Given the velocity potential by f(x,y) of a flow, find the velocity V=?f of the field and its value V(P) at P. Where f(x,y)=3x^2-8x+y^3,P:(-2,3).
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Given f(xy) = 3√(r^2 + y^2), we can rewrite this as f(xy) = 3(r^2 + y^2)^(1/2). Taking the partial derivative with respect to x, we get: ∂f/∂x = 0 Taking the partial derivative with respect to y, we get: ∂f/∂y = 3y/(r^2 + y^2)^(1/2) Show more…
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