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

1. Problem Description

We are asked to find the gradient vector and the equation of the tangent plane of a given function $f(x, y)$ at a given point $p$. We will solve problem number

1

1. The function is $f(x, y) = x^2y - xy^2$ and the point is $p = (-2, 3)$.

2. Solution Steps

First, we need to find the gradient vector of $f(x, y)$, which is given by $\nabla f(x, y) = \langle f_x, f_y \rangle$, where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively.

$f(x, y) = x^2y - xy^2$

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

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

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

$f_x(-2, 3) = 2(-2)(3) - (3)^2 = -12 - 9 = -21$

$f_y(-2, 3) = (-2)^2 - 2(-2)(3) = 4 + 12 = 16$

So, the gradient vector at $p$ is $\nabla f(-2, 3) = \langle -21, 16 \rangle$.

Next, we find the value of the function at the point $p$:

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

Now, we find the equation of the tangent plane. The equation of the tangent plane to the surface $z = f(x, y)$ at the point $(x_0, y_0)$ is given by:

$z - f(x_0, y_0) = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$

In our case, $(x_0, y_0) = (-2, 3)$ and $f(-2, 3) = 30$.

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

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

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

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

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

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

3. Final Answer

The gradient vector at $p = (-2, 3)$ is $\langle -21, 16 \rangle$.

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