Points: 0.56 of 1 Save A function $$ f $$ and a point $$ P $$ are given. Let $$ \theta $$ correspond to the direction of the directional derivative. Complete parts (a) through (e). $$ f(x, y)=16-3 x^{2}-2 y^{2}, P(2,3) $$ $$ \square $$ (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 angle(s) is/are associated with the direction of zero change? The angle(s) associated with the directions of zero change is/are $$ \frac{3 \pi}{4}, \frac{7 \pi}{4} $$. (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 $$ \mathrm{g}(\theta) $$. $$ g(\theta)=-12(\cos \theta+\sin \theta) $$ 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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UpStudy AI · Accepted Answer

We start by computing the gradient of  
$$
f(x,y)=16-3x^2-2y^2.
$$
The partial derivatives are  
$$
f_x(x,y)=-6x,\quad f_y(x,y)=-4y.
$$
At the point $$P(2,3)$$ this gives  
$$
\nabla f(2,3)=\langle -6(2),-4(3)\rangle=\langle -12,-12\rangle.
$$

The directional derivative in the direction corresponding to the angle $$\theta$$ is given by  
$$
g(\theta)=\nabla f(2,3)\cdot\langle\cos\theta,\sin\theta\rangle=-12\cos\theta-12\sin\theta=-12(\cos\theta+\sin\theta).
$$

To maximize $$g(\theta)$$ we write the sum $$\cos\theta+\sin\theta$$ in the form of a single cosine function. Note that  
$$
\cos\theta+\sin\theta=\sqrt{2}\cos\left(\theta-\frac{\pi}{4}\right).
$$
Thus,  
$$
g(\theta)=-12\sqrt{2}\cos\left(\theta-\frac{\pi}{4}\right).
$$

Since the factor $$-12\sqrt{2}$$ is negative, the directional derivative $$g(\theta)$$ is maximized when  
$$
\cos\left(\theta-\frac{\pi}{4}\right) \text{ is minimized.}
$$
The minimum value of the cosine function is $$-1$$. Therefore, we require  
$$
\cos\left(\theta-\frac{\pi}{4}\right)=-1.
$$

This occurs when  
$$
\theta-\frac{\pi}{4}=\pi \quad (\text{mod } 2\pi).
$$
Taking the principal solution between $$0$$ and $$2\pi$$,  
$$
\theta-\frac{\pi}{4}=\pi \quad\Longrightarrow\quad \theta=\frac{5\pi}{4}.
$$

The maximum value of $$g(\theta)$$ is then  
$$
g\left(\frac{5\pi}{4}\right)=-12\sqrt{2}\cos\left(\frac{5\pi}{4}-\frac{\pi}{4}\right)=-12\sqrt{2}\cos(\pi)=-12\sqrt{2}(-1)=12\sqrt{2}.
$$

Thus, the value of $$\theta$$ that maximizes the directional derivative is

$$
\theta=\frac{5\pi}{4}
$$
The value of $$ \theta $$ that maximizes the directional derivative is $$ \frac{5\pi}{4} $$.

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