Show that |x(t)|d/dt*|x(t)| = x(t)x`(t)

[Smazmbazm]

Homework Statement

Show that |x(t)| d/dt |x(t)| = x(t) \cdot x`(t)

The Attempt at a Solution

I don't know what they want to see here but my go at this is to take x(t) = 2t. Differentiating this gives x`(t) = 2.

Taking the normal of both |x(t)| and |x`(t)| gives 2t and 2, respectively.

So 2t * 2 = 4t.

Then just do the same with x(t) and x`(t) to get 4t again?

I don't know if this will suffice as a proper answer

 

[Bacle2]

I assume they expect you to show the result for any path x(t), not just x(t)=2t .

I think working with the cases (for the absolute value) ; when is x(t)>0 , when is it <0 ,=0 ,

etc. will be helpful.

 

[Infrared]

Hint: Use the fact that \frac{d}{dt} |t|=\frac{t}{|t|} with the chain rule to compute your derivative in the two cases x(t) &gt; 0 and x(t) &lt; 0 . You have to consider the case x(t)=0 carefully.