Correct option is (b) \(\frac{7}{\sqrt{5}} \)
The best explanation: Here \(\overrightarrow{AB} = (2-3) i ̂ + (3-2) j ̂ = – i ̂ + j ̂ \)
Directional derivative of φ(x,y) toward \(\overrightarrow{AB} \) is
\(∇ φ . \widehat{(AB)} = (\hat{i} \frac{∂φ}{∂x}+ \hat{j}\frac{∂φ}{∂y}) .(\frac{- i ̂ + j ̂ }{\sqrt{2}}) = 3\sqrt{2}\)
\(– \frac{∂φ}{∂x} + \frac{∂φ}{∂y} = 6 …..(i) \)
Directional derivative at A(3,2) towards C(1,0) is
\(∇ φ . \widehat{(AC)} = (i ̂ \frac{∂φ}{∂x}+ j ̂ \frac{∂φ}{∂y}) . \frac{-2 i ̂-2j ̂ }{\sqrt{8}} = \sqrt{8} \)
\(– 2 \frac{∂φ}{∂x} -2 \frac{∂φ}{∂y} = 8 \,or\, \frac{∂φ}{∂x} + \frac{∂φ}{∂y} = 4 …..(ii) \)
From (i) & (ii), \(∇ φ = -5\hat{i} + \hat{j} \)
Hence directional derivative at A(3,2) towards D(2,4) is
\(∇ φ . \widehat{AD} = (-5\hat{i} + \hat{j}) . \left(\frac{-5\hat{i} + \hat{j}}{\sqrt{5}}\right) = \frac{7}{\sqrt{5}}.\)