# Find the magnitude of velocity at point $$(2,3)$$ for a two-dimensional flow field whose stream function is given by $$\psi = 2x^{2}-y^{2}$$.

Find the magnitude of velocity at point $$(2,3)$$ for a two-dimensional flow field whose stream function is given by $$\psi = 2x^{2}-y^{2}$$.

Asked on 5th May 2021 in
The stream function for the flow field is $\psi =2x^{2}-y^{2}$Velocity component $$u$$ and $$v$$ of the two-dimensional flow field is $u=\frac{\partial \psi }{\partial y},v=-\frac{\partial \psi }{\partial x}$Therefore,$u=\frac{\partial \psi }{\partial y}=\frac{\partial }{\partial y}\left ( 2x^{2}-y^{2} \right ) =\frac{\partial }{\partial y}\left ( 2x^{2} \right )-\frac{\partial }{\partial y}\left ( y^{2} \right )=0-2y=-2y$$v=-\frac{\partial \psi }{\partial x} =-\frac{\partial }{\partial x}\left ( 2x^{2}-y^{2} \right ) =-\frac{\partial }{\partial x}\left ( 2x^{2} \right )+\frac{\partial }{\partial x}\left ( y^{2} \right )=-4x+0=-4x$ Therefore, at point $$(2,3)$$,$u=-2y=-2\left ( 3 \right )=-6\\v=-4x=-4\left ( 2 \right )=-8$Velocity, $$V=-6i-8j$$
Magnitude of velocity = $\sqrt{u^{2}+v^{2}}=\sqrt{\left ( -6 \right )^{2}+\left ( -8\right)^{2}}=\sqrt{36+64}=10$