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This can be foound in N. L. White - Matroid Applications page 2.

Unidimensional case

Author considers a set of points $P = \{ p_i | i \in \mathbb{N} \}$. In this case points are to be considered on the real line, so it is really simple, no components, just scalars.

Author considers that two points, arbitrarily chosen in the set, say $p_i,p_j \in P$. Then he wants to evaluate their distance $|| p_i - p_j ||$. No problems till now. All the calculations are done in order to find a good relation and see that, if those points are connected using a rigid bar and they move with a certain speed then their distance as they move must not change. This is possible using derivates.

He also says that the distance and its square must not change. Makes sense of course.

Than he writes the following.

$\frac{d}{dt}[p_i(t)-p_j(t)]^2 = [p_i(t)-p_j(t)][p'_i(t)-p'_j(t)] = 0$

What's this, can you explain?

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If the square of the distance is constant, then the derivative of the square of the distance is zero. Then the author uses the product rule incorrectly; it should be $2 (p_i(t) - p_j(t))(p_i'(t) - p_j'(t)) = 0$. –  Qiaochu Yuan Feb 24 '12 at 1:28
    
ah ok just omitting... of course, sorry I am so specific and strict sometimes... thanks Yuan... post yuor answer I`ll set it as the correct one :) –  Andry Feb 24 '12 at 1:35

1 Answer 1

Note that $(p_{i}(t)-p_{j}(t))^2$ is constant in relation to time then it's derivative is zero, that is,

$$\frac{d}{dt}(p_{i}(t)-p_{j}(t))^2=0$$

Or evaluating the derivative with the chain rule

$$2(p_{i}(t)-p_{j}(t))(p_{i}'(t)-p_{j}'(t))=0$$

What is the same ( in the sense that one equation holds if and only if the other does) that

$$(p_{i}(t)-p_{j}(t))(p_{i}'(t)-p_{j}'(t))=0.$$

Note: Perhaps in the book there is a typing mistake in the first equality but what is important is that $(p_{i}(t)-p_{j}(t))(p_{i}'(t)-p_{j}'(t))=0$.

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