Question Bank • QB_EM_ADV_029 NAT • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Partial derivatives

Question Statement:

If $z = f(x, y)$ where $x = r \cos\theta$ and $y = r \sin\theta$, given: $\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 = k \left[ \left( \frac{\partial z}{\partial r} \right)^2 + \frac{1}{r^2} \left( \frac{\partial z}{\partial \theta} \right)^2 \right]$ The value of $k$ is ________ (answer in integer).

Answer Key:

1

Step-by-Step Solution & Explanation

By the multivariable chain rule: $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \cos\theta + \frac{\partial z}{\partial y} \sin\theta$ $\frac{\partial z}{\partial \theta} = -r \frac{\partial z}{\partial x} \sin\theta + r \frac{\partial z}{\partial y} \cos\theta \implies \frac{1}{r} \frac{\partial z}{\partial \theta} = -\frac{\partial z}{\partial x} \sin\theta + \frac{\partial z}{\partial y} \cos\theta$ Squaring both expressions and adding: $\left( \frac{\partial z}{\partial r} \right)^2 + \frac{1}{r^2} \left( \frac{\partial z}{\partial \theta} \right)^2 = \left( \frac{\partial z}{\partial x} \right)^2 (\cos^2\theta + \sin^2\theta) + \left( \frac{\partial z}{\partial y} \right)^2 (\sin^2\theta + \cos^2\theta) + 0$ $= \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2$ Comparing with the given equation yields $k = 1$.
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