The correct choice is (a) \(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{dx}\)
The explanation: For an airfoil with x – component of velocity as V∞ + u^‘ and y – component of the velocity as v^‘, the surface boundary condition is
\(\frac {df}{dx} = \frac {v^{‘}}{V_∞ + u^{‘}}\) = tanθ
Since it is a thin airfoil, the perturbation vector u^‘ is very small in comparison to the freestream velocity V∞, resulting in \(\frac {df}{dx} = \frac {v^{‘}}{V_∞}\) = θ (Where tanθ ~ θ for small angles). Expressing the perturbation v^‘ in terms of velocity potential we get
v^‘ = \(\frac {∂ϕ}{∂x}\)
Substituting this in the above equation:
\(\frac {df}{dx} = \frac {\frac {∂ϕ}{∂x}}{V_∞}\) = θ
\(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{dx}\)