Question

The temperature of a point in space is given by T = x^2 + y^2 – z. An insect located at a point (1, 1, 2) desire to fly in such a direction such that it will get warm as soon as possible. In what direction it should move?

(a) \(\frac{-2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3} \)

(b) \(\frac{2\hat{i}}{3}+\frac{-2\hat{j}}{3}+\frac{-\hat{k}}{3} \)

(c) \(\frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{\hat{k}}{3} \)

(d) \(\frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3} \)

I got this question in quiz.

Question is from Directional Derivative topic in portion Vector Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (d) \(\frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3} \)

Easy explanation: Here the insect will get warm quickly if it moves normal to the given

Surface in space.

 As \(T = x^2+y^2-z \)

\(∴ ∇ T = (\frac{∂}{∂x}\hat{i} + \frac{∂}{∂x}\hat{j} + \frac{∂}{∂x}\hat{k})(x^2+y^2-z) \)

\( = 2x\hat{i} + 2y\hat{j} – \hat{k} \)

Now \( ∇ T |_{(1,1,2)} = 2\hat{i} + 2\hat{j} – \hat{k} \)

∴ required direction\( = \frac{∇ T}{|∇ T|} \)

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

\(= \frac{2\hat{i} +2\hat{j} – \hat{k}}{3} \)

\(= \frac{2\hat{i}}{3}+\frac{2\hat{j}}{3}+\frac{-\hat{k}}{3}. \)