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We are tasked with finding the directional derivative of the function \( f(u, v) = u^2 e^{-v} \) at the point \( (3, 0) \) in the direction of the vector \( \mathbf{v} = 3\mathbf{i} + 4\mathbf{j} \).

### Step-by-Step Solution

1. **Gradient of \( f(u, v) \):**
   The gradient of \( f \), denoted as \( \nabla f(u, v) \), is the vector of partial derivatives with respect to \( u \) and \( v \).

   \[
   \frac{\partial f}{\partial u} = 2u e^{-v}
   \]
   \[
   \frac{\partial f}{\partial v} = -u^2 e^{-v}
   \]

   Therefore, the gradient is:
   \[
   \nabla f(u, v) = \left( 2u e^{-v}, -u^2 e^{-v} \right)
   \]

2. **Gradient at the point \( (3, 0) \):**
   Substituting \( u = 3 \) and \( v = 0 \) into the gradient:

   \[
   \nabla f(3, 0) = \left( 2(3) e^{0}, -(3)^2 e^{0} \right) = (6, -9)
   \]

3. **Unit vector in the direction of \( \mathbf{v} \):**
   The direction vector \( \mathbf{v} = 3\mathbf{i} + 4\mathbf{j} \) needs to be converted into a unit vector by dividing it by its magnitude.

   The magnitude of \( \mathbf{v} \) is:
   \[
   |\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
   \]

   So, the unit vector is:
   \[
   \mathbf{u_v} = \left( \frac{3}{5}, \frac{4}{5} \right)
   \]

4. **Directional derivative:**
   The directional derivative of \( f \) in the direction of \( \mathbf{v} \) at the point \( (3, 0) \) is the dot product of \( \nabla f(3, 0) \) and the unit vector \( \mathbf{u_v} \):

   \[
   D_{\mathbf{v}} f(3, 0) = \nabla f(3, 0) \cdot \mathbf{u_v}
   \]

   \[
   D_{\mathbf{v}} f(3, 0) = (6, -9) \cdot \left( \frac{3}{5}, \frac{4}{5} \right)
   \]

   Compute the dot product:
   \[
   D_{\mathbf{v}} f(3, 0) = 6 \times \frac{3}{5} + (-9) \times \frac{4}{5} = \frac{18}{5} - \frac{36}{5} = \frac{-18}{5}
   \]

Thus, the directional derivative is:
\[
D_{\mathbf{v}} f(3, 0) = \frac{-18}{5}
\]

Would you like more details or have any questions?

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Here are 5 related questions:
1. How do you compute the gradient of a multivariable function?
2. What is the significance of the directional derivative in geometry?
3. How do you find the unit vector of a given direction vector?
4. What is the relationship between the gradient and the directional derivative?
5. In what contexts are directional derivatives commonly used in physics or engineering?

**Tip:** Always remember to normalize the direction vector when computing the directional derivative to ensure it's unit length.

Find the directional derivative of f(u, v) = u^2 * e^(-v) at the point (3, 0) in the direction of the vector v = 3i + 4j.

Learn how to compute the directional derivative of f(u, v) = u² * e^(-v) at the point (3, 0) in the direction of the vector v = 3i + 4j. The problem involves calculating the gradient of the function, normalizing the direction vector, and performing the dot product for the final solution.

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