In a flow, velocity vector is given by \(\vec{V} = -y^2\hat{i} - 6x\hat{j}\) . The equation of streamline passing through the point (1,1) will be:

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SJVN ET Mechanical 2019 Official Paper

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Answer 1

Concept:

For a 2D flow, the equation of a streamline is given by:

\( \frac{dx}{u} = \frac{dy}{v} \)

Given:

\( u = -y^2, \quad v = -6x \)

So,

\( \frac{dx}{-y^2} = \frac{dy}{-6x} \Rightarrow \frac{dx}{y^2} = \frac{dy}{6x} \)

Cross-multiplying:

\( 6x \, dx = y^2 \, dy \)

Integrating both sides:

\( \int 6x \, dx = \int y^2 \, dy \Rightarrow 3x^2 = \frac{y^3}{3} + C \)

Multiply by 3:

\( 9x^2 - y^3 = C \)

Using point (1,1):

\( 9(1)^2 - (1)^3 = 8 \Rightarrow C = 8 \)