We are tasked with finding the equation of the tangent plane to the surface given by the function:

$z = 3x - (2y^2 + 2x^2) + 2$

at the point $(2, -5, -50)$.

Step 1: Compute the partial derivatives

The general equation of a tangent plane to a surface $z = f(x, y)$ at a point $(x_0, y_0, z_0)$ is given by:

$z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$

Where:

• $f_x$ is the partial derivative of $f(x, y)$ with respect to $x$,
• $f_y$ is the partial derivative of $f(x, y)$ with respect to $y$.

We need to compute these partial derivatives.

Partial derivative with respect to $x$:

$f(x, y) = 3x - (2y^2 + 2x^2) + 2$ Taking the derivative with respect to $x$: $f_x = \frac{\partial}{\partial x}(3x - 2y^2 - 2x^2) = 3 - 4x$

Partial derivative with respect to $y$:

$f_y = \frac{\partial}{\partial y}(3x - 2y^2 - 2x^2) = -4y$

Step 2: Evaluate the partial derivatives at the point $(2, -5)$

• $f_x(2, -5) = 3 - 4(2) = 3 - 8 = -5$
• $f_y(2, -5) = -4(-5) = 20$

Step 3: Use the point and the partial derivatives to form the equation of the tangent plane

The equation of the tangent plane is:

$z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$

Substitute $(x_0, y_0, z_0) = (2, -5, -50)$, $f_x(2, -5) = -5$, and $f_y(2, -5) = 20$:

$z - (-50) = -5(x - 2) + 20(y + 5)$

Step 4: Simplify the equation

$z + 50 = -5(x - 2) + 20(y + 5)$

Expand both sides:

$z + 50 = -5x + 10 + 20y + 100$

Combine like terms:

$z + 50 = -5x + 20y + 110$

Finally, subtract 50 from both sides:

$z = -5x + 20y + 60$

Thus, the equation of the tangent plane is:

$z = -5x + 20y + 60$

Let me know if you'd like more details or have any questions!

Here are some related questions you can explore:

1. What is the geometric interpretation of the tangent plane?
2. How do we compute the gradient of a multivariable function?
3. How would the tangent plane change if we moved to a different point on the surface?
4. How do we find normal vectors to surfaces using partial derivatives?
5. What are the conditions for a surface to have a horizontal tangent plane?

Tip: The gradient vector $(f_x, f_y)$ points in the direction of the steepest ascent at a given point on the surface.