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QB_EM_ADV_029
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).
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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