15, 2012

dumbQuestion

Hello,

I often encounter some confusion.

For example, we have the equation for a circle of radius one given by:

x2+y2=1

Now we can't express this as a function in R2 because it's a relation. however, we can still find derivatives like dy/dx or dx/dy using implicit differentiation.

Something that gets me though is, how does this implicit function relate to a function of two varaibles? For example, we could create the function

z=x2+y2

Now for example if we let z=1, won't this be the circle of radius one, lying in the plane z=1?

How do the derivatives of these two things relate to each other? For example, in multivariable equations, we wouldn't just say "what's the derivative of z" because that doesn't make sense, we'd think, "well, in terms of what variable?". If we wanted dz/dx we'd take the partial derivative of z in terms of x. But for example, if we go back to the 2D relation x2+y2=1, if we want to find the derivative of y in terms of x, we use implicit differentiation and go:

d/dx(x2+y2)=d/dx(1)

2x+2y(dy/dx)=0 --> (dy/dx)=(-x/y)

But if we now go back to our 3D equation z=x2+y2, say we want the derivative of y in terms of x when z = 1. Would this end up being the same thing?

2. Sep 15, 2012

Bipolarity

Partial derivatives hold either x or y to be constant so you can see how the height of the surface Z varies with only a single variable.

In your example of a circle, it is simply Z assigned to be a constant 1, in which case the partial derivatives of Z has no significance, since Z is no longer a function of the other two.

However, if you take $z^{2} = x^{2} + y^{2}$ and solve it for y, you get $y^{2} = z^{2} - x^{2}$

In 3 dimensions, this is still the exact same surface, but it has y as the subject. Now you can find partial derivatives of y with respect to x holding z constant!
So you can find

$$\frac{∂y}{∂x} = \frac{-2x}{\sqrt{z^{2}-x^{2}}}$$

This basically tells you how y changes with respect to x, holding z to be a constant. In your case, z=1, so we set z=1.

With some algebra, we get

$$\frac{∂y}{∂x} = \frac{-x}{y}$$ when z=1

BiP