Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    During my mechanics course the professor had an equation like:

    \displaystyle \frac{F}{m}=\frac{dv}{dt} he then split the fraction up like this and integrated \displaystyle \int \frac{F}{m} ~dt=\int dv like so to find a formula for \displaystyle v. (Can't remember if this was the expression but it is irrelevant.)

    I was wondering how you can do such a thing (what is really going on and what justifications have to be made in order to do such a thing)?

    I remember treating the derivative as a fraction before at A-Level and not knowing what I was actually doing so I have decided to try and fix that.

    I have know the formal definitions of derivatives, riemann integrals and limits (and just now seeing the fundamental theorem of calculus.)

    I haven't taken any modules in differential equations and I'm not really sure if it is beyond my scope at the moment.

    Could anyone explain or give me a rough idea of what is actually going on or point me in the direction of some resources I can look at?

    Cheers.
    Offline

    13
    ReputationRep:
    I think this is called the variable separable method of integration. (I learnt about it a while ago). Pretty useful in C4 and FP2-3. Sorry I can't help more!


    Posted from TSR Mobile
    Offline

    13
    ReputationRep:
    (Original post by alex2100x)
    Could anyone explain or give me a rough idea of what is actually going on or point me in the direction of some resources I can look at?

    Cheers.
    From what I can tell, he's simply integrating both sides of the equation with respect to t. Think back to implicit differentiation: because both sides are equal, their differentials must be equal, too. This is the same concept, but with integrals instead.
    The integral of \displaystyle \frac{F}{m} with respect to t is just \displaystyle \int \frac{F}{m} ~dt.
    As for the right hand side of the equation, it's pretty much the same story, except the dt essentially cancels out.

    Look at that bit as a fraction, and consider what's actually going on in terms of gradients/areas. Put simply, the fraction's got dt on the bottom and by integrating, you're multiplying by dt, so those terms cancel out because, obviously, dt/dt is just 1.

    That's probably the simplest way of looking at it.
    Offline

    11
    ReputationRep:
    It can be justified by considering that:

    \displaystyle \int \frac{dv}{dt} dt = v + c

    since we have just differentiated then integrated the result of that differentiation - these are inverse operations (up to the constant c) so we end up where we started. Also:

    \displaystyle \int dv = \int 1 dv = v+c =  \int \frac{dv}{dt} dt

    so these expressions are equivalent
    • Thread Starter
    Offline

    2
    ReputationRep:
    Ok thanks I guess it relies on FTOC to justify that differentiating and integrating are inverses of each other.

    cheers.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.