# Find the equation of the streamline which is passing through the point $$\left ( 2,6 \right )$$.

A velocity field is given by $$u=\frac{y}{x^{3}+y}$$ and $$v=\frac{-x}{x^{3}+y}$$. Find the equation of the streamline which is passing through the point $$\left ( 2,6 \right )$$.

Asked on 15th April 2021 in
Differntial equation of a streamline in a 2-D flow is $\frac{dy}{dx}=\frac{v}{u}$$\Rightarrow \frac{dy}{dx}=\frac{\frac{-x}{x^{3}+y}}{\frac{y}{x^{3}+y}}$$\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$$\Rightarrow ydy=-xdx$ On integrating
$\int ydy = -\int xdx$$\Rightarrow \frac{y^{2}}{2}+c_{1}=\frac{-x^{2}}{2}+c_{2}$$\Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}=c_{2}-c_{1}$$\Rightarrow x^{2}+y^{2} = 2\left ( c_{2}-c_{1} \right )$$\Rightarrow x^{2}+y^{2}=C$ For a streamline passing through the point $$(2,6)$$,$2^{2}+6^{2}=C$$\Rightarrow C=40$ Therefore, the equation of the streamline passing through the point $$(2,6)$$ is$x^{2}+y^{2}=40$This streamline represents a circle with its center at the origin and radius $$\sqrt{40}$$ units.