Question #296971

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


1
Expert's answer
2022-02-14T14:33:22-0500

ϕ=x2y2\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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS