Why Does the Expression -dy/dx/dx/dy Calculate the Slope at a Point?

[ktpr2]

I recently wrote a program to check implicit differentiation and the key bit I saw in other programs was the logic amounting to:

-\frac{ \frac{dy}{dx}} {\frac{dx}{dy}}

when given a valid (x,y) point on the curve. Can someone explain why the above generates the slope at point (x,y)?

 

[learningphysics]

I recently wrote a program to check implicit differentiation and the key bit I saw in other programs was the logic amounting to:

-\frac{ \frac{dy}{dx}} {\frac{dx}{dy}}

when given a valid (x,y) point on the curve. Can someone explain why the above generates the slope at point (x,y)?

Are you sure you don't mean something like:
\frac{ \frac{dy}{dt}} {\frac{dx}{dt}}

This is just the operation of the chain rule:
dy/dt = (dy/dx) (dx/dt)

so dy/dx = (dy/dt)/(dx/dt)

 

[ktpr2]

Well that's the code snippet I used. To take the derivative of x in terms of y seems kinda heady. The actual code was:

(-der1(eqn,x))/(der1(eqn,y)), which means, according to my ti-86 manual, returns the value of the first derivative ... with respect to variable for the current variable value. This is provided, as a point on the curve.

So I suppose I'm actually just taking the derivative of the expression in terms of a constant. I'm not quite sure what that would mean or if it's nothing more than normal division?

EDIT - Actually, it's probably the derivative of the expression with the value for x or y plugged in. I should probably research dx/dy because that's what's going on here.

 

[dextercioby]

In the way you've written it,i.e.that ration,i hope u see that it makes no sense whatsoever.

Daniel.

 

[ktpr2]

Well, that's the program code and it works just fine. I do believe my teacher said you can take dy/dx, just that we won't be doing it anytime soon. I'll talk to him about it.

 

[Jameson]

Why are you trying to find dx/dy? I'm confused on what the purpose of your program is. It seems there is some practical purpose, so I just don't see where dx/dy fits in. The code in the first post would find \frac{d^2y}{dx^2} if I'm not mistaken.

Jameson

 

[StatusX]

It sounds like you mean:

\frac{d y}{d x} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}

Is this right? If so, this is the equation for the slope of the contour of constant f passing through some point (x,y) in a 2D contour plot of f(x,y). This formula works because if you change x by some amount, you change f by df/dx*dx. Then to get back on the contour, you must change y by enough to get the opposite change, df/dy*dy=-df/dx*dx. This gives you your formula.

 

[ktpr2]

statusX- yeah that seems like what's going on.

Jameson -THe program merely checks your answer for implict differentation; you find a point on the curve via TI 8X's Solver function, and run it and it'll give you the slope and ask you to type in your answer. It compares and tells you if you're right or not. (http://www.ticalc.org/pub/86/basic/math/imp.zip )

 

[jdavel]

ktpr,

What are you entering into the calculator to get dx/dy? I'm guessing it's something like "der(y,y)" where you have defined that first "y" to be your function. But that would give you dy/dy which is 1 and would explain why your ratio comes out to be simply dy/dx.

Am I close?