To solve this, we need to find two things:

1. A unit vector in the direction in which the function $f(x, y) = 4x^3 y^2$ increases most rapidly at the point $P(-1, 1)$.
2. The rate of change of $f$ at $P$ in that direction.

Step 1: Compute the Gradient of $f$

The gradient of $f$, denoted as $\nabla f(x, y)$, gives the direction of the greatest rate of increase of $f$. The gradient is the vector of partial derivatives with respect to $x$ and $y$:

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

For $f(x, y) = 4x^3 y^2$:

• The partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = 12x^2 y^2$

• The partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = 8x^3 y$

Thus, the gradient is: $\nabla f(x, y) = (12x^2 y^2, 8x^3 y)$

Step 2: Evaluate the Gradient at Point $P(-1, 1)$

Now, we evaluate the gradient at the point $P(-1, 1)$:

$\nabla f(-1, 1) = \left( 12(-1)^2 (1)^2, 8(-1)^3 (1) \right)$ $\nabla f(-1, 1) = (12, -8)$

Step 3: Find the Unit Vector in the Direction of the Gradient

The direction of greatest increase is the direction of the gradient vector. To obtain the unit vector, we divide the gradient vector by its magnitude.

First, calculate the magnitude of $\nabla f(-1, 1)$: $|\nabla f(-1, 1)| = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}$

Now, the unit vector $\mathbf{u}$ in the direction of the gradient is: $\mathbf{u} = \frac{\nabla f(-1, 1)}{|\nabla f(-1, 1)|} = \frac{(12, -8)}{4\sqrt{13}} = \left( \frac{12}{4\sqrt{13}}, \frac{-8}{4\sqrt{13}} \right)$ $\mathbf{u} = \left( \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right)$

Step 4: Find the Rate of Change of $f$ in the Direction of $\mathbf{u}$

The rate of change of $f$ in the direction of the unit vector $\mathbf{u}$ is given by the directional derivative, which is the dot product of the gradient $\nabla f(-1, 1)$ and the unit vector $\mathbf{u}$:

$D_{\mathbf{u}} f = \nabla f(-1, 1) \cdot \mathbf{u}$

Performing the dot product:

$D_{\mathbf{u}} f = (12, -8) \cdot \left( \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right)$ $D_{\mathbf{u}} f = \left( 12 \times \frac{3}{\sqrt{13}} \right) + \left( -8 \times \frac{-2}{\sqrt{13}} \right)$ $D_{\mathbf{u}} f = \frac{36}{\sqrt{13}} + \frac{16}{\sqrt{13}} = \frac{52}{\sqrt{13}} = 4\sqrt{13}$

Final Answer:

• The unit vector in the direction in which $f$ increases most rapidly at $P(-1, 1)$ is: $\mathbf{u} = \left( \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right)$
• The rate of change of $f$ at $P(-1, 1)$ in that direction is: $D_{\mathbf{u}} f = 4\sqrt{13}$

Would you like a detailed explanation of any step? Or any further questions on related topics?

Here are some follow-up questions for you:

1. What does the gradient of a function tell us in multivariable calculus?
2. How would you calculate the rate of change of a function in an arbitrary direction?
3. Can the directional derivative be negative? If so, what would that indicate?
4. How do the concepts of gradient and directional derivative relate to the concept of tangent planes?
5. How would you approach a problem that asks for the rate of change in a direction other than the gradient?

Tip: When working with gradients, remember that the direction of the gradient always points in the direction of the greatest rate of increase of the function, and the magnitude of the gradient gives you the rate of increase in that direction.