Why Can't the Power Rule Be Used for the Derivative of Velocity Squared?

[sin2beta]

Homework Statement

This really isn't a specific homework problem. It is just something that I've never actually known the reason for. An example is in the derivation of:

[tex]\frac{dT}{dt} = F\dot v [\tex]

In order to arrive at it, I replace T with [tex]1/2mv^2[\tex] and assume m is constant and then I have to do the product rule on v^2 and I'm never sure why I can't use the power rule. Any explanation is appreciated.

On a side note, this makes sense to me when I substitute momentum in for one of the velocities and then do the product rule. I just don't understand why I can't leave it as velocity and use the power rule. Thanks!

 

[CompuChip]

Well, v = v(t) is a function of t, and you are differentiating with respect to t.
So when you differentiate v(t)^2 you use the chain rule with u = v(t) and get
$$\frac{d(v(t))^2}{dt} = \frac{du}{dv} \frac{dv(t)}{dt} = (2v) (v'(t))$$
where du/dv = d(v^2)/dv = 2 v, just like differentiating
$$(x^2 + 6)^2$$
with respect to x first gives you 2(x^2 + 6) by the power rule and then another factor of 2x because of the chain rule.

So, using v'(t) = a(t), and F = m v(t)
$$\frac{dT}{dt} = \frac12 m (2 v a) = m v a = F v$$
[If you want to treat v(t) as a vector, you need to be a little more careful as you will get some dot products]

 

[nlightner]

The reason why you cannot use the power rule when deriving the derivative of velocity squared is because velocity is a function of time, and therefore its value is changing over time. The power rule can only be used when the variable is independent of time. In this case, the velocity is dependent on time and therefore requires the use of the product rule. This is because the derivative of a product of two functions is not simply the product of the individual derivatives, but rather a more complex formula involving both functions and their derivatives.

In the equation you provided, the derivative of [tex]1/2mv^2[\tex] is [tex]mv\dot v[\tex], which is then further simplified to [tex]F\dot v[\tex] using the product rule. This is because the derivative of [tex]v^2[\tex] is [tex]2v[\tex], but since v is a function of time, we must also take into account the derivative of v, which is [tex]\dot v[\tex].

Substituting momentum in for one of the velocities does not change the fact that velocity is still a function of time, and therefore the product rule must still be used. I hope this explanation helps clarify the reasoning behind using the product rule when deriving the derivative of velocity squared.