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zryv0621

Mathematics

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The velocity of a wave of length L, in deep water is v=Ksqrt. L/C C/L where K and C are known positive constants. What is the length of the wave that gives the minimum velocity?

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Explanation & Answer

Since v = K + (L/C + C/L)^(1/2), differentiating (with respect to L) yields
dv/dL = 0 + (1/2)(L/C + C/L)^(-1/2) * (1/C - C/L^2)
= (1/C - C/L^2) / [2(L/C + C/L)^(1/2)].

Setting dv/dL = 0 (specifically, its numerator), we obtain

1/C - C/L^2 = 0.
  C/L^2 = 1/C
  L^2 = C^2
  L = C, since L > 0.

Note that for 0 < L < C, we have that dv/dL < 0 (try L = C/2 for instance),
and for L > C, we have that dv/dL > 0 (try L = 2C for instance).

Since we have a local minimum at L = C by the First Derivative Test.
Moreover, this yields the (global) minimum, being the only critical point (with L > 0).


Anonymous
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