The gradient is a vector that combines all these rates of change and points in the direction of the steepest ascent of the function.....give me numerical example on direction 

Sure! Let's go through a numerical example to illustrate how the gradient vector points in the direction of the steepest ascent.

### Example:
Consider the function \( f(x, y) = 3x^2 + 2y^2 \).

1. **Calculate the partial derivatives:**
   - The partial derivative of \( f \) with respect to \( x \) is:
     \[
     \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^2 + 2y^2) = 6x
     \]
   - The partial derivative of \( f \) with respect to \( y \) is:
     \[
     \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(3x^2 + 2y^2) = 4y
     \]

2. **Compute the gradient vector:**
   The gradient of \( f(x, y) \) is given by:
   \[
   \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (6x, 4y)
   \]
   
   So, the gradient vector is \( \nabla f(x, y) = (6x, 4y) \).

3. **Evaluate the gradient at a specific point:**
   Let's evaluate the gradient at the point \( (x, y) = (1, 2) \):
   \[
   \nabla f(1, 2) = (6(1), 4(2)) = (6, 8)
   \]
   
   So, at the point \( (1, 2) \), the gradient vector is \( (6, 8) \).

4. **Direction of the gradient:**
   - The gradient vector \( (6, 8) \) points in the direction where the function \( f(x, y) \) increases the fastest from the point \( (1, 2) \).
   - The direction of this vector can be found by computing its unit vector:
     \[
     \text{Unit vector} = \frac{(6, 8)}{\sqrt{6^2 + 8^2}} = \frac{(6, 8)}{\sqrt{36 + 64}} = \frac{(6, 8)}{\sqrt{100}} = \frac{(6, 8)}{10} = \left(\frac{3}{5}, \frac{4}{5}\right)
     \]
   - This unit vector \( \left(\frac{3}{5}, \frac{4}{5}\right) \) indicates the direction in which the function increases most rapidly at the point \( (1, 2) \).

### Interpretation:
- The gradient vector \( (6, 8) \) points in the direction of the steepest ascent of the function \( f(x, y) \) at the point \( (1, 2) \).
- The unit vector \( \left(\frac{3}{5}, \frac{4}{5}\right) \) gives the exact direction of this ascent.

Would you like to explore more about gradients or this particular example?

Here are 5 questions you might find useful:
1. How would the gradient change if we evaluated it at a different point, say \( (2, 1) \)?
2. What does it mean if the gradient vector at a point is zero?
3. How can we use the gradient to find the maximum or minimum of a function?
4. What happens to the gradient direction when we apply constraints to the function?
5. How can the concept of gradients be applied in physics, such as in the context of potential energy?

**Tip:** Always normalize the gradient to find the direction of steepest ascent as a unit vector, especially when comparing directions at different points.

Explore how gradients and steepest ascent are crucial in function analysis. Learn through a numerical example how the gradient vector points in the direction of the steepest ascent of a function. Suitable for advanced mathematics enthusiasts and students.

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