Tuesday, February 5, 2013

second-order differentiation of speed with respect to wind speed using the chain rule

I was reminded of some issues in differential calculus when working on the previous blog post.

I had an equation for power as a function of speed and wind speed:

p(s, sw)

Since I assume that power is constant, it is an implicit equation: it defines contours in (s, sw) space of constant power. In the case of power, solving s as a function of sw may be a considerable challenge. But The Chain Rule comes to the rescue.

With the Chain Rule I can write:

dP = (∂ P / ∂ s) ds + (∂P / ∂ sw) dsw .

But I'm assuming constant P, so dP = 0. I can then write:

(∂ P / ∂ s) ds = −(∂ P / ∂ sw) dsw .

And with re-arranging terms I get what I want:

ds / dsw = −(∂ P / ∂ sw) / (∂ P / ∂ s) .

This was all fairly simple (although that minus sign confused me when I first saw it as an undergraduate). The tricky bit for me was the second derivative. But there's no reason for it to be complicated, either. If I write the first derivative as a function r ≣ ds / dsw :

d2s / d sw2 = d r / d sw .

So I apply the chain rule as before with the derivative function r:

dr = (∂ r / ∂ sw) dsw + (∂ r / ∂ s) ds .

But what I care about is dr / dsw, so I divide all terms by dsw:

d2s / d sw2 = dr / dsw = (∂ r / ∂ sw) + (∂ r / ∂ s) / r .

Note the substitution ds / dsw = 1/r. This equation is all in terms of r ≣ ds / dsw, and so I can evaluate it, since I already evaluated r. So I solve the first and second derivative of speed with respect to wind speed without ever actually evaluating a closed-form solution.

It took me several hours before I had clarity on this. Note the first term is a partial derivative with respect to the wind speed. The second term is a partial derivative with respect to speed multiplied by a full derivative of speed with respect to wind speed. The first term is due to, at a constant rider speed, the effect of wind speed changes on the rate of change of rider speed with respect to wind speed (this takes a bit of thought to wrap your head around). The second term is the effect of rider speed on the dependence of wind speed on rider speed multiplied by the effect on rider speed of wind speed. So it's adding two critical components, each ignoring the other, to come up with the net effect.

What I posted yesterday included the first term but omitted the second term. That's more complicated. I'll fix yesterday's post when I get it right.

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