A function $$ f $$ and a point $$ P $$ are given. Let $$ \theta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ \mathbf{e} $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed. Use a comma to separate answers as needed.) What angles are associated with the direction of zero change? $$ \frac{\pi}{3}, \frac{4 \pi}{3} $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed. Use a comma to separate answers as needed.) c. Write the directional derivative at P as a function of $$ \theta $$; call this function $$ \mathrm{g}(\theta) $$. $$ g(\theta)=\frac{2 \sqrt{3} \cos \theta-2 \sin \theta}{\sqrt{11}} $$ (Type an exact answer, using radicals as needed.) d. Find the value of $$ \theta $$ that maximizes $$ g(\theta) $$ and find the maximum value. What value of $$ \theta $$ maximizes $$ g(\theta) $$ ? $$ \theta=\square $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed.)

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We start with the given function and point:
$$
f(x,y)=\sqrt{3+2x^2+2y^2},\quad P\bigl(\sqrt{3},-1\bigr).
$$
At the point $$ P $$, the gradient is computed by differentiating $$ f(x,y) $$. Write
$$
f(x,y)=\sqrt{3+2x^2+2y^2}=(3+2x^2+2y^2)^{\frac{1}{2}},
$$
so by the chain rule,
$$
\nabla f(x,y)=\frac{1}{2}(3+2x^2+2y^2)^{-\frac{1}{2}} \cdot (4x,4y)=\frac{(2x,2y)}{\sqrt{3+2x^2+2y^2}}.
$$

At $$ P $$ we have:
$$
f(P)=\sqrt{3+2(\sqrt{3})^2+2(-1)^2}=\sqrt{3+2\cdot3+2}=\sqrt{11},
$$
and thus
$$
\nabla f(P)=\frac{(2\sqrt{3},-2)}{\sqrt{11}}.
$$

The directional derivative in the direction given by the unit vector
$$
\mathbf{u}=(\cos\theta,\sin\theta)
$$
is given by
$$
g(\theta)=D_{\mathbf{u}}f(P)=\nabla f(P)\cdot\mathbf{u}=\frac{2\sqrt{3}\cos\theta-2\sin\theta}{\sqrt{11}}.
$$

To find the value of $$\theta$$ that maximizes $$g(\theta)$$, recall that for any expression of the form
$$
A\cos\theta+B\sin\theta,
$$
the maximum value is attained when the unit vector $$(\cos\theta,\sin\theta)$$ points in the direction of $$(A,B)$$. In our case, we compare with:
$$
A=2\sqrt{3}\quad \text{and}\quad B=-2.
$$

Thus the maximum is reached when
$$
(\cos\theta, \sin\theta)=\frac{1}{\sqrt{(2\sqrt{3})^2+(-2)^2}}(2\sqrt{3},-2).
$$
We compute the magnitude:
$$
\sqrt{(2\sqrt{3})^2+(-2)^2}=\sqrt{12+4}=\sqrt{16}=4.
$$
The direction of the vector $$(2\sqrt{3}, -2)$$ is given by
$$
\theta=\arctan\left(\frac{-2}{2\sqrt{3}}\right)=\arctan\left(-\frac{1}{\sqrt{3}}\right).
$$

Since $$ 2\sqrt{3}>0 $$ and $$-2<0$$ the vector is in the fourth quadrant. The basic angle $$ \alpha $$ satisfying
$$
\tan\alpha=\frac{1}{\sqrt{3}}
$$
is
$$
\alpha=\frac{\pi}{6}.
$$
Being in the fourth quadrant we have
$$
\theta=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}.
$$

Thus, the value of $$\theta$$ that maximizes $$g(\theta)$$ is:
$$
\theta=\frac{11\pi}{6}.
$$
The angle that maximizes $$ g(\theta) $$ is $$ \frac{11\pi}{6} $$.

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