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We are given the function \( f(x, y) = x \sin y \) and asked to determine the nature of the critical point at \( (0, 0) \).

### Steps to determine the nature of the critical point:

1. **Find the partial derivatives**:
   To find the critical points and their nature, we start by computing the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \).

   - \( f_x(x, y) = \frac{\partial}{\partial x} (x \sin y) = \sin y \)
   - \( f_y(x, y) = \frac{\partial}{\partial y} (x \sin y) = x \cos y \)

2. **Check the critical points**:
   The critical points occur when both partial derivatives are zero:

   - \( f_x(x, y) = 0 \) implies \( \sin y = 0 \), so \( y = n\pi \) where \( n \in \mathbb{Z} \) (integer values).
   - \( f_y(x, y) = 0 \) implies \( x \cos y = 0 \). This gives two possibilities:
     - \( x = 0 \), or
     - \( \cos y = 0 \), meaning \( y = (2n+1)\frac{\pi}{2} \).

   For the point \( (0,0) \), we have \( x = 0 \) and \( y = 0 \), so this satisfies both conditions, confirming that \( (0,0) \) is a critical point.

3. **Second derivative test**:
   We now use the second derivative test to classify the critical point. First, compute the second partial derivatives:

   - \( f_{xx}(x, y) = 0 \)
   - \( f_{yy}(x, y) = -x \sin y \)
   - \( f_{xy}(x, y) = \cos y \)

   At \( (0, 0) \), the second partial derivatives are:

   - \( f_{xx}(0, 0) = 0 \)
   - \( f_{yy}(0, 0) = 0 \)
   - \( f_{xy}(0, 0) = 1 \)

   The **Hessian determinant** is given by:
   \[
   D = f_{xx}f_{yy} - (f_{xy})^2 = (0)(0) - (1)^2 = -1
   \]
   Since \( D < 0 \), the critical point at \( (0, 0) \) is a **saddle point**.

### Conclusion:
The critical point \( (0, 0) \) is a **saddle point**.

Would you like any further details or explanations?

Here are some related questions to deepen your understanding:
1. How do you calculate the Hessian matrix for a multivariable function?
2. What is the significance of the sign of the Hessian determinant?
3. Can a critical point with \( D < 0 \) ever be anything other than a saddle point?
4. How does the second derivative test differ from finding critical points in single-variable calculus?
5. What happens if the Hessian determinant is zero?

**Tip:** When using the second derivative test, always check the sign of the Hessian determinant to classify the critical point.

The function f(x,y) = x sin y has a critical point at (0,0). What sort of critical point is it?

Learn how to determine the nature of the critical point at (0,0) for the function f(x,y) = x sin y. This problem involves calculating partial derivatives, applying the second derivative test, and understanding the Hessian determinant. The critical point is classified as a saddle point based on the test.

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