Question

The velocity of a wave of length \(L\) in deep water is

\(v = K\sqrt {\frac{L}{C} + \frac{C}{L}} \)

Where \(K\) and \(C\) are known positive constants. What is the length of the wave that gives the minimum velocity?

Short answer

Thus, the velocity of the given wave is minimum in deep water when \(L = C\).

Step-by-step solution

Reduce the given function.

The function \(v\) is minimum when the quantity under the \(\sqrt {\frac{L}{C} + \frac{C}{L}} \) is minimum as the value of \(K\) is given to be a constant.

Let \(x = \frac{L}{C}\), then \(\frac{1}{x} = \frac{C}{L}\).

Thus, minimize \(f\left( x \right) = x + \frac{1}{x}\) in order to minimize the wave velocity \(v\).

Obtain minimum value of wave velocity.

Differentiate \(f\left( x \right)\) with respect to \(x\) gives,

\(\frac{{df}}{{dx}} = 1 - \frac{1}{{{x^2}}}\)

For critical points, \(\frac{{{\bf{df}}}}{{{\bf{dt}}}}{\bf{ = 0}}\) thus,

\(\begin{aligned}{l}1 - \frac{1}{{{x^2}}} = 0\\{x^2} = 1\\x = \pm 1\end{aligned}\)

Differentiate \(\frac{{df}}{{dx}}\) with respect to \(x\) gives,

\(\frac{{{d^2}f}}{{d{t^2}}} = \frac{2}{{{x^3}}} > 0\)

Thus, \(f\left( x \right)\) have minimum value.

At \(x = 1\) the value of the second derivative is as follows,

\(\begin{aligned}{c}\frac{{{d^2}f}}{{d{t^2}}} = \frac{2}{{{x^3}}}\\ = 2 > 0\end{aligned}\)

Thus, the value of \(f\left( x \right)\) is minimum when \(x = 1\) that is \(\frac{L}{C} = 1\).

Hence, the velocity of the given wave is minimum in deep water when \(L = C\).