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In the book "Sea Loads on Ships and Offshore Structures" by Faltinsen, page 60 there is an identity:

$$ \sin(\omega t - kR \cos(\theta)) \cos(\theta) d\theta $$

They claim that under long wavelengths (much larger than the radius $R$, so $kR$ is small), that we can approximate the above by:

$$ \cos(\theta) d\theta \sin(\omega t) - kR \cos^2(\theta) d\theta \cos(\omega t) $$

I see how - if you set $kR$ zero - you get:

$$ \sin(\omega t)\cos(\theta) d\theta $$

And if you then set $\omega t$ zero you get (via the sine near zero approximation):

$$\lim_{x\rightarrow0}\sin(x) = x$$

$$ \Rightarrow \sin(-kR \cos(\theta)) \cos(\theta) d\theta = -kR \cos^2(\theta) d\theta $$

But I'm confused as to what makes it valid to take compose these approximations together: what is it that makes them linear? I can picture a unit circle with the sine function, but won't the linear approximation be invalid near angles of $n\pi$?

  • Composing two linear approximations in sequence is exactly the meaning of the chain rule. – Ian Sep 30 '16 at 19:31
  • @Ian are you implying that we can apply the chain rule on this equation? –  Sep 30 '16 at 19:35

1 Answers1

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$$f(x+h)\approx f(x)+f'(x)h.$$

In your case $f(x)=\sin(x)$, $x=\omega t$ and $h=-kR\cos(\omega)$.

Alex R.
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  • That makes a lot of sense! But, $f'(x)$ would be derived with respect to x, so that should give just $\cos(\omega t - kR\cos(\theta)) \cdot 1$, or is it derived with respect to another variable? –  Sep 30 '16 at 19:46
  • To avoid confusion, write $f(z)=\sin(z)$ and calculate $f'(z)$, then plug in $z=x$. – Alex R. Sep 30 '16 at 19:48
  • Oh I was just looking at it wrongly. Thanks! –  Sep 30 '16 at 19:52