A function $$ f $$ and a point $$ P $$ are given. Let $$ heta $$ correspond to the direction of the directional derivative. Complete parts a. through $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ a. Find the gradient and evaluate it at $$ P $$. The gradient at $$ P $$ is $$ \left.\left\lvert\, \frac{2 \sqrt{3}}{\sqrt{11}} ight.,-\frac{2}{\sqrt{11}} ight) $$. (Type exact answers, using radicals as needed.) 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? $$ \square $$ (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 $$ e $$. $$ f(x, y)=\sqrt{3+2 x^{2}+2 y^{2}}, P(\sqrt{3},-1) $$ a. Find the gradient and evaluate it at $$ P $$. The gradient at $$ P $$ is $$ \left.\left\lvert\, \frac{2 \sqrt{3}}{\sqrt{11}}ight.,-\frac{2}{\sqrt{11}}ight) $$. (Type exact answers, using radicals as needed.) 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? $$ \square $$ (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

Given the function
$$
f(x,y)=\sqrt{3+2x^2+2y^2},
$$
its gradient is
$$
\nabla f(x,y)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right).
$$

Because
$$
f(x,y)=(3+2x^2+2y^2)^{1/2},
$$
using the chain rule we have
$$
\frac{\partial f}{\partial x}=\frac{1}{2}(3+2x^2+2y^2)^{-1/2}\cdot(4x)=\frac{2x}{\sqrt{3+2x^2+2y^2}},
$$
and similarly,
$$
\frac{\partial f}{\partial y}=\frac{2y}{\sqrt{3+2x^2+2y^2}}.
$$

At the point $$P(\sqrt{3},-1)$$ we have
$$
3+2(\sqrt{3})^2+2(-1)^2 = 3+2\cdot 3+2\cdot 1=3+6+2=11.
$$

Thus, the gradient at $$P$$ is
$$
\nabla f(\sqrt{3},-1)=\left(\frac{2\sqrt{3}}{\sqrt{11}},\frac{2(-1)}{\sqrt{11}}\right)
=\left(\frac{2\sqrt{3}}{\sqrt{11}},-\frac{2}{\sqrt{11}}\right).
$$

---

The directional derivative is maximized in the direction of the gradient and minimized in the opposite direction. To find the angles (with respect to the positive $$x$$-axis) we consider the components of the gradient.

The gradient vector at $$P$$ is
$$
\mathbf{G}=\left(\frac{2\sqrt{3}}{\sqrt{11}},-\frac{2}{\sqrt{11}}\right).
$$
Its direction is given by
$$
\theta =\arctan\frac{y\text{-component}}{x\text{-component}} = \arctan\frac{-\frac{2}{\sqrt{11}}}{\frac{2\sqrt{3}}{\sqrt{11}}} = \arctan\left(-\frac{1}{\sqrt{3}}\right).
$$

Since $$\arctan\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}$$ and the $$x$$-component is positive while the $$y$$-component is negative, the direction is in the fourth quadrant. Converting $$-\frac{\pi}{6}$$ to an angle between 0 and $$2\pi$$ gives
$$
-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}.
$$
Thus, the direction of maximum increase is along
$$
\theta_{\text{max inc}}=\frac{11\pi}{6}.
$$

---

For the direction of maximum decrease, we use the fact that it is opposite to the gradient. To obtain the opposite vector we add $$\pi$$ to the angle:
$$
\theta_{\text{max dec}}=\frac{11\pi}{6}+\pi = \frac{11\pi}{6}+\frac{6\pi}{6}=\frac{17\pi}{6}.
$$
Since $$\frac{17\pi}{6}$$ exceeds $$2\pi$$, subtract $$2\pi$$:
$$
\frac{17\pi}{6}-\frac{12\pi}{6}=\frac{5\pi}{6}.
$$
Thus, the direction of maximum decrease is along
$$
\theta_{\text{max dec}}=\frac{5\pi}{6}.
$$

---

Additionally, the directions of zero change (i.e. the level curves of $$ f $$) are those perpendicular to the gradient. These occur at angles obtained by adding or subtracting $$\frac{\pi}{2}$$ from $$\theta_{\text{max inc}}$$.

1. Adding $$\frac{\pi}{2}$$:
   $$
   \frac{11\pi}{6}+\frac{\pi}{2}=\frac{11\pi}{6}+\frac{3\pi}{6}=\frac{14\pi}{6}=\frac{7\pi}{3}.
   $$
   Since $$\frac{7\pi}{3}=2\pi+\frac{\pi}{3}$$, this direction is equivalent to
   $$
   \frac{\pi}{3}.
   $$

2. Subtracting $$\frac{\pi}{2}$$:
   $$
   \frac{11\pi}{6}-\frac{\pi}{2}=\frac{11\pi}{6}-\frac{3\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3}.
   $$

Thus, the directions (angles) of zero change are
$$
\theta_{\text{zero change}}=\frac{\pi}{3} \quad \text{and} \quad \frac{4\pi}{3}.
$$

---

Summary of answers:

- Maximum increase direction: $$\theta=\frac{11\pi}{6}$$.
- Maximum decrease direction: $$\theta=\frac{5\pi}{6}$$.
- Zero change directions: $$\theta=\frac{\pi}{3}$$ and $$\theta=\frac{4\pi}{3}$$.
Maximum increase direction: $$ \theta = \frac{11\pi}{6} $$ Maximum decrease direction: $$ \theta = \frac{5\pi}{6} $$ Zero change directions: $$ \theta = \frac{\pi}{3} $$ and $$ \theta = \frac{4\pi}{3} $$

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Calculus Questions