Kinematics - Acceleration equation confustion

[help1please]

Kinematics -- Acceleration equation confustion

I have been following a derivation and can't quite wrap my head round what is happening. I don't know if the person who wrote it has made a mistake somewhere or I am making the mistake.

I am informed that

$$a = \frac{dv}{dt}$$

Fine. Then I am told

$$a= \frac{dv}{dx}\frac{dx}{dt}$$

(which is odd, one moment it is dv/dt now it seems to be dv/dx)

and that I should convince myself that this is the acceleration through a cross multiplication... but I did it and I got

$$\frac{dvdt}{dx dx}$$

This surely isn't acceleration, have I made a mistake somewhere? And if the person I am following is wrong, what is the correct way to state this...?

They further said that

$$\frac{dx}{dt}$$

is just velocity, so they end up with

$$a = v \frac{dv}{dx}$$

Can someone help?

 

[grzz]

... $$a= \frac{dv}{dx}\frac{dx}{dt}$$

(which is odd, one moment it is dv/dt now it seems to be dv/dx) ...

No.

The acceleration a does NOT seem to be dv/dx.

It would seem that a is dv/dx if we have a = dv/dx and we do not have that because we have instead that

a = (dv/dx)(dx/dt).

 

[ehild]

Assume that the acceleration is function of the position x: a=a(x(t)), a composite function of t. You get the derivative with respect to t if you derive with respect to x first then derive x with respect to t.

da(x(t))/dt=(da/dx)*(dx/dt).

ehild

 

[oli4]

Hi help1please,
you just messed up your cross multiplication
dv/dx * dx/dt is not dvdt/dxdx but dvdx/dtdx as you can see, you could just as well simplify by the 'artificially' injected dx/dx (=1) and get back to a=dv/dt

Cheers...

 

[CAF123]

Acceleration is (dv/dt) but by application of the chain rule this can be written as a = (dv/dx)(dx/dt) because we are effectively cancelling out the dx top and bottom.
(I say effectively because dx is really a differential operator)

In your last eqn, (dx/dt) = v so we get a = v (dv/dx)

This sort of chain rule application can be important if you want to find the velocity of a particle as a function of position, acceleration as function of velocity etc.., although this requires some extra mathematics.(Separable diff eqns)

 

[CWatters]

You probably understand it by now but..

a = dv/dt

Multiply top and bottom by dx

= (dv/dt)(dx/dx)

rearrange

= (dv/dx)(dx/dt)

 

[Saitama]

Fine. Then I am told

$$a= \frac{dv}{dx}\frac{dx}{dt}$$

(which is odd, one moment it is dv/dt now it seems to be dv/dx)

That's simply some manipulation (multiplying numerator and denominator by same thing), scratch out dx from numerator and denominator, you get the same expression as before.

 

[help1please]

Hi help1please,
you just messed up your cross multiplication
dv/dx * dx/dt is not dvdt/dxdx but dvdx/dtdx as you can see, you could just as well simplify by the 'artificially' injected dx/dx (=1) and get back to a=dv/dt

Cheers...

I didn't mess up anything, it was work I was following and noticed something was wrong, that's all. But thanks!