Explanation of the breaking up of a derivative

[jamesd2008]

If you have dv/dt you say to yourself its the derivative of v with respect to t. But in an example of deriving the first kinematic equation for constant acceleration you go from a=dv/dt, to dv=a.dt and then you integrate this equation to give you the velocity. i.e v=u+1/2a(tsquared), using initial conditions. In this sense what is the dv part and what is the dt part? I hope you understand as finding it hard to put into words. Is dv and dt both derivatives?

Thanks
James

 

[mathman]

dv/dt is the derivative. dv and dt and called differentials, and have meaning only in terms of carrying out integration.

 

[jamesd2008]

Hi thanks for the reply could explain further, for example,

see attached word doc sorry can't type in here with the symbols so put it in a word document.

Thanks in advance
James

 

[slider142]

Hi thanks for the reply could explain further, for example,

see attached word doc sorry can't type in here with the symbols so put it in a word document.

Thanks in advance
James

I do not understand where you believe the last equation you have in your document should come from.

 

[jamesd2008]

I just mean that when you integrate the differential dt you get t so should the last equation not include a extra t term from the differential dt. As on the other side of the equation dx becomes x. Thanks for taking the time to look at it james

 

[slider142]

At the current level, treating the derivative as if it were a fraction is simply an aid to manipulation and is not rigorously defined by what you have learned so far. In fact, it is simply an application of the fundamental theorem of calculus. That is, given f = dv/dt, the fundamental theorem tells us that \int f = v(t) + C where C is an arbitrary constant of integration. As you can see, "multiplying both sides by dt" is unnecessary.
The proper presentation of dv and dt as separate entities called differentials will be covered in a course on differential equations, or differential geometry.

 

[jamesd2008]

Thanks slide again for taking the time to look at this. Will let it go and just except that's the way it is. James

 

[slider142]

I just mean that when you integrate the differential dt you get t so should the last equation not include a extra t term from the differential dt. As on the other side of the equation dx becomes x. Thanks for taking the time to look at it james

dx is not the entity that becomes x. In the integral
\int dx
you are integrating the function between the integral sign and the dx symbol. This is the identity function 1, whose primitive with respect to x is x ( + C).
Similarly, the primitive of at with respect to t is at²/2. The dt at your level is only telling you which variable you are integrating over and is not an active participant in the integral.

 

[jamesd2008]

ok so if there is no function to be integrated with respect to x then the identity function of 1 is used.

You have been a great help
Thanks
James

 

[HallsofIvy]

Be careful with your terminology here. The identity function is f(x)= x. You mean the constant function, f(x)= 1.