dy/dx problem

Eliot Rosewater

I doing a problem and Ive come across this issue. I'm scaling variables so as to clean up the equation into something simpler. The problem is that one of the variables Im scaling is the bottom variable in dy/dx, ie the x. I drew up this very contrived example to illustrate what Im on about:



See the way I subsumed the a into the bottom of the dy/dx in (3), and then changed it into dy/dt in (5) given that z=at by (4). Is that actually correct procedure?


Btw in the actual problem doing this gets the correct answer.

Michael Collins

You have got the correct answer, like you say, but your method isn't right. It doesn't work in general (i.e. for different kinds of scaling), although it will work for scaling variables in a linear fashion, like you have done.

The mathematically sound way to do it would be to use the chain rule:

[latex] \displaystyle \frac{dy}{dx} = a^2x + ae^{2ax} [/latex]

Let [latex] \displaystyle z = ax [/latex]

[latex] \displaystyle \frac{dy}{dz}=\frac{dy}{dx}\frac{dx}{dz} [/latex]

[latex] \displaystyle \frac{dx}{dz} = \frac{d}{dz} \left(\frac{z}{a}\right) = \frac{1}{a} [/latex]

[latex] \displaystyle \frac{dy}{dz}= \left(a^2x + ae^{2ax}\right)\left(\frac{1}{a}\right) = ax + e^{2ax} = z + e^{2z} [/latex]

Eliot Rosewater

Excellent, cheers for that Micheal Collins.

Incidentally, how do you get the maths equations into images (I assume you using Latex)?

Michael Collins

No problem.

Eliot Rosewater wrote: »

Incidentally, how do you get the maths equations into images (I assume you using Latex)?

Just wrap LaTeX tags around each line of the usual LaTeX code, i.e.

(latex) LaTeX Code here! (/latex)

Using square brackets above intead of parentheses. Including the \displaystyle primitive usually makes things look better too

(latex)\displaystyle LaTeX Code here! (/latex)

Just another note on what you were doing, you weren't too far away from a method that is used, namely, differentials.

Using differentials you can do calculations like

[latex] \displaystyle d(x^2) = 2xdx [/latex]
[latex] \displaystyle d(x^3) = 3x^2dx [/latex]
[latex] \displaystyle d(\sin x) = (\cos x)dx [/latex]

etc

So in your example you wanted to get dz on the bottom
so to figure out what dz is in terms of dx, just use differentials:

[latex] \displaystyle dz = d(ax) = adx [/latex]

So

[latex] \displaystyle \frac{dy}{dz} = \frac{dy}{adx} = \left(\frac{1}{a}\right) \left(\frac{dy}{dx}\right)[/latex]

which is what you had.

But suppose you wanted to scale x by squaring it, then you don't get the right answer by just replacing a with x above.

Using differntials you get

Letting [latex] \displaystyle z = x^2 [/latex]

[latex] \displaystyle dz = d(x^2) = 2xdx [/latex]

[latex] \displaystyle \frac{dy}{dz} = \frac{dy}{2xdx} = \left(\frac{1}{2x}\right) \left(\frac{dy}{dx}\right)[/latex]

Eliot Rosewater

Cheer again!

For some reason I never mastered manipulation of derivatives and the like in secondary school. I'm in first year Maths Science and doing a course in Maths Modeling, where derivatives play a huge part, so Im learning steadily


Thanks for the Latex tip too. Im a beginner, and Im actually using Lyx out of pure convenience for the movement, but learning to hand code would be a goal somewhere down the line. We have a stats assignment thats to be done in Lyx but I figure that it will be good practice to do all the assignments in Latex. It looks excellent.