# 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 $$\left ( 2,6 \right )$$.

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A streamline is a path traced by imaginary particles moving with the flow. At every point along the path velocity is tangent to the path. Bernoulli’s equation which relates pressure and velocity can be used along the streamline.

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.

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