Let's break this down step-by-step:

(a) Find the gradient of $f(x, y) = \sin(2x + 3y)$.

The gradient of a scalar field $f(x, y)$ is a vector containing its partial derivatives with respect to $x$ and $y$. In other words:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

1. Partial derivative of $f(x, y)$ with respect to $x$: $\frac{\partial f}{\partial x} = \frac{d}{dx} \left( \sin(2x + 3y) \right) = \cos(2x + 3y) \cdot \frac{d}{dx} (2x + 3y) = 2 \cos(2x + 3y)$

2. Partial derivative of $f(x, y)$ with respect to $y$: $\frac{\partial f}{\partial y} = \frac{d}{dy} \left( \sin(2x + 3y) \right) = \cos(2x + 3y) \cdot \frac{d}{dy} (2x + 3y) = 3 \cos(2x + 3y)$

Thus, the gradient of $f(x, y)$ is:

$\nabla f(x, y) = \left( 2 \cos(2x + 3y), 3 \cos(2x + 3y) \right)$

(b) Evaluate the gradient at the point $P(-3, 2)$.

Now, substitute $x = -3$ and $y = 2$ into the gradient:

$\nabla f(-3, 2) = \left( 2 \cos(2(-3) + 3(2)), 3 \cos(2(-3) + 3(2)) \right)$

First, compute the argument of the cosine:

$2(-3) + 3(2) = -6 + 6 = 0$

So, we have:

$\nabla f(-3, 2) = \left( 2 \cos(0), 3 \cos(0) \right)$

Since $\cos(0) = 1$, the gradient at $P(-3, 2)$ is:

$\nabla f(-3, 2) = \left( 2(1), 3(1) \right) = (2, 3)$

(c) Find the rate of change of $f$ at $P$ in the direction of the vector $u = \frac{1}{2}(3i - j)$.

The rate of change of $f$ in the direction of a unit vector $u$ is given by the directional derivative:

$D_u f(x, y) = \nabla f(x, y) \cdot \hat{u}$

where $\hat{u}$ is the unit vector in the direction of $u$. First, we normalize $u$:

$u = \frac{1}{2}(3i - j) = \left( \frac{3}{2}, -\frac{1}{2} \right)$

The magnitude of $u$ is:

$|u| = \sqrt{\left( \frac{3}{2} \right)^2 + \left( -\frac{1}{2} \right)^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$

Thus, the unit vector $\hat{u}$ is:

$\hat{u} = \frac{u}{|u|} = \frac{\left( \frac{3}{2}, -\frac{1}{2} \right)}{\frac{\sqrt{10}}{2}} = \left( \frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}} \right)$

Now, compute the directional derivative:

$D_u f(-3, 2) = (2, 3) \cdot \left( \frac{3}{\sqrt{10}}, -\frac{1}{\sqrt{10}} \right)$

$D_u f(-3, 2) = 2 \cdot \frac{3}{\sqrt{10}} + 3 \cdot \left( -\frac{1}{\sqrt{10}} \right)$

$D_u f(-3, 2) = \frac{6}{\sqrt{10}} - \frac{3}{\sqrt{10}} = \frac{3}{\sqrt{10}}$

Thus, the rate of change of $f$ at $P(-3, 2)$ in the direction of $u$ is:

$D_u f(-3, 2) = \frac{3}{\sqrt{10}}$

Summary:

• (a) $\nabla f(x, y) = \left( 2 \cos(2x + 3y), 3 \cos(2x + 3y) \right)$
• (b) $\nabla f(-3, 2) = (2, 3)$
• (c) $D_u f(-3, 2) = \frac{3}{\sqrt{10}}$

Let me know if you'd like to explore any part of this further! Here are some related questions for you to practice:

1. What is the gradient of $g(x, y) = e^{x^2 + y^2}$?
2. How do you compute the directional derivative of a function given two points?
3. If $u$ were a unit vector, how would the rate of change change in part (c)?
4. What is the maximum rate of change of a function at a given point?
5. How do you calculate the gradient in 3-dimensional space?

Tip: The gradient vector always points in the direction of the steepest ascent for a function at any given point.