# 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.

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,

\[9y^{2}-5x^{2}=-44\]\[\Rightarrow 9y^{2}-5x^{2}+44=0\]