The directional derivative of the function f(x, y) = x² + y² along a line directed from (0, 0) to (1, 1), evaluated at the point x = 1, y = 1 is

The correct answer is 2√2.

Step-by-step Explanation:

Step 1: Understand the formula for the directional derivative
The directional derivative of a function $f(x,y)$ at a given point in the direction of a unit vector $\hat{u}$ is given by the dot product of the gradient vector of the function at that point and the unit vector:
$D_{\hat{u}}f=\nabla f\cdot \hat{u}$

Step 2: Find the gradient of the function f(x, y)
The given function is:
$f(x,y)=x^2+y^2$
First, we find the partial derivatives with respect to $x$ and $y$:
$\frac{\partial f}{\partial x}=2x$
$\frac{\partial f}{\partial y}=2y$
Thus, the gradient vector $\nabla f(x,y)$ is:
$\nabla f(x,y)=2x\hat{i}+2y\hat{j}$

Step 3: Evaluate the gradient at the given point (1, 1)
Substituting $x=1$ and $y=1$ into the gradient expression:
$\nabla f(1,1)=2\left(1\right)\hat{i}+2\left(1\right)\hat{j}=2\hat{i}+2\hat{j}$

Step 4: Determine the unit vector in the specified direction
The line is directed from the point $A(0,0)$ to the point $B(1,1)$. The direction vector $\overrightarrow{v}$ is:
$\overrightarrow{v}=(1-0)\hat{i}+(1-0)\hat{j}=\hat{i}+\hat{j}$
Next, calculate the magnitude of $\overrightarrow{v}$:
$\left|\overrightarrow{v}\right|=\sqrt{1^2+1^2}=\sqrt{2}$
The unit vector $\hat{u}$ in this direction is:
$\hat{u}=\frac{\overrightarrow{v}}{\left|\overrightarrow{v}\right|}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

Step 5: Compute the directional derivative
Now, take the dot product of the gradient vector at $(1,1)$ and the unit vector $\hat{u}$:
$D_{\hat{u}}f=\nabla f(1,1)\cdot \hat{u}$
$D_{\hat{u}}f=(2\hat{i}+2\hat{j})\cdot \left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$
$D_{\hat{u}}f=\frac{(2\cdot 1)+(2\cdot 1)}{\sqrt{2}}=\frac{4}{\sqrt{2}}$
Simplifying the expression by rationalizing the denominator:
$D_{\hat{u}}f=\frac{4\sqrt{2}}{2}=2\sqrt{2}$

Therefore, the evaluated directional derivative is 2√2.