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? $$ \frac{\partial \rho}{\partial y}=x $$ (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? $$ \rho^{\prime}(t)=16 $$ cos $$ 2 t $$ (Type an expression using $$ t $$ as the variable.) b. At what point(s) on the edge of the plate is the density a maximum? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. O. AThe density on the edgetyf the plate has a maximum value at the point(s) $$ \square $$. (Type an ordered pair. Type an exact answer. Use a comma to separate answers as needed.) B. The density on the edge of the plate has no maximum value.

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**Step 1. Compute the derivative of the density with respect to $$ t $$.**

The density is given by  
$$
\rho(x,y)= 4 + xy,
$$  
and its partial derivatives are  
$$
\frac{\partial \rho}{\partial x}= y,\quad \frac{\partial \rho}{\partial y}= x.
$$

On the edge of the plate,  
$$
x= 4\cos t,\quad y= 4\sin t.
$$

Differentiate these with respect to $$ t $$:
$$
\frac{dx}{dt}= -4\sin t,\quad \frac{dy}{dt}= 4\cos t.
$$

The rate of change of the density is given by the chain rule:
$$
\frac{d\rho}{dt}=\frac{\partial \rho}{\partial x}\frac{dx}{dt}+\frac{\partial \rho}{\partial y}\frac{dy}{dt}.
$$
Substitute the values:
$$
\frac{d\rho}{dt}= y(-4\sin t)+x(4\cos t).
$$

Replace $$ x $$ and $$ y $$ by their parametric expressions:
$$
\frac{d\rho}{dt}= (4\sin t)(-4\sin t)+(4\cos t)(4\cos t)=-16\sin^2 t+16\cos^2 t.
$$

Factor the common factor:
$$
\frac{d\rho}{dt}= 16(\cos^2 t-\sin^2 t).
$$

Recall the double-angle identity:
$$
\cos^2 t-\sin^2 t=\cos 2t.
$$

Thus, the derivative is:
$$
\rho'(t)= 16\cos 2t.
$$

**Step 2. Find the points on the edge where the density is maximum.**

To locate the maximum density on the edge, substitute the parametric equations into the density:
$$
\rho(t)= 4+xy= 4+ (4\cos t)(4\sin t)=4+ 16\cos t\sin t.
$$

Use the identity $$ 2\cos t\sin t = \sin 2t $$ to obtain:
$$
\rho(t)= 4+ 8\sin 2t.
$$

The maximum of $$ \rho(t) $$ occurs when $$ \sin 2t $$ is maximized. Since
$$
\sin 2t \le 1,
$$
the maximum occurs when
$$
\sin 2t=1.
$$

Solve for $$ t $$:
$$
2t= \frac{\pi}{2}+2\pi k\quad \Rightarrow \quad t=\frac{\pi}{4}+\pi k,\quad k\in\mathbb{Z}.
$$

Since $$ 0\le t\le 2\pi $$, the relevant values are:
$$
t=\frac{\pi}{4}\quad \text{and} \quad t=\frac{5\pi}{4}.
$$

Now compute the corresponding points on the edge.

For $$ t=\frac{\pi}{4} $$:
$$
x= 4\cos\frac{\pi}{4}= 4\cdot\frac{\sqrt{2}}{2}=2\sqrt{2},\quad y= 4\sin\frac{\pi}{4}=2\sqrt{2}.
$$
The point is $$ \left(2\sqrt{2},\,2\sqrt{2}\right) $$.

For $$ t=\frac{5\pi}{4} $$:
$$
x= 4\cos\frac{5\pi}{4}= 4\left(-\frac{\sqrt{2}}{2}\right)=-2\sqrt{2},\quad y= 4\sin\frac{5\pi}{4}=-2\sqrt{2}.
$$
The point is $$ \left(-2\sqrt{2},\,-2\sqrt{2}\right) $$.

**Final Answers:**

a. The rate of change of the density with respect to $$ t $$ on the edge is  
$$
\rho'(t)=16\cos 2t.
$$

b. The density on the edge of the plate has a maximum value at the points  
$$
\left(2\sqrt{2},\,2\sqrt{2}\right) \quad \text{and} \quad \left(-2\sqrt{2},\,-2\sqrt{2}\right).
$$
a. The rate of change of the density with respect to $$ t $$ on the edge is $$ \rho'(t)=16\cos 2t $$. b. The density on the edge of the plate has a maximum value at the points $$ \left(2\sqrt{2},\,2\sqrt{2}\right) $$ and $$ \left(-2\sqrt{2},\,-2\sqrt{2}\right) $$.

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