Question

Find the directional derivative of φ = xy^2 + yz^3 at (1, -1, 1), towards the point (2, 1, -1).

(a) \( \frac{5}{3} \)

(b) \( \frac{-5}{3} \)

(c) \( \frac{7}{3} \)

(d) \( \frac{1}{3} \)

This question was addressed to me during an interview.

The above asked question is from Directional Derivative topic in section Vector Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) \( \frac{5}{3} \)

The best explanation: \(\frac{dφ}{dx}  = y^2,   \frac{dφ}{dy} =2xy+z^3, \frac{dφ}{dz} = 3yz^3 \)

\(∇ φ = y^2 \hat{i} + (2xy+z^3)  \hat{j} + 3yz^3\hat{k} ̂\)

\([∇ φ]_{(1,-1,1)}  = \hat{i} – \hat{j} – 3\hat{k} ̂\)

Now \(a ̅ \) is along the line joining(1,-1,1) and (2,1,-1)

\(a ̅ =(2-1) \hat{i} + (1+1) \hat{j} + (-1-1) \hat{k} =  \hat{i} + 2\hat{j} – 2\hat{k} \)

\(a ̂ = \frac{(i ̂  +2 j ̂  -2\hat{k})}{\sqrt{(1+4+4)}}  = \frac{1}{3} (i ̂ +2 j ̂ -2\hat{k})\)

∴ Directional derivative = \(∇ φ. a ̂ = (i ̂ – j ̂ -3\hat{k}) . \frac{1}{3} (i ̂ +2 j ̂ -2\hat{k})\)

\(= \frac{1}{3} (1-2+6) = \frac{5}{3} .\)