(1 point) Find an equation of the tangent plane to the surface 𝑧=3𝑥−(2𝑦2+2𝑥2)+2
 at the point (2,−5,−50)


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.

Find an equation of the tangent plane to the surface 𝑧=3𝑥−(2𝑦^2+2𝑥^2)+2 at the point (2,−5,−50)

Learn how to find the equation of the tangent plane to the surface z = 3x - (2y^2 + 2x^2) + 2 at the point (2, -5, -50). This solution involves computing partial derivatives and applying the tangent plane theorem, suitable for university-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation