A function $$ f $$ and a point $$ P $$ are given. Let $$ heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ c. Write the directional derivative at $$ P $$ as a function of $$ heta $$; call this function $$ g( heta) $$. (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.) g( $$ heta)=\frac{2 \sqrt{3} \cos heta-2 \sin heta}{\sqrt{11}} $$ (Type an exact answer, using radicals as needed.) d. Find the value of $$ heta $$ that maximizes $$ g( heta) $$ and find the maximum value. What value of $$ heta $$ maximizes $$ g( heta) $$ ? $$ heta=\frac{11 \pi}{6} $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed.) What is the maximum value? $$ g( heta)=\square $$ (Type an exact answer, using radicals as needed.)

Question

A function $$ f $$ and a point $$ P $$ are given. Let $$ 	heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ c. Write the directional derivative at $$ P $$ as a function of $$ 	heta $$; call this function $$ g(	heta) $$. (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.) g( $$ 	heta)=\frac{2 \sqrt{3} \cos 	heta-2 \sin 	heta}{\sqrt{11}} $$ (Type an exact answer, using radicals as needed.) d. Find the value of $$ 	heta $$ that maximizes $$ g(	heta) $$ and find the maximum value. What value of $$ 	heta $$ maximizes $$ g(	heta) $$ ? $$ 	heta=\frac{11 \pi}{6} $$ (Type any angles in radians between 0 and $$ 2 \pi $$. Type an exact answer, using $$ \pi $$ as needed.) What is the maximum value? $$ g(	heta)=\square $$ (Type an exact answer, using radicals as needed.)

UpStudy AI · Accepted Answer

We first compute the gradient of  
$$
f(x,y)=\sqrt{3+2x^2+2y^2}.
$$

The partial derivatives are obtained via the chain rule. Write  
$$
f(x,y)=(3+2x^2+2y^2)^{1/2}.
$$

The partial with respect to $$x$$ is
$$
f_x(x,y)=\frac{1}{2}(3+2x^2+2y^2)^{-1/2}\cdot(4x)=\frac{2x}{\sqrt{3+2x^2+2y^2}},
$$
and with respect to $$y$$
$$
f_y(x,y)=\frac{1}{2}(3+2x^2+2y^2)^{-1/2}\cdot(4y)=\frac{2y}{\sqrt{3+2x^2+2y^2}}.
$$

At the point $$P(\sqrt{3},-1)$$, the value of the inner expression is
$$
3+2(\sqrt{3})^2+2(-1)^2=3+2\cdot 3+2\cdot 1=3+6+2=11.
$$
Thus, the gradient at $$P$$ is
$$
\nabla f(P)=\left(\frac{2\sqrt{3}}{\sqrt{11}},\, \frac{-2}{\sqrt{11}}\right).
$$

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

Now, the maximum value of the directional derivative is given by the magnitude of the gradient:
$$
\|\nabla f(P)\|=\sqrt{\left(\frac{2\sqrt{3}}{\sqrt{11}}\right)^2+\left(\frac{-2}{\sqrt{11}}\right)^2}=\sqrt{\frac{12+4}{11}}=\sqrt{\frac{16}{11}}=\frac{4}{\sqrt{11}}.
$$

The directional derivative achieves this maximum when the unit direction vector is in the direction of $$\nabla f(P)$$. The angle corresponding to the gradient is
$$
\theta=\arctan\left(\frac{-2}{2\sqrt{3}}\right)=\arctan\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}.
$$
Expressed in the interval $$[0, 2\pi)$$, we add $$2\pi$$ to obtain
$$
\theta=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}.
$$

Thus, the maximum value of $$g(\theta)$$ is
$$
\boxed{\frac{4}{\sqrt{11}}}.
$$
The directional derivative at point $$ P(\sqrt{3}, -1) $$ is given by: $$ g(\theta) = \frac{2\sqrt{3} \cos \theta - 2 \sin \theta}{\sqrt{11}} $$ The angle that maximizes $$ g(\theta) $$ is $$ \theta = \frac{11\pi}{6} $$, and the maximum value is $$ \frac{4}{\sqrt{11}} $$.

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