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The density of a thin circular plate of radius 4 is given by \( \rho(x, y)=4+x y \). The edge of the plate is described by the parametric equations \( x=4 \cos t, y=4 \sin t \), for \( 0 \leq t \leq 2 \pi \). a. Find the rate of change of the density with respect to 1 on the edge of the plate. b. At what point(s) on the edge of the plate is the density a maximum? \( \frac{\partial \rho}{\partial x}=y \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=-4 \sin t \)

Ask by Brewer Stewart. in the United States
Mar 17,2025

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**a. Rate of Change of Density Along the Edge** The rate of change of the density \(\rho\) with respect to time \(t\) along the edge of the plate is: \[ \frac{d \rho}{d t} = 16 \cos 2t \] **b. Points of Maximum Density** The density reaches its maximum at the points: \[ (2\sqrt{2}, 2\sqrt{2}) \quad \text{and} \quad (-2\sqrt{2}, -2\sqrt{2}) \]

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To analyze the rate of change of the density along the edge of the plate, we note that \( \frac{d\rho}{dt} \) can be computed using the chain rule. First, we find \( \frac{d\rho}{dt} = \frac{\partial \rho}{\partial x} \frac{dx}{dt} + \frac{\partial \rho}{\partial y} \frac{dy}{dt} \). Since you've given \( \frac{\partial \rho}{\partial x} = y \) and the derivatives \( \frac{dx}{dt} = -4 \sin t \) and \( \frac{dy}{dt} = 4 \cos t \), we need \( \frac{\partial \rho}{\partial y} \). We calculate \( \frac{\partial \rho}{\partial y} = 4 + x \). Thus, plugging the values back we get: \[ \frac{d\rho}{dt} = y(-4 \sin t) + (4 + x)(4 \cos t). \] To find where the density is maximized, we can differentiate \( \rho(x, y) \) with respect to \( t \). Setting \( x = 4 \cos t \) and \( y = 4 \sin t \), we substitute into \( \rho \) to get: \[ \rho(4 \cos t, 4 \sin t) = 4 + 4 \cos t \cdot 4 \sin t = 4 + 16 \cos t \sin t. \] To find the maxima, we'd take the derivative with respect to \( t \) and set it to zero, solving it will yield critical points for \( t \) between \( 0 \) and \( 2\pi \). --- Did you know that the study of density in this context has historical significance in physics, particularly in understanding fluid mechanics and material strength? The calculus of variables like density has been pivotal from ancient Greek times, right through to present-day engineering applications! As for real-world applications, understanding density gradients is crucial in fields like meteorology, where variations in density in the atmosphere can lead to weather patterns and phenomena like storms or air pressure systems. It’s not just about plates; it’s a life-and-death matter for pilots and meteorologists!

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