Part 1 of 8 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 $$ \langle\square, \square $$. (Type exact answers, using radicals as needed.)

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Part 1 of 8 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 $$ \langle\square, \square $$. (Type exact answers, using radicals as needed.)

UpStudy AI · Accepted Answer

**Step 1. Write the function in a convenient form**

We have
$$
f(x,y)=\sqrt{3+2x^2+2y^2}=(3+2x^2+2y^2)^{\frac{1}{2}}.
$$

**Step 2. Find the general partial derivatives**

We differentiate using the chain rule.

For the partial derivative with respect to $$x$$:
$$
f_x(x,y)=\frac{1}{2}(3+2x^2+2y^2)^{-\frac{1}{2}}\cdot (4x)=\frac{2x}{\sqrt{3+2x^2+2y^2}}.
$$

For the partial derivative with respect to $$y$$:
$$
f_y(x,y)=\frac{1}{2}(3+2x^2+2y^2)^{-\frac{1}{2}}\cdot (4y)=\frac{2y}{\sqrt{3+2x^2+2y^2}}.
$$

Thus, the gradient of $$f$$ is
$$
\nabla f(x,y)=\left\langle\frac{2x}{\sqrt{3+2x^2+2y^2}},\,\frac{2y}{\sqrt{3+2x^2+2y^2}}\right\rangle.
$$

**Step 3. Evaluate the gradient at $$P(\sqrt{3}, -1)$$**

Substitute $$x=\sqrt{3}$$ and $$y=-1$$ into the gradient formula.

First, compute the value under the square root:
$$
3+2(\sqrt{3})^2+2(-1)^2=3+2(3)+2(1)=3+6+2=11.
$$

Then,
$$
f_x(\sqrt{3}, -1)=\frac{2\sqrt{3}}{\sqrt{11}},\quad f_y(\sqrt{3}, -1)=\frac{2(-1)}{\sqrt{11}}=-\frac{2}{\sqrt{11}}.
$$

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

**Final Answer:**

$$
\langle \frac{2\sqrt{3}}{\sqrt{11}},\, -\frac{2}{\sqrt{11}} \rangle.
$$
The gradient at point $$ P(\sqrt{3}, -1) $$ is $$ \left\langle \frac{2\sqrt{3}}{\sqrt{11}},\, -\frac{2}{\sqrt{11}} \right\rangle $$.

Part 1 of 8
A function and a point are given. Let correspond to the direction of the directional derivative. Complete parts a.
through .

a. Find the gradient and evaluate it at .
The gradient at is .
(Type exact answers, using radicals as needed.)

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Part 1 of 8 A function and a point are given. Let correspond to the direction of the directional derivative. Complete parts a. through . a. Find the gradient and evaluate it at . The gradient at is . (Type exact answers, using radicals as needed.)