To find the directional derivative of $f(x, y, z) = x^2 + 5y^2 + 3z^2$ at the point $(1,1,1)$ in the direction of the vector $\vec{v} = 3\vec{i} + 4\vec{j} + 5\vec{k}$, follow these steps:

Step 1: Compute the Gradient of $f$

The gradient of $f(x, y, z)$ is given by: $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$ Calculate each partial derivative:

1. $\frac{\partial f}{\partial x} = 2x$
2. $\frac{\partial f}{\partial y} = 10y$
3. $\frac{\partial f}{\partial z} = 6z$

Thus, the gradient $\nabla f = (2x, 10y, 6z)$.

Step 2: Evaluate the Gradient at $(1,1,1)$

Substitute $x = 1$, $y = 1$, and $z = 1$: $\nabla f(1, 1, 1) = (2 \cdot 1, 10 \cdot 1, 6 \cdot 1) = (2, 10, 6)$

Step 3: Normalize the Direction Vector

The given direction vector is $\vec{v} = 3\vec{i} + 4\vec{j} + 5\vec{k}$. First, find its magnitude: $|\vec{v}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$ Now, the unit vector in the direction of $\vec{v}$ is: $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \left( \frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, \frac{5}{5\sqrt{2}} \right) = \left( \frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$

Step 4: Compute the Directional Derivative

The directional derivative of $f$ at $(1,1,1)$ in the direction of $\hat{u}$ is: $D_{\hat{u}} f = \nabla f \cdot \hat{u}$ Substitute $\nabla f(1,1,1) = (2, 10, 6)$ and $\hat{u} = \left( \frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$: $D_{\hat{u}} f = 2 \cdot \frac{3}{5\sqrt{2}} + 10 \cdot \frac{4}{5\sqrt{2}} + 6 \cdot \frac{1}{\sqrt{2}}$ Calculate each term separately:

1. $2 \cdot \frac{3}{5\sqrt{2}} = \frac{6}{5\sqrt{2}}$
2. $10 \cdot \frac{4}{5\sqrt{2}} = \frac{40}{5\sqrt{2}} = \frac{8}{\sqrt{2}}$
3. $6 \cdot \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}}$

So, $D_{\hat{u}} f = \frac{6}{5\sqrt{2}} + \frac{8}{\sqrt{2}} + \frac{6}{\sqrt{2}}$ Combine terms over a common denominator: $D_{\hat{u}} f = \frac{6 + 8 \cdot 5 + 6 \cdot 5}{5\sqrt{2}} = \frac{6 + 40 + 30}{5\sqrt{2}} = \frac{76}{5\sqrt{2}}$ Thus, the directional derivative is: $D_{\hat{u}} f = \frac{76}{5\sqrt{2}}$

Would you like further details on this solution, or do you have any specific questions?

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Here are some additional questions based on this problem:

1. What is the importance of finding the directional derivative in real-world applications?
2. How does the choice of direction vector affect the magnitude of the directional derivative?
3. Can the directional derivative be zero? If so, under what conditions?
4. How would the directional derivative change if the direction vector were normalized differently?
5. What happens if the point of evaluation is at the origin?

Tip: When working with directional derivatives, always normalize the direction vector to ensure accurate results.