find the critical points in f(x,y) = xsiny and their type

To find the critical points of the function \( f(x, y) = x \sin y \), we need to follow these steps:

### Step 1: Compute the Partial Derivatives
The critical points occur where the partial derivatives of \( f(x, y) \) with respect to both \( x \) and \( y \) are equal to zero.

- First, find the partial derivative of \( f(x, y) \) with respect to \( x \):
  \[
  f_x(x, y) = \frac{\partial}{\partial x}(x \sin y) = \sin y
  \]

- Next, find the partial derivative of \( f(x, y) \) with respect to \( y \):
  \[
  f_y(x, y) = \frac{\partial}{\partial y}(x \sin y) = x \cos y
  \]

### Step 2: Set Partial Derivatives Equal to Zero
Now, set both partial derivatives equal to zero and solve for \( x \) and \( y \):

1. \( f_x(x, y) = \sin y = 0 \)
2. \( f_y(x, y) = x \cos y = 0 \)

#### Solve \( f_x(x, y) = 0 \):
\[
\sin y = 0 \implies y = n\pi \quad \text{(where \( n \in \mathbb{Z} \))}
\]

#### Solve \( f_y(x, y) = 0 \):
For \( x \cos y = 0 \), we have two cases:
- If \( x = 0 \), then this equation is satisfied for any \( y \).
- If \( x \neq 0 \), then \( \cos y = 0 \implies y = \frac{\pi}{2} + m\pi \quad \text{(where \( m \in \mathbb{Z} \))} \).

### Step 3: Analyze the Critical Points
We now have two possible cases for critical points:
1. **\( x = 0 \) for any \( y \)**: This gives a line of critical points along the \( y \)-axis.
2. **\( y = n\pi \) for any \( x \)**: These correspond to lines where \( y \) is a multiple of \( \pi \), along which the function behaves critically.

### Step 4: Classify the Critical Points
To classify these critical points, we use the second derivative test. We compute the second-order partial derivatives to form the Hessian matrix.

- \( f_{xx} = \frac{\partial^2}{\partial x^2}(x \sin y) = 0 \)
- \( f_{yy} = \frac{\partial^2}{\partial y^2}(x \sin y) = -x \sin y \)
- \( f_{xy} = f_{yx} = \frac{\partial^2}{\partial x \partial y}(x \sin y) = \cos y \)

The Hessian matrix \( H \) is:
\[
H = \begin{pmatrix}
f_{xx} & f_{xy} \\
f_{xy} & f_{yy}
\end{pmatrix}
= \begin{pmatrix}
0 & \cos y \\
\cos y & -x \sin y
\end{pmatrix}
\]

The determinant of the Hessian is:
\[
\text{det}(H) = (0)(-x \sin y) - (\cos y)^2 = -\cos^2 y
\]

Since \( \text{det}(H) = -\cos^2 y \leq 0 \), the determinant is non-positive for all \( y \), which means the critical points are saddle points.

### Conclusion:
- **The critical points** are \( (0, y) \) for any \( y \), and along the lines where \( y = n\pi \).
- **All these critical points are saddle points**, as the determinant of the Hessian is non-positive.

Would you like more details or have any questions?

Here are some related questions:
1. How do we interpret the meaning of saddle points in this context?
2. What are the properties of the Hessian matrix, and how do they relate to critical points?
3. How can we visualize the function \( f(x, y) = x \sin y \) and its critical points?
4. How does the second derivative test classify critical points in more complex functions?
5. What happens if we introduce constraints in this type of optimization problem?

**Tip:** The Hessian determinant helps classify critical points; a negative determinant indicates a saddle point, while a positive one indicates a local maximum or minimum.

Find the critical points in f(x,y) = x sin y and their type

Explore the step-by-step process of finding the critical points for the function f(x, y) = x sin y. Learn how to use partial derivatives and the Hessian matrix to classify the critical points as saddle points using the second derivative test. Suitable for undergraduate calculus students.

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