Question

If \(v=3xy\), the magnitude of the velocity vector at \((2,-2)\) is ____

• A. \(4\sqrt{2}\)
• B. \(6\sqrt{2}\)
• C. \(12\)
• D. \(0\)

Hint:

For a velocity potential \(\phi\), \[ \boxed{ u=\frac{\partial\phi}{\partial x}, \qquad v=\frac{\partial\phi}{\partial y} } \] Velocity magnitude: \[ \boxed{ V=\sqrt{u^2+v^2}. } \]

Answer 1

The Correct Option is B

Step 1: Treat the given function as a potential and take its gradient.
With $\phi = 3xy$, the velocity field is the gradient of this potential, \[ \vec{V} = \left(\frac{\partial \phi}{\partial x},\ \frac{\partial \phi}{\partial y}\right) = (3y,\ 3x) \]
Step 2: Plug in the point $(2,-2)$. \[ u = 3(-2) = -6, \qquad v = 3(2) = 6 \]
Step 3: Find the magnitude as a vector length. \[ |\vec{V}| = \sqrt{u^2+v^2} = \sqrt{(-6)^2+6^2} = \sqrt{72} = 6\sqrt{2} \]
which is the same as treating $(-6,6)$ as a right triangle with equal legs of length 6 and taking its hypotenuse.
\[ \boxed{6\sqrt{2}} \]