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 \( t \) on the edge of the plate, b. At what point(s) on the edge of the plate is the density a maximum? First, find some intermediate derivatives. \( \frac{\partial \rho}{\partial x}=y \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\mathrm{dx}}{\mathrm{dt}}=\square \) (Type an expression using \( t \) as the variable.)

To find the rate of change of the density with respect to \( t \) on the edge of the plate, we need to combine the chain rule with our density function, \( \rho(x, y) = 4 + xy \). 1. Express the density in terms of \( t \): - Since \( x = 4 \cos t \) and \( y = 4 \sin t \): \[ \rho(t) = 4 + (4 \cos t)(4 \sin t) = 4 + 16 \cos t \sin t = 4 + 8 \sin(2t) \] 2. Differentiate \( \rho(t) \) with respect to \( t \): \[ \frac{d\rho}{dt} = \frac{d}{dt}(4 + 8 \sin(2t)) = 16 \cos(2t) \] Now for the partial derivatives: - For the given \( \rho(x, y) \), we already have: \[ \frac{\partial \rho}{\partial x} = y \] - Since \( y = 4 \sin t \), this gives: \[ \frac{\partial \rho}{\partial x} = 4 \sin t \] Next, we compute \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = \frac{d}{dt}(4 \cos t) = -4 \sin t \] Now we have all the required intermediate derivatives: 1. \( \frac{\partial \rho}{\partial x} = 4 \sin t \) 2. \( \frac{dx}{dt} = -4 \sin t \)