The problem asks us to find the gradient vector of a given function $f(x, y)$ at a given point $p$, and then find the equation of the tangent plane at the point $p$. We will solve problem 11: $f(x, y) = x^2y - xy^2$ and $p = (-2, 3)$.

1. Problem Description

The problem asks us to find the gradient vector of a given function $f(x, y)$ at a given point $p$, and then find the equation of the tangent plane at the point $p$. We will solve problem 11: $f(x, y) = x^2y - xy^2$ and $p = (-2, 3)$.

2. Solution Steps

First, we need to find the gradient vector of $f(x, y)$. The gradient vector is given by:

$\nabla f(x, y) = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle$

We compute the partial derivatives:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y - xy^2) = 2xy - y^2$

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2y - xy^2) = x^2 - 2xy$

So the gradient vector is:

$\nabla f(x, y) = \langle 2xy - y^2, x^2 - 2xy \rangle$

Now, we evaluate the gradient vector at the point $p = (-2, 3)$:

$\nabla f(-2, 3) = \langle 2(-2)(3) - (3)^2, (-2)^2 - 2(-2)(3) \rangle = \langle -12 - 9, 4 + 12 \rangle = \langle -21, 16 \rangle$

Next, we need to find the value of the function $f(x, y)$ at the point $p = (-2, 3)$:

$f(-2, 3) = (-2)^2(3) - (-2)(3)^2 = (4)(3) - (-2)(9) = 12 + 18 = 30$

Finally, we find the equation of the tangent plane at the point $p = (-2, 3)$. The equation of the tangent plane is given by:

$z - f(x_0, y_0) = \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0)$

$z - 30 = -21(x - (-2)) + 16(y - 3)$

$z - 30 = -21(x + 2) + 16(y - 3)$

$z - 30 = -21x - 42 + 16y - 48$

$z = -21x + 16y - 42 - 48 + 30$

$z = -21x + 16y - 60$

The equation of the tangent plane is $21x - 16y + z + 60 = 0$.

3. Final Answer

Gradient vector at $p=(-2, 3)$: $\langle -21, 16 \rangle$

Equation of the tangent plane at $p=(-2, 3)$: $z = -21x + 16y - 60$ or $21x - 16y + z + 60 = 0$.