# Find the equation of the streamline passing through the point $$(5,3)$$ at t =3 seconds.

A velocity field is given by $$V=3yt\hat{i}+5x\hat{j}$$. Find the equation of the streamline passing through the point $$(5,3)$$ at $$t =3$$. Units of $$x,y$$ are in meters and time-$$t$$ is in seconds.

Asked on 5th May 2021 in

A streamline in a fluid flow

Velocity field is $$V=\left ( 3yt\hat{i}+5x\hat{j} \right )$$
In a streamline flow, velocity is tangent to the flow. $$V$$ and $$dr$$ is in the same direction. Therefore,
$V\times dr=0$Since, two vectors are in the same direction, their cross product is $$0$$.
$V\times dr = 0$$\Rightarrow \left ( 3yt\hat{i}+5x\hat{j} \right )\times \left ( dx\hat{i}+dy\hat{j} \right )=0$
$\Rightarrow \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 3yt& 5x &0 \\ dx&dy & 0 \end{vmatrix}=0$
$\Rightarrow \hat{k}\left ( 3ytdy-5xdx \right )=0$
At, time $$t = 3$$
$\Rightarrow 9y\left ( 3 \right )dy -5xdx=0$$\Rightarrow 9ydy -5xdx=0$$\Rightarrow 9ydy=5xdx$
On integrating
$\Rightarrow \int 9ydy=\int 5xdx$
$\Rightarrow 9\int ydy=5\int xdx$
$\Rightarrow 9\frac{y^{2}}{2}+c_{1}=\frac{5x^{2}}{2}+c_{2}$
$\Rightarrow \frac{9y^{2}+2c_{1}}{2}=\frac{5x^{2}+2c_{2}}{2}$
$\Rightarrow 9y^{2}+2c_{1}=5x^{2}+2c_{2}$
$\Rightarrow 9y^{2}-5x^{2}=2c_{2}-2c_{1}$
$\Rightarrow 9y^{2}-5x^{2}=C$

At, point $$(5,3)$$, since,
$9y^{2}-5x^{2}=C$$\Rightarrow C= 9\left ( 3 \right )^{2}-5\left ( 5 \right )^{2}$$\Rightarrow C=\left ( 9\times 9 \right )-\left ( 5\times 25 \right )$$\Rightarrow C=81-125=-44$Therefore, the equation of the streamline is,
$9y^{2}-5x^{2}=-44$$\Rightarrow 9y^{2}-5x^{2}+44=0$

Answered on 13th May 2021.