A function $$ f $$ and a point $$ P $$ are given. Let $$ heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ b. Find the angles $$ heta $$ (with respect to the positive $$ x $$-axis) associated with the directions of maximum increase, maximum decrease, and zero change. What angles are associated with the direction of maximum increase? $$ \frac{11 \pi}{6} $$ (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 maximum decrease? $$ \frac{5 \pi}{6} $$ (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? $$ \square \quad \square $$ (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.)

Question

A function $$ f $$ and a point $$ P $$ are given. Let $$ 	heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ b. Find the angles $$ 	heta $$ (with respect to the positive $$ x $$-axis) associated with the directions of maximum increase, maximum decrease, and zero change. What angles are associated with the direction of maximum increase? $$ \frac{11 \pi}{6} $$ (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 maximum decrease? $$ \frac{5 \pi}{6} $$ (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? $$ \square \quad \square $$ (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.)

UpStudy AI · Accepted Answer

We begin by finding the gradient of 
$$
f(x,y)=\sqrt{3+2x^2+2y^2}.
$$
Write 
$$
g(x,y)=3+2x^2+2y^2,
$$
so that 
$$
f(x,y)=\sqrt{g(x,y)}.
$$
Then the gradient is given by
$$
\nabla f(x,y)=\frac{1}{2\sqrt{g(x,y)}}\nabla g(x,y).
$$
Since
$$
\nabla g(x,y)=\left(4x,\, 4y\right),
$$
we have
$$
\nabla f(x,y)=\left(\frac{4x}{2\sqrt{3+2x^2+2y^2}},\,\frac{4y}{2\sqrt{3+2x^2+2y^2}}\right)
=\left(\frac{2x}{\sqrt{3+2x^2+2y^2}},\,\frac{2y}{\sqrt{3+2x^2+2y^2}}\right).
$$

At the point $$P(\sqrt{3},-1)$$, we first compute the denominator:
$$
\sqrt{3+2(\sqrt{3})^2+2(-1)^2} = \sqrt{3+2(3)+2(1)} = \sqrt{11}.
$$
Thus,
$$
\nabla f(\sqrt{3},-1)=\left(\frac{2\sqrt{3}}{\sqrt{11}},\,\frac{-2}{\sqrt{11}}\right).
$$

The direction of the gradient (i.e. maximum increase) is given by the vector $$\left(\frac{2\sqrt{3}}{\sqrt{11}},\,\frac{-2}{\sqrt{11}}\right)$$. Notice that the $$x$$-component is positive and the $$y$$-component is negative, so this vector lies in the fourth quadrant. Its direction angle $$\theta$$ (from the positive $$x$$-axis) is such that
$$
\tan \theta=\frac{-2}{2\sqrt{3}}=-\frac{1}{\sqrt{3}},
$$
which corresponds to an angle of
$$
\theta = \frac{11\pi}{6}.
$$
This agrees with the given answer for maximum increase.

The direction for maximum decrease is found by taking the opposite of the gradient, so its angle is
$$
\frac{11\pi}{6}+\pi=\frac{11\pi}{6}+\frac{6\pi}{6}=\frac{17\pi}{6}.
$$
Reducing modulo $$2\pi$$ (since $$\frac{17\pi}{6}-\frac{12\pi}{6}=\frac{5\pi}{6}$$), we have the angle for maximum decrease as
$$
\frac{5\pi}{6}.
$$

For directions of zero change, the directional derivative must vanish. This happens when the direction (a unit vector) is perpendicular to the gradient. In terms of angles, if the gradient has an angle of $$\frac{11\pi}{6}$$, then the directions perpendicular to it are obtained by adding and subtracting $$\frac{\pi}{2}$$. That is, the angles are
$$
\frac{11\pi}{6}+\frac{\pi}{2} \quad \text{and} \quad \frac{11\pi}{6}-\frac{\pi}{2}.
$$
First, compute:
$$
\frac{11\pi}{6}+\frac{\pi}{2}=\frac{11\pi}{6}+\frac{3\pi}{6}=\frac{14\pi}{6}=\frac{7\pi}{3}.
$$
Since $$\frac{7\pi}{3}$$ exceeds $$2\pi$$ (because $$2\pi=\frac{6\pi}{3}$$), we subtract $$2\pi$$:
$$
\frac{7\pi}{3}-2\pi=\frac{7\pi}{3}-\frac{6\pi}{3}=\frac{\pi}{3}.
$$
Next, compute:
$$
\frac{11\pi}{6}-\frac{\pi}{2}=\frac{11\pi}{6}-\frac{3\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3}.
$$

Thus, the directions associated with zero change have angles
$$
\frac{\pi}{3} \quad \text{and} \quad \frac{4\pi}{3}.
$$
The angles associated with the direction of maximum increase are $$ \frac{11\pi}{6} $$. The angles for maximum decrease are $$ \frac{5\pi}{6} $$. The angles for zero change are $$ \frac{\pi}{3} $$ and $$ \frac{4\pi}{3} $$.

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