Find An Equation Of The Tangent Plane To The Surface Z=4y^2 - 2x^2 At The Point (4, -2, -16).
Understanding how to find the equation of a tangent plane to a surface at a specific point is a fundamental concept in multivariable calculus. This process involves calculating the gradient of the surface's defining function and then using it to formulate the tangent plane equation. In this article, we will walk through each step in detail, illustrating how to derive the tangent plane equation for the surface \( Z = 4y^2 - 2x^2 \) at the point \( (4, -2, -16) \). This guide aims to provide clarity and depth, making it accessible for students and enthusiasts alike.
Understanding the Surface and Its Equation
Surface Equation Overview
The surface in question is defined implicitly by the function:\[
Z = 4y^2 - 2x^2
\]
This is a quadratic surface involving both \( x \) and \( y \), defining a three-dimensional shape in space. The goal is to find the equation of the tangent plane at a specific point on this surface.
Point of Interest
The point provided is:\[
( x0, y0, z_0 ) = ( 4, -2, -16 )
\]
Before proceeding, it's prudent to verify that this point indeed lies on the surface. Substituting \( x = 4 \) and \( y = -2 \) into the surface equation:
\[
Z = 4(-2)^2 - 2(4)^2 = 4(4) - 2(16) = 16 - 32 = -16
\]
The calculated \( Z \) matches the given \( z_0 = -16 \). Therefore, the point lies on the surface, confirming the appropriateness of the point for tangent plane calculation.
Mathematical Foundations for Finding the Tangent Plane
Implicit Function and Gradient Vector
In multivariable calculus, the tangent plane to a surface at a point can be found if the surface is expressed as an implicit function:\[
F(x, y, z) = 0
\]
Rearranged from the given surface equation:
\[
F(x, y, z) = 4y^2 - 2x^2 - z = 0
\]
The gradient vector of \( F \), denoted as \( \nabla F \), points in the direction of the greatest rate of increase of \( F \), and it is perpendicular to the surface at the point.
Calculating the gradient:
\[
\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)
\]
This vector will serve as the normal vector to the tangent plane.
Formula for the Tangent Plane
The general equation of the tangent plane at a point \( (x0, y0, z_0) \) is:\[
\nabla F (x0, y0, z0) \cdot \left( (x - x0), (y - y0), (z - z0) \right) = 0
\]
or explicitly,
\[
\left. \frac{\partial F}{\partial x} \right|{(x0, y0, z0)} (x - x0) + \left. \frac{\partial F}{\partial y} \right|{(x0, y0, z0)} (y - y0) + \left. \frac{\partial F}{\partial z} \right|{(x0, y0, z0)} (z - z_0) = 0
\]
This formula provides a straightforward method to derive the tangent plane once the gradient is known.
Calculating Partial Derivatives of the Surface Function
Partial Derivatives of \( F(x, y, z) \)
Given:\[
F(x, y, z) = 4y^2 - 2x^2 - z
\]
Calculate each partial derivative:
- With respect to \( x \):
\[
\frac{\partial F}{\partial x} = -4x
\]
- With respect to \( y \):
\[
\frac{\partial F}{\partial y} = 8y
\]
- With respect to \( z \):
\[
\frac{\partial F}{\partial z} = -1
\]
Evaluate these derivatives at the point \( (4, -2, -16) \):
- \( \frac{\partial F}{\partial x} \big|_{(4, -2, -16)} = -4(4) = -16 \)
- \( \frac{\partial F}{\partial y} \big|_{(4, -2, -16)} = 8(-2) = -16 \)
- \( \frac{\partial F}{\partial z} \big|_{(4, -2, -16)} = -1 \)
The gradient vector at the point is therefore:
\[
\nabla F (4, -2, -16) = (-16, -16, -1)
\]
Formulating the Equation of the Tangent Plane
Applying the Gradient to the Point
Using the formula:\[
-16 (x - 4) + (-16)(y + 2) + (-1)(z + 16) = 0
\]
Note: Since \( y0 = -2 \), \( y - y0 = y + 2 \); similarly for \( x \), \( x - x0 = x - 4 \); and for \( z \), \( z - z0 = z + 16 \).
Simplifying the Equation
Distribute:\[
-16x + 64 - 16y - 32 - z - 16 = 0
\]
Combine like terms:
\[
-16x - 16y - z + (64 - 32 - 16) = 0
\]
\[
-16x - 16y - z + 16 = 0
\]
Rearranged:
\[
-16x - 16y - z = -16
\]
To make the equation more standard, multiply through by \( -1 \):
\[
16x + 16y + z = 16
\]
This is the equation of the tangent plane to the surface at the point \( (4, -2, -16) \).
Final Equation of the Tangent Plane
\[
\boxed{
16x + 16y + z = 16
}
\]
This linear equation describes the tangent plane at the specified point. It provides a precise geometric representation of how the surface behaves in the immediate vicinity of \( (4, -2, -16) \).
Conclusion and Additional Insights
Understanding how to find the tangent plane to a surface is essential for analyzing local behavior in multivariable calculus. The process involves:
- Confirming the point lies on the surface.
- Expressing the surface as an implicit function.
- Calculating the gradient vector of the function at the point.
- Using the gradient vector as the normal vector to the tangent plane.
- Formulating the tangent plane equation using the point-normal form.
This method is broadly applicable to various surfaces, including spheres, cylinders, and more complex shapes. Mastery of these techniques enhances spatial reasoning and enables deeper insights into the geometry of multivariable functions.
Whether you're studying for exams, working on engineering projects, or simply exploring the fascinating world of calculus, understanding how to derive tangent planes equips you with a powerful tool to analyze and visualize three-dimensional surfaces effectively.
If you'd like further assistance with related topics such as normal vectors, derivatives of multivariable functions, or applications in physics and engineering, numerous resources and tutorials are available online to deepen your understanding.
