To find the direction in which the maximum rate of change of the function $f(x, y) = 2x \sin(xy)$ occurs at the point $(5, 3)$, we need to compute the gradient of $f(x, y)$, since the gradient vector points in the direction of the steepest ascent.

Step 1: Compute the partial derivatives of $f(x, y)$

The gradient vector is given by: $\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

1. Partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 2x \sin(xy) \right)$ Applying the product rule: $\frac{\partial f}{\partial x} = 2 \sin(xy) + 2x \cdot y \cos(xy)$

2. Partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( 2x \sin(xy) \right)$ This involves differentiating $\sin(xy)$ with respect to $y$: $\frac{\partial f}{\partial y} = 2x \cdot x \cos(xy) = 2x^2 \cos(xy)$

Step 2: Evaluate the gradient at the point $(5, 3)$

Now, substitute $x = 5$ and $y = 3$ into the partial derivatives:

1. Evaluating $\frac{\partial f}{\partial x}$ at $(5, 3)$: $\frac{\partial f}{\partial x} = 2 \sin(5 \times 3) + 2 \times 5 \times 3 \cos(5 \times 3)$ $= 2 \sin(15) + 30 \cos(15)$

2. Evaluating $\frac{\partial f}{\partial y}$ at $(5, 3)$: $\frac{\partial f}{\partial y} = 2 \times 5^2 \cos(5 \times 3) = 50 \cos(15)$

Step 3: Calculate the gradient vector at $(5, 3)$

Thus, the gradient vector at $(5, 3)$ is: $\nabla f(5, 3) = \left( 2 \sin(15) + 30 \cos(15), 50 \cos(15) \right)$

Step 4: Find the unit vector in the direction of the gradient

The maximum rate of change occurs in the direction of the gradient. To find the unit vector, we normalize the gradient vector.

The magnitude of the gradient vector is: $|\nabla f(5, 3)| = \sqrt{\left( 2 \sin(15) + 30 \cos(15) \right)^2 + \left( 50 \cos(15) \right)^2}$

The unit vector is: $\hat{u} = \frac{\nabla f(5, 3)}{|\nabla f(5, 3)|}$

Would you like me to calculate the final numerical values for this expression?

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Relative Questions:

1. What is the geometric interpretation of the gradient vector?
2. How does the direction of the gradient relate to the level curves of the function?
3. Why is the unit vector important in the context of maximum rate of change?
4. How would the gradient change if the function were modified to $f(x, y) = 2x \cos(xy)$?
5. What is the physical significance of the gradient in real-world applications?

Tip:

The gradient always points in the direction of the steepest increase of the function, and its magnitude tells you how steep that increase is.