Problem 2 - Calculus Early Transcendentals

Question

I-6 Find an equation of the tangent plane to the given surface at the specified point. $$ z=3(x-1)^{2}+2(y+3)^{2}+7, \quad(2,-2,12) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = 6x + 4y + 8 \).

Step 1: Understand the Problem

The problem asks for the equation of the tangent plane to the surface defined by \( z = 3(x-1)^2 + 2(y+3)^2 + 7 \) at the specific point \((2, -2, 12)\). A tangent plane has the general form: \( z = z_0 + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \), where \( f_x \) and \( f_y \) are the partial derivatives of \( z \) with respect to \( x \) and \( y \), respectively.

Step 2: Find the Partial Derivatives

Calculate the partial derivative of \( z \) with respect to \( x \): \[ f_x = \frac{\partial}{\partial x}[3(x-1)^2 + 2(y+3)^2 + 7] = 6(x-1) \] Now calculate the partial derivative of \( z \) with respect to \( y \): \[ f_y = \frac{\partial}{\partial y}[3(x-1)^2 + 2(y+3)^2 + 7] = 4(y+3) \]

Step 3: Evaluate the Partial Derivatives at the Given Point

Substitute \((x_0, y_0) = (2, -2)\) into the partial derivatives. For \( f_x \), we get: \[ f_x(2, -2) = 6(2-1) = 6 \] For \( f_y \), we get: \[ f_y(2, -2) = 4(-2+3) = 4 \]

Step 4: Write the Equation of the Tangent Plane

Using the general formula for the tangent plane, insert \( z_0 = 12 \), \( x_0 = 2 \), \( y_0 = -2 \), \( f_x(2, -2) = 6 \), and \( f_y(2, -2) = 4 \): \[ z = 12 + 6(x-2) + 4(y+2) \] This simplifies to: \[ z = 12 + 6x - 12 + 4y + 8 \] \[ z = 6x + 4y + 8 \] Hence, the equation of the tangent plane is \( z = 6x + 4y + 8 \).

Key Concepts

Partial Derivatives

Partial derivatives help us analyze how functions change when we tweak one variable while keeping others constant. Think of it as slicing through a surface to see how the elevation shifts.
In calculus, partial derivatives are crucial for functions of several variables. For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted as \( f_x \), tells us how \( f \) changes as \( x \) changes, holding \( y \) constant. Similarly, \( f_y \) is the partial derivative with respect to \( y \).

• For our specific surface \( z = 3(x-1)^2 + 2(y+3)^2 + 7 \), calculating \( f_x \) gives us \( 6(x-1) \).
• For \( f_y \), it results in \( 4(y+3) \).

These derivatives are the slopes of the surface in the direction of \( x \) and \( y \). They're stepping stones in finding the tangent plane.

Calculus

Calculus is the mathematics of change. It is all about finding rates of change (derivatives) and areas under curves (integrals). When you have a function of multiple variables, like in our example, calculus helps us explore how one quantity affects another, such as changes on a curved surface.
The equation of a tangent plane, \( z = z_0 + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \), is a key concept here.

• It approximates the surface near a specific point.
• By plugging in partial derivatives' results, we fine-tune this approximation.

This equation is akin to the equation of a line (y = mx + b) but for several dimensions, giving a flat surface that 'hugs' the curve exactly at a single point.

Surface Analysis

Surface analysis involves studying the shape and features of a surface, similar to understanding a landscape from bird's-eye view. By using calculus tools like partial derivatives, we dive deeper into these surfaces.
To analyze a surface, you often calculate how steep it is in different directions. Here, partial derivatives come into play, showing how quickly \( z \) changes when \( x \) or \( y \) changes.

• For instance, at the point \((2, -2)\), \( f_x = 6 \) implies that a unit change in \( x \) results in a change of \( 6 \) in \( z \).
• Similarly, \( f_y = 4 \) suggests that a unit change in \( y \) changes \( z \) by \( 4 \).

Using this data, we write the equation of a tangent plane, providing a flat 'snapshot' of the surface, making it easier to visualize and understand near that point.