# Help understanding math on inclined plane example

1. Sep 8, 2009

### jimz

I'm trying to follow a very simple example example in a text (Marion/Thornton example 2.1) and I think my rusty calculus is tripping me up and I'm just being stupid.

I understand how to derive the acceleration down the plane:

$$\ddot{x}=g\sin\Theta$$

but next they say 'we can find the velocity of the block after it moves from rest a distance x0 down the plane by multiplying by 2x' and integrating.

$$2\dot{x}\ddot{x}=2\dot{x}g\sin\Theta$$

the next steps confuse me...

$$\frac{d}{dt}(\dot{x}^2)=2g\sin\Theta\frac{dx}{dt}$$

What happens to the 2 on the left? What am I doing wrong here?
$$2\dot{x}\ddot{x}=2\dot{x}\frac{d}{dt}\dot{x}=2\frac{d}{dt}(\dot{x}^2)$$

Next, the limits of integration are chosen:

$$\int_{0}^{v^2_0}d(\dot{x}^2)=2g\sin\Theta\int_{0}^{x_0}dx$$

I have no idea what is happening on the left... where do they pull v^2 0 from? What happens to time? What is going on here again.. last time I had calculus was years ago. Thanks.

2. Sep 8, 2009

### kyiydnlm

2VdV/dt = dV^2/dt

3. Sep 9, 2009

### Redbelly98

Staff Emeritus
They are using the chain rule from calculus.

du/dt = (du/dv) (dv/dt) , where u is a function of v -- i.e. u(v)

Letting u=v2, note that (du/dv) = 2v. So applying the chain rule:

d(v2)/dt = (du/dv) (dv/dt)
= (2v) (dv/dt)
= 2 v dv/dt​

.

As for the limits on the integral: note that "v2" is the variable that is being integrated. The limits must be in terms of that variable. And since v2 varies from 0 to vo2, those are the limits of integration.

Hope that helps. And welcome to Physics Forums!