Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all points \((x, y)\) at which the tangent plane to the graph of \(z=x^{2}-6 x+2 y^{2}-10 y+2 x y\) is horizontal.

Step-by-Step Solution

Answer

The point is \((1, 2)\).

Step 1: Understand the Problem

To find the points where the tangent plane to the surface is horizontal, we need to find points where the partial derivatives of the function with respect to both \(x\) and \(y\) are zero. At these points, the slopes in all directions in the plane will be zero.

Step 2: Calculate Partial Derivatives

First, compute the partial derivatives of the function \(z=f(x,y)=x^2-6x+2y^2-10y+2xy\). The partial derivative with respect to \(x\) is \(f_x=2x-6+2y\). The partial derivative with respect to \(y\) is \(f_y=4y-10+2x\).

Step 3: Set Partial Derivatives to Zero

Set the partial derivatives equal to zero to find critical points:1. \(f_x = 2x - 6 + 2y = 0\) 2. \(f_y = 4y - 10 + 2x = 0\)

Step 4: Solve the System of Equations

Solve the system of equations derived from setting the partial derivatives to zero:1. From \(2x - 6 + 2y = 0\), rearrange to get \(x + y = 3\).2. From \(4y - 10 + 2x = 0\), rearrange to get \(x + 2y = 5\). Now, solve these two simultaneous equations.

Step 5: Find the Solution

Solve the equations: 1. Substitute \(x = 3 - y\) into \(x + 2y = 5\): \[(3 - y) + 2y = 5\] \[3 + y = 5\] \[y = 2\]2. Substitute \(y = 2\) back into \(x = 3 - y\): \[x = 3 - 2 = 1\].Thus, the point \((x, y)\) is \((1, 2)\).

Key Concepts

Partial Derivatives

When dealing with multivariable functions like the one presented in this exercise, understanding partial derivatives is essential. A partial derivative represents the rate of change of a function concerning one variable while keeping the other variables constant.
In mathematical terms, for a function of two variables, say \( z = f(x, y) \), the partial derivative with respect to \( x \) (denoted as \( f_x \)) assesses how \( z \) changes as \( x \) changes, while \( y \) is held constant. Similarly, \( f_y \) examines changes in \( z \) with variations in \( y \), holding \( x \) consistent.
In the given problem, calculating the partial derivatives \( f_x = 2x - 6 + 2y \) and \( f_y = 4y - 10 + 2x \) is vital to determining the conditions where the tangent plane is horizontal.

Critical Points

Critical points in the context of functions with two variables, like \( f(x, y) \), are the points where the partial derivatives equal zero. This means there is no change in the function value concerning either direction.
These points often help identify regions on a surface where a tangent plane is horizontal, or the function has potential maximum, minimum, or saddle points.
Finding these critical points involves:

• Calculating the partial derivatives for the function.
• Setting these derivatives equal to zero, as demonstrated in the solution with equations \( f_x = 0 \) and \( f_y = 0 \).

By solving the resulting equations, we identify the points of interest for our analysis.

System of Equations

A system of equations is a collection of two or more equations with a shared set of unknowns. In this scenario, solving these equations allows us to find where the conditions of horizontal tangency occur.
We derived the equations \( f_x = 0 \) as \( 2x - 6 + 2y = 0 \) and \( f_y = 0 \) as \( 4y - 10 + 2x = 0 \).
The steps to solve such a system often involve:

• Rearranging each equation to a simpler form, such as solving for one variable in terms of another.
• Substituting one equation into another, reducing the system to a single equation with one unknown.
• Solving the reduced equation to find values for the variables.

For this exercise, solving that system revealed that the point \((1, 2)\) holds a horizontal tangent plane.

Tangent Plane

In mathematics, the concept of a tangent plane is analogous to understanding tangent lines in calculus, but in a 3-dimensional space. It deals with planar approximations of surfaces and provides insight into slopes and the natural behavior of surfaces at particular points.
For the function \( z = f(x, y) \), the tangent plane can be described at a specific point \((x_0, y_0)\), where both partial derivatives have been computed.
The equation of this plane is typically given in linear expansion:

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

In our problem, since both partial derivatives are zero at the critical point \((1, 2)\), the tangent plane there is horizontal, meaning it has an equation like \( z = ext{constant} \), indicating no slope in any direction.