Find the directional derivative at the point 𝑃(1,2,3) to the function
𝑓 = 𝑥*2 − 𝑦*2 + 2𝑧*2
in the direction of the line 𝑃𝑄 where 𝑄 is the
point (5,0,4). In what direction it will be maximum? Also, find the
magnitude of this maximum

To find the directional derivative of the function \( f(x, y, z) = x^2 - y^2 + 2z^2 \) at the point \( P(1, 2, 3) \) in the direction of the line \( PQ \) where \( Q(5, 0, 4) \), we follow these steps:

1. **Calculate the gradient of \( f \) at \( P \).**
2. **Determine the direction vector from \( P \) to \( Q \).**
3. **Normalize the direction vector.**
4. **Compute the directional derivative using the gradient and the unit direction vector.**
5. **Identify the direction in which the derivative is maximum and find the magnitude of this maximum.**

### Step 1: Calculate the Gradient of \( f \)

The gradient of \( f(x, y, z) \) is given by:

\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]

For \( f(x, y, z) = x^2 - y^2 + 2z^2 \):

\[ \frac{\partial f}{\partial x} = 2x \]
\[ \frac{\partial f}{\partial y} = -2y \]
\[ \frac{\partial f}{\partial z} = 4z \]

Thus,

\[ \nabla f = (2x, -2y, 4z) \]

At the point \( P(1, 2, 3) \):

\[ \nabla f|_P = (2 \cdot 1, -2 \cdot 2, 4 \cdot 3) = (2, -4, 12) \]

### Step 2: Determine the Direction Vector from \( P \) to \( Q \)

The direction vector \( \vec{PQ} \) is:

\[ \vec{PQ} = Q - P = (5 - 1, 0 - 2, 4 - 3) = (4, -2, 1) \]

### Step 3: Normalize the Direction Vector

The magnitude of \( \vec{PQ} \) is:

\[ |\vec{PQ}| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21} \]

The unit direction vector \( \mathbf{u} \) is:

\[ \mathbf{u} = \frac{\vec{PQ}}{|\vec{PQ}|} = \left( \frac{4}{\sqrt{21}}, \frac{-2}{\sqrt{21}}, \frac{1}{\sqrt{21}} \right) \]

### Step 4: Compute the Directional Derivative

The directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{u} \) is given by:

\[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} \]

\[ D_{\mathbf{u}} f = (2, -4, 12) \cdot \left( \frac{4}{\sqrt{21}}, \frac{-2}{\sqrt{21}}, \frac{1}{\sqrt{21}} \right) \]

\[ D_{\mathbf{u}} f = 2 \cdot \frac{4}{\sqrt{21}} + (-4) \cdot \frac{-2}{\sqrt{21}} + 12 \cdot \frac{1}{\sqrt{21}} \]

\[ D_{\mathbf{u}} f = \frac{8}{\sqrt{21}} + \frac{8}{\sqrt{21}} + \frac{12}{\sqrt{21}} \]

\[ D_{\mathbf{u}} f = \frac{8 + 8 + 12}{\sqrt{21}} = \frac{28}{\sqrt{21}} = \frac{28 \sqrt{21}}{21} = \frac{4 \sqrt{21}}{3} \]

### Step 5: Maximum Directional Derivative

The directional derivative is maximum in the direction of the gradient \( \nabla f \).

The magnitude of the maximum directional derivative is the magnitude of the gradient \( \nabla f \):

\[ |\nabla f| = \sqrt{2^2 + (-4)^2 + 12^2} = \sqrt{4 + 16 + 144} = \sqrt{164} = 2\sqrt{41} \]

### Summary

- The directional derivative of \( f \) at \( P(1, 2, 3) \) in the direction of the line \( PQ \) is \( \frac{4 \sqrt{21}}{3} \).
- The maximum directional derivative is in the direction of the gradient \( \nabla f \).
- The magnitude of this maximum directional derivative is \( 2\sqrt{41} \).

Would you like further details or have any questions about this solution?

Here are 5 more questions you might be interested in:

1. How do you compute the gradient of a scalar field?
2. What are some applications of directional derivatives?
3. How do you find the unit vector in a given direction?
4. Can you explain the concept of a gradient vector field?
5. How do you determine if a function is differentiable?

Tip: When computing gradients, always remember to evaluate the partial derivatives accurately to ensure the correct direction and magnitude of the vector.

Learn how to calculate the directional derivative at point P(1,2,3) to function f(x, y, z) = x^2 - y^2 + 2z^2 in the direction of line PQ(5,0,4). Find the maximum direction and its magnitude with detailed step-by-step solutions.

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