UPSC Maths 2025 Paper 1 Q6c-i — Solution

10 marks · Section B

Question

Find the absolute value of the directional derivative of $\phi(x, y, z) = x^2 y^2 z^2$ at the point $(1, 1, -1)$ in the direction of the tangent to the curve $x = e^t$ , $y = 2\sin t + 1$ , $z = t - \cos t$ , at $t = 0$ .

Technique

The directional derivative is $\nabla\phi\cdot\hat{n}$ , where $\hat n$ is the unit tangent to the curve, $\hat n = \dfrac{\mathbf r'(t)}{|\mathbf r'(t)|}$ .

Solution

Gradient of $\phi$ . $\nabla\phi = \bigl(2xy^2z^2,\; 2x^2yz^2,\; 2x^2y^2z\bigr).$ At $(1,1,-1)$ (note $z^2=1$ , $z=-1$ ): $\nabla\phi\big|_{(1,1,-1)} = (2,\;2,\;-2).$

Tangent to the curve at $t=0$ . With $\mathbf r(t) = (e^t,\;2\sin t + 1,\;t-\cos t)$ , $\mathbf r'(t) = \bigl(e^t,\; 2\cos t,\; 1+\sin t\bigr),\qquad \mathbf r'(0) = (1,\;2,\;1).$ (Check the point: $\mathbf r(0) = (1,1,-1)$ , which is exactly the given point — consistent.)

Magnitude $|\mathbf r'(0)| = \sqrt{1^2+2^2+1^2} = \sqrt6$ , so the unit tangent is $\hat n = \frac{1}{\sqrt6}(1,\,2,\,1).$

Directional derivative. $D_{\hat n}\phi = \nabla\phi\cdot\hat n = \frac{(2)(1) + (2)(2) + (-2)(1)}{\sqrt6} = \frac{2 + 4 - 2}{\sqrt6} = \frac{4}{\sqrt6} = \frac{4}{\sqrt6}\cdot\frac{\sqrt6}{\sqrt6} = \frac{4\sqrt6}{6} = \frac{2\sqrt6}{3}.$

This is already positive; its absolute value is $\dfrac{2\sqrt6}{3} \approx 1.633$ .

Answer

$\boxed{\;\bigl|D_{\hat n}\phi\bigr| = \frac{4}{\sqrt6} = \frac{2\sqrt6}{3} \approx 1.633.\;}$