Correct Answer - Option 3 :

\(2\sqrt 5 \)
**Explanation:**

\(\vec V = \left( { - {x^2} + 3y} \right)\hat i + 2xy\hat j\)

u = - x^{2} + 3y; v = 2xy

\({a_x} = u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( { - 2x} \right) + 2xy\left( 3 \right)\)

a_{x} = [-(1)^{2 }+ 3 (-1)] (-2(1))+6(1)(-1)

a_{1 }= 8 – 6

a_{x} = 2

\({a_y} = u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( {2y} \right) + \left( {2xy} \right)\left( {2x} \right)\)

\({a_y} = \left[ { - {{\left( 1 \right)}^2} + 3\left( { - 1} \right)} \right]\left[ {2\left( { - 1} \right)\left] + \right[4{{\left( 1 \right)}^2}\left( { - 1} \right)} \right]\)

a_{y} = 8 – 4

a_{y} = 4

\(a = \sqrt {a_x^2 + a_y^2\;} = \sqrt {{2^2} + {4^2}} = \sqrt {20} \)

\(= 2\sqrt 5 \)