Question Bank • QB_SUBEXP_S1_004 MSQ • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Total differentiation

Question Statement:

Let $z = f(x, y)$ be a differentiable function where $x = r \cos \theta$ and $y = r \sin \theta$. Which of the following expressions are mathematically CORRECT?

Options:

  • (A) $\left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial \theta}\right)^2 = \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2$
  • (B) $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \cos \theta + \frac{\partial z}{\partial y} \sin \theta$
  • (C) $\frac{\partial z}{\partial \theta} = -r \frac{\partial z}{\partial x} \sin \theta + r \frac{\partial z}{\partial y} \cos \theta$
  • (D) $\frac{\partial^2 z}{\partial r^2} + \frac{1}{r} \frac{\partial z}{\partial r} + \frac{1}{r^2} \frac{\partial^2 z}{\partial \theta^2} = \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}$

Answer Key:

A, B, C, D

Step-by-Step Solution & Explanation

By the chain rule of partial differentiation: $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} = z_x \cos \theta + z_y \sin \theta$. (Statement B True) $\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x}(-r \sin \theta) + \frac{\partial z}{\partial y}(r \cos \theta)$. (Statement C True) Sum of squares: $(\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2 = (z_x \cos \theta + z_y \sin \theta)^2 + (-z_x \sin \theta + z_y \cos \theta)^2 = z_x^2(\cos^2 \theta + \sin^2 \theta) + z_y^2(\sin^2 \theta + \cos^2 \theta) = z_x^2 + z_y^2$. (Statement A True) Transforming the Laplacian $\nabla^2 z$ to polar coordinates yields $\nabla^2 z = \frac{\partial^2 z}{\partial r^2} + \frac{1}{r}\frac{\partial z}{\partial r} + \frac{1}{r^2}\frac{\partial^2 z}{\partial \theta^2} = \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2}$. (Statement D True)
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