Finding the intergral function (dy/dt = a(q-y))

miniradman

1. The problem statement, all variables and given/known data
I was working on an assignment and when I got my draft back, my teacher said I've made some errors working this out, however I'm not sure what I did wrong...

Find the intergral function of:
dy/dt=a(q-y) where t ≥0, y(0)=0 a and q are constants.


2. Relevant equations



3. The attempt at a solution

[itex]\frac{dy}{dt}[/itex]=a(q-y)

[itex]\frac{dy}{}[/itex]=a(q-y)dt

[itex]\frac{dy}{(q-y)}[/itex]=adt

[itex]\int\frac{dy}{(q-y)}[/itex]=[itex]\int adt[/itex]

ln(q-y)=at+c

q-y=e^at+e^c where: e^c is a constant so let it = A

q-y=Ae^at

-y=Ae^at-q

[itex]\frac{-y}{-1}[/itex]= -(Ae^at) [itex]\frac{ (-q)}{-1}[/itex]

y= -Ae^at+ q

∴ y= -Ae^at+ q

Can anyone see any mistakes?

failexam

Your mistake was in converting from the first to the next statement. Try and think where the mistake is.

Hint: What is the derivative of ln[f(x)]?

miniradman

I know the derivative of ln(x) to be

[itex]\frac{1}{x}[/itex]

failexam

The derivative of ln[f(x)] is f'(x)/f(x), where f'(x) is the derivative of f(x) with respect to x.

Substitute f(x) = x into this formula to convince yourself that the formula I wrote is a generalisation of the derivative of ln(x).

Then try to work out the derivative of ln(q-y), where q is a constant.

miniradman

wow... why must I divide the derivative function by the original function?

failexam

You mean 'why must I divide the derivative function by the original function to obtain the derivative of ln[f(x)]?

I think you are asking why f'(x)/f(x) has to be the derivative of ln[f(x)]. This is a very theoretical question, and as such it is unlikely to come up in your tests. But just so you know, that formula can be obtained by considering the mother of all formulas you have ever studied in differentiation. It is this:

[itex]\frac{dg}{dx} = \underbrace{lim}_{h \rightarrow 0} \frac{g(x+h) - g(x)}{(x+h) - (x)}[/itex], where [itex]\frac{dg}{dx}[/itex] is the derivative of a function g(x) with respect to x.

This is the fundamental definition of the differentiation of a function and all other formulas that you have learnt can be obtained by using this mother of all formulas. For instance, to obtain the derivative of ln[f(x)], substitute g(x) = ln[f(x)] in that fundamental definition and work out the result. The result has to be f'(x)/f(x).

On a more practical note, d/dx { ln[f(x)] } = f'(x)/f(x) means that d/dx { ln(x) } = ????

Try to work out the answer and then find d/dy { ln(q-y)}.

miniradman

I've always had trouble deriving by first principles, could I use the chain rule?

let ln(q-y)=ln u?

where u = (q-y) where q is constant

then

[itex]\frac{d}{dy}[/itex] = [itex]\frac{d}{du}[/itex] x [itex]\frac{du}{dy}[/itex] ?

failexam

Oh no! Don't try to find d/dy { ln(q-y) } using first principles. I elaborated on first principles just to answer your question on why you must divide the derivative function by the original function to obtain the derivative of ln[f(x)].

Anyway, by the looks of your last post, you appear comfortable with the topic of differentiation, so try the chain rule.

And you can check that you will get the same answer if you use d/dx { ln[f(x)] } = f'(x)/f(x).

miniradman

so.

d/du = 1/u

du/dy = -1?

d/dy = 1/(q-y) x -1

d/dy = -1/q-y

errr... this can't be right, can it?

RoshanBBQ

Do you know how to do integration by substitution?

andrien

that would not be ln(q-y). it will be -

miniradman

Nope, I've only learn the basic rules for intergrating trigonometric and basic polynomials (i.e. f(x)). Also the basic rule for intergrating 1/x which equals lnx

Also the method that I used in my opening post... which I guess hasn't gotten me anywhere =D

RoshanBBQ

Consider
[tex]\int \frac{1}{q-y}dy[/tex]
Let
[tex]u = q-y[/tex]
Then
[tex]\frac{du}{dy} = -1\rightarrow dy = -du[/tex]

We then substitute using q-y = u and dy = -du
[tex]\int \frac{1}{u} \left (-du \right )=-\int \frac{1}{u} du[/tex]
Once you accomplish the integration, treating u just like you'd treat x, you replace any u in your answer with q-x.

miniradman

Hmmm, why in this case are we deriving du/dy? instead of dy/du?

RoshanBBQ

It doesn't matter which one you do (if you can do both easily). When you do a U-substitution, though, you will write

u = f(y)

So it makes sense to differentiate with respect to y. You then get du/dy = f'(y).

miniradman

Ok mate, you've lost me
If:
[tex]\int \frac{1}{q-y}dy[/tex]
Let
[tex]u = q-y[/tex]

then

[tex]\int \frac{1}{u}dy[/tex] right?

So why doesn't...

d = lnu y

because [itex]\int dy[/itex] = y and [itex]\int\frac{1}{u}[/itex] = ln u ?

Where is my mistake?

vela

Other than your abuse of notation, what you wrote is correct. So if f(y) = ln (q-y), you have that f'(y) = -1/(q-y). You can use this information to solve
$$\int\frac{1}{q-y}\,dy = \int -f'(y)\,dy = \ ?$$