Question #296971

For the velocity potential function, ϕ = x ^ 2 - y ^ 2 find the velocity components at the point (4, 5)


Expert's answer

ϕ=x2−y2\phi=x^2-y^2

Point (4,5)

We know that

ux=dϕdt=2xu_x=\frac{d\phi}{dt}=2x

uy=dϕdt=−2yu_y=\frac{d\phi}{dt}=-2y

ux=2×2=4u_x=2\times2=4

uy=−2×5=−10u_y=-2\times5=-10

Velocity


v=ux2+uy2=42+(−10)2=16+100=116v=\sqrt{u_x^2+u_y^2}=\sqrt{4^2+(-10)^2}=\sqrt{16+100}=\sqrt{116}

∣v∣=10.77|v|=10.77


LATEST TUTORIALS
APPROVED BY CLIENTS