Why does this work? (Integration) (1 Viewer)

[QZP]

I find higher level maths (algebra) to lack common sense to me; instead becoming sought of a rote process. My foundations start to break down.

e.g. For growth and decay (from 3U), finding the equation of Q given

dQ/dt = kQ

So, dQ/kQ = dt ---------(1)

integral (dQ/kQ) = integral (dt) ---------(2)

1/k . ln(Q) = t

... so forth and making Q the subject.

Questions:
* In (1), I treat the dy/dx operator as a fraction (this works since it arises from rise/run in first principle - I GET THIS).
BUT WHEN I MAKE IT SUCH THAT
dQ/ kQ = dt; what the hell does it mean to just have a dQ on itself? It's like dy = 3. What!?? dy/dx = 1 makes sense to me. Not dy = dx. What does the fundamentals speak about this? What happened to everything I learnt about differentiation (gradients)?

* In (2), I apply the integration operator to both sides. BUT WHY CAN I APPLY IT WITHOUT RESPECT TO ANYTHING? Usually if I had, say, a = b
Then I can apply the integral operator on both WITH RESPECT TO time, say. Not just slap on the integral sign without respect to anything as such in (2)




Am I making sense to anyone? Math doesn't appeal to common sense to me anymore.

 

[RealiseNothing]

1) Think of it like this



2) You ARE integrating with respect to something.

 

[QZP]

1) Think of it like this



2) You ARE integrating with respect to something.

Thanks for 1. Makes more sense that way.

But for 2 I still disagree and my problem is the way I view the integral operation.

For dy = dx, I (would) treat dy,dx as integrands and then integrate them both with respect to something common.
E.g. integral (dy).dt = integral (dx).dt
But this doesn't make sense at all... and it doesn't make sense (to me personally) that I can apply the integral operation on both sides with the result being they are integrated with different respects... because an operation should be.... consistent. Is there another way of seeing this? Or do I just have to accept it.

 

[seanieg89]

Another way of thinking about such questions:

Another way of seeing it is that the chain rule tells you that the derivative of an inverse is the inverse of the derivative.

So, assuming Q is locally invertible, we can write t as a function of Q such that dt/dQ=1/(dQ/dt)=1/(kQ). This is something we can easily solve to give us an expression for t in terms of Q, which we then find the inverse of to find Q in terms of t.

Note that this isn't entirely precise either, as it glosses over the issue of potential division by zero and requires an assumption of local invertibility. It is satisfactory as a high school level solution though, without needing to actually carefully examine these issues.


About the method in your OP:

The splitting up of dQ/dt by "multiplying by dt" in your OP should probably just be viewed as a notational convenience for now. The dQ and the dt are actually legitimate objects called differential forms, but the understanding of these guys is far beyond high school maths.

Do you understand why your line (1) is true if you replace the ds with deltas and equality with approximate equality? (*)

If you really want to convince yourself that the OP method "works", then you can use an approximation of your two integrals by Riemann sums, and show that these approximations are close together using the observation (*) (so the integrals must be close together).

 

[QZP]

Another way of thinking about such questions:

Another way of seeing it is that the chain rule tells you that the derivative of an inverse is the inverse of the derivative.

So, assuming Q is locally invertible, we can write t as a function of Q such that dt/dQ=1/(dQ/dt)=1/(kQ). This is something we can easily solve to give us an expression for t in terms of Q, which we then find the inverse of to find Q in terms of t.

Note that this isn't entirely precise either, as it glosses over the issue of potential division by zero and requires an assumption of local invertibility. It is satisfactory as a high school level solution though, without needing to actually carefully examine these issues.


About the method in your OP:

The splitting up of dQ/dt by "multiplying by dt" in your OP should probably just be viewed as a notational convenience for now. The dQ and the dt are actually legitimate objects called differential forms, but the understanding of these guys is far beyond high school maths.

Do you understand why your line (1) is true if you replace the ds with deltas and equality with approximate equality? (*)

If you really want to convince yourself that the OP method "works", then you can use an approximation of your two integrals by Riemann sums, and show that these approximations are close together using the observation (*) (so the integrals must be close together).

Replying chronologically:
a) Yeah I know of your top method (most common way I think to solve the question but I was just exploring this "4u" method)
b) "The dQ and the dt are actually legitimate objects called differential forms" - this is something that sounds very interesting I guess I'll just have to accept it
c) "Do you understand why your line (1) is true if you replace the ds with deltas and equality with approximate equality?" Because d (small delta) is the limit as a quantity --> 0 leading to an exact value, rather than big delta (triangle) which is just a "small quantity" and thus leads to an approximate answer?

Thanks for the awesome post

 

[Trebla]

Regarding (2), another way to think of it is as

 

[seanieg89]

c)

Yep. This approximate equality comes from just the definition of the derivative, letting h be small (so we are close to our limit) and multiplying out by h.

So this equation tells you approximately how Q changes given an small change in x. (*)

If we are making a sizable change in x, this is the same as making many successive small changes in x. Each of these we have a handle of because of (*).

If we add up lots of these small changes, we get an equality of Riemann sums which when we pass to the limit is the same thing as the integral equation that follows from your formal line (1).