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? \( \frac{\partial \rho}{\partial x}=y \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=-4 \) sin \( t \) (Type an expression using \( t \) as the variable.) \( \frac{\partial \rho}{\partial y}=x \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d y}{d t}=4 \) cos \( t \) (Type an expression using \( t \) as the variable.) a. What is the rate of change of the density with respect to \( t \) on the edge of the plate? \( \rho^{\prime}(t)=\square \) (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 can use the chain rule. The density function is given as \( \rho(x, y) = 4 + xy \), and we need to express this in terms of \( t \) by substituting \( x(t) = 4 \cos t \) and \( y(t) = 4 \sin t \). Using the chain rule, we find: \[ \frac{d\rho}{dt} = \frac{\partial \rho}{\partial x} \frac{dx}{dt} + \frac{\partial \rho}{\partial y} \frac{dy}{dt} \] Given: \[ \frac{\partial \rho}{\partial x} = y \quad \text{and} \quad \frac{\partial \rho}{\partial y} = x \] Also, we have: \[ \frac{dx}{dt} = -4 \sin t \quad \text{and} \quad \frac{dy}{dt} = 4 \cos t \] Now substituting \( y \) and \( x \): \[ \rho^{\prime}(t) = y(-4 \sin t) + x(4 \cos t) \] Substituting \( y = 4 \sin t \) and \( x = 4 \cos t \): \[ \rho^{\prime}(t) = (4 \sin t)(-4 \sin t) + (4 \cos t)(4 \cos t) \] \[ = -16 \sin^2 t + 16 \cos^2 t \] \[ = 16(\cos^2 t - \sin^2 t) \] So the final expression is: \[ \rho^{\prime}(t) = 16(\cos^2 t - \sin^2 t) \]