Expert: Paul Klarreich - 2/5/2007

Question
Hello,

Here is something I've been wondering about:

Given a curve in parametric form, x = x(t), y = y(t), dx/dt = x-dot, dy/dt = y-dot, the derivative of y with respect to x is said to be y-dot/x-dot.

On the other hand, if the same curve is given in the form F(x,y) = C (or F(x,y) - C = 0), through implicit differentiation with respect to x we get DF/Dx + DF/Dy.dy/dx = 0, and hence dy/dx = y' = -(DF/Dx)/(DF/Dy) (the capital Ds denoting partial derivatives, as I hope is clear from context).

It turns out, then, that y-dot/x-dot = -(DF/Dx)/(DF/Dy). Is this true, and if so, why? (If not, why not?)

Thanks a lot,

Martin

Answer
Questioner:   Martin
Category:  Advanced Math
 
Subject:  Derivatives
Question:  Hello,

Here is something I've been wondering about:

Given a curve in parametric form, x = x(t), y = y(t), dx/dt = x-dot, dy/dt = y-dot, the derivative of y with respect to x is said to be y-dot/x-dot.

>> Not only is it said, it IS.

On the other hand, if the same curve is given in the form F(x,y) = C (or F(x,y) - C = 0), through implicit differentiation with respect to x we get DF/Dx + DF/Dy.dy/dx = 0,

and hence dy/dx = y' = -(DF/Dx)/(DF/Dy) (the capital Ds denoting partial derivatives, as I hope is clear from context).

It turns out, then, that y-dot/x-dot = -(DF/Dx)/(DF/Dy). Is this true, and if so, why?

(If not, why not?)

Thanks a lot,

Martin
......................................
Hi, Martin,

How about the chain rule?  Suppose you have an equation:  F(x,y) = C, and your x and y are parametrized in terms of t.  [Hmmm.  Is it 'parameterized'?]  

Then we really have F(t) = C, and the Chain Rule says:(I shall adopt your notation: D means partial derivative.)

dF   DF dx   DF dy
-- = -- -- + -- --
dt   Dx dt   Dy dt

But since F(t) = C, dF/dt = 0.  Then

DF dx   DF dy
-- -- + -- -- = 0
Dx dt   Dy dt

and I think your conclusion follows from this.