Correct Answer - Option 3 :
\(2\sqrt 5 \)
Explanation:
\(\vec V = \left( { - {x^2} + 3y} \right)\hat i + 2xy\hat j\)
u = - x2 + 3y; v = 2xy
\({a_x} = u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( { - 2x} \right) + 2xy\left( 3 \right)\)
ax = [-(1)2 + 3 (-1)] (-2(1))+6(1)(-1)
a1 = 8 – 6
ax = 2
\({a_y} = u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( {2y} \right) + \left( {2xy} \right)\left( {2x} \right)\)
\({a_y} = \left[ { - {{\left( 1 \right)}^2} + 3\left( { - 1} \right)} \right]\left[ {2\left( { - 1} \right)\left] + \right[4{{\left( 1 \right)}^2}\left( { - 1} \right)} \right]\)
ay = 8 – 4
ay = 4
\(a = \sqrt {a_x^2 + a_y^2\;} = \sqrt {{2^2} + {4^2}} = \sqrt {20} \)
\(= 2\sqrt 5 \)