(Vector Calculus)

A student asked me the following:
$$\frac{d}{dt}\left[\vec{R}\cdot\vec{R}\right]=\frac{d\vec{R}}{dt}\cdot\vec{R}+\vec{R}\cdot\frac{d\vec{R}}{dt}=2\vec{R}\cdot\frac{d\vec{R}}{dt}=2\vec{R}\cdot\vec{V}.$$However (from dot-product properties),
$$\frac{d}{dt}\left[\vec{R}\cdot\vec{R}\right]=\frac{d}{dt}\left[\left|\vec{R}\right|^2\right]=2\left|\vec{R}\right|\cdot\frac{d\vec{R}}{dt}=2\left|\vec{R}\right|\cdot\vec{V}.$$What's the catch here?

Question

(Vector Calculus)

A student asked me the following:
$$\frac{d}{dt}\left[\vec{R}\cdot\vec{R}\right]=\frac{d\vec{R}}{dt}\cdot\vec{R}+\vec{R}\cdot\frac{d\vec{R}}{dt}=2\vec{R}\cdot\frac{d\vec{R}}{dt}=2\vec{R}\cdot\vec{V}.$$However (from dot-product properties),
$$\frac{d}{dt}\left[\vec{R}\cdot\vec{R}\right]=\frac{d}{dt}\left[\left|\vec{R}\right|^2\right]=2\left|\vec{R}\right|\cdot\frac{d\vec{R}}{dt}=2\left|\vec{R}\right|\cdot\vec{V}.$$What's the catch here?

amistre64 · Answer

its prolly in the 2(R.R')

amistre64 · Answer

or not :)

jamesj · Answer

Note that in the second method it is
$$2 |R| \frac{d |R|}{dt}$$
Note also this isn't a dot product, just regular multiplication of real numbers or real-valued functions.

across · Answer

I was thinking perhaps I'm performing the chain rule incorrectly, as there's an absolute value, if I may, given these are vectors, in the expression.

anonymous · Answer

On the second line you can't have a scalar (|R|) dotted with a vector (V).

across · Answer

I know.

across · Answer

Apparently,
$$2\vec{R}\cdot\vec{V}=2\left | \vec{R} \right |\frac{d\left | \vec{R} \right |}{dt}?$$

jamesj · Answer

Let's do an example: uniform circular motion,  R = (cos t, sin t)

Then 2R.R' = 2(cos t, sin t).(-sin t, cos t) = 0

and 2|R| d|R|/dt = 2 times 1 times 0 = 0, as |R| = 1 for all t


So the two formulae are consistent.

jamesj · Answer

Yes.

jamesj · Answer

It makes sense physical this way.  Write R is polar coordinates (or spherical polars if you're in R^3)

Then the dot product tells you how much the velocity is in the direction of the radial displacement, as the spherical terms dot product to zero

phi · Answer

I'm thinking R and V are orthogonal?

jamesj · Answer

and then the other expression is exactly the radial displacement times the rate of change of the radial displacement.  So the two interpretations are the same thing.

jamesj · Answer

In the example above, of course there is no change in radial displacement, so both terms are zero.

phi · Answer

In the second equation, once you do R dot R don't you get a scalar, and the derivative is zero.

jamesj · Answer

no, because it's a function of t