Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all points on the surface $$ z=x^{2}-2 x y-y^{2}-8 x+4 y $$ where the tangent plane is horizontal.

Step-by-Step Solution

Answer

The point is (3, -1, -14).

Step 1: Define the Tangent Plane Condition

The tangent plane is horizontal if its normal vector has no vertical component, meaning the gradient of the surface with respect to \(x\) and \(y\) has a zero \(z\)-component. Mathematically, this means the partial derivative with respect to \(z\) should equal zero, but the surface equation \( z = f(x, y) \) gives us: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1 \right) \). Therefore, the gradient \( abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = (0, 0) \).

Step 2: Calculate Partial Derivatives

Find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\):\(\frac{\partial f}{\partial x} = 2x - 2y - 8\)\(\frac{\partial f}{\partial y} = -2x - 2y + 4\)

Step 3: Set Partial Derivatives to Zero

Set each partial derivative equal to zero and solve the system of equations:1. \(2x - 2y - 8 = 0\)2. \(-2x - 2y + 4 = 0\)This simplifies to the system:1. \(x - y = 4\)2. \(x + y = 2\)

Step 4: Solve the System of Equations

Solve the equations simultaneously:From \(x - y = 4\) and \(x + y = 2\), add the equations:\(2x = 6 \Rightarrow x = 3\)Substitute \(x = 3\) back into \(x - y = 4\):\(3 - y = 4 \Rightarrow y = -1\)

Step 5: Find the Corresponding z-value

Substitute \(x = 3\) and \(y = -1\) back into the original equation to find \(z\):\(z = (3)^2 - 2(3)(-1) - (-1)^2 - 8(3) + 4(-1)\)\(= 9 + 6 - 1 - 24 - 4\)\(= -14\)

Step 6: Conclusion

The point \((x, y, z)\) on the surface where the tangent plane is horizontal is \((3, -1, -14)\).

Key Concepts

Partial Derivatives

When working with functions of multiple variables, such as the function given by the surface equation \( z = x^2 - 2xy - y^2 - 8x + 4y \), we need to use partial derivatives. A partial derivative describes how the function changes as one variable changes, while keeping the other variables constant.

For the surface in question, we calculate the partial derivatives with respect to \( x \) and \( y \) which help us understand the rate of change on the surface along these directions:

• The partial derivative \( \frac{\partial f}{\partial x} \) measures the rate of change of \( f \) as only \( x \) is varied: \( 2x - 2y - 8 \).
• The partial derivative \( \frac{\partial f}{\partial y} \) analyzes the change as we vary \( y \): \( -2x - 2y + 4 \).

Understanding these derivatives is essential for identifying where the tangent plane to the surface can be horizontal.

Gradient

The gradient is an important vector in multivariable calculus that indicates the direction of the steepest ascent of a function. For our surface equation, the gradient is particularly crucial because it helps in determining where the tangent plane can be horizontal.

The gradient \( abla f \) of the function \( f(x, y) \) is given by the vector composed of the partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \].

In our solution, the gradient is \( (2x - 2y - 8, -2x - 2y + 4) \). To have a horizontal tangent plane, we need the gradient to be \( (0, 0) \), meaning there is no change in the surface along any direction in the \( xy \)-plane.

System of Equations

A system of equations arises from the requirement that the gradient must equal the zero vector \( (0,0) \) for horizontal tangency. In our exercise, this condition translates into two equations based on the partial derivatives:

• \(2x - 2y - 8 = 0\)
• \(-2x - 2y + 4 = 0\)

Solving these simultaneously lets us find the values of \( x \) and \( y \) that guarantee a horizontal tangent plane. We solve this system by conveniently adding and subtracting the equations to eliminate variables, making it easier to find specific values for \( x \) and \( y \).

This is a key step in determining the desired points on the surface.

Horizontal Tangent Plane

A horizontal tangent plane to a surface is one that is perfectly flat, with no inclination in any direction. In mathematical terms, this means that both components of the surface's gradient are zero, resulting in no slope in any direction in the \( xy \)-plane.

In our exercise, we established that at the point where \( x = 3 \) and \( y = -1 \), the gradient \( abla f \) becomes \( (0, 0) \). This zero gradient indicates that the tangent plane at this point is horizontal.

It's important to also find the corresponding \( z \)-value by substituting \( x = 3 \) and \( y = -1 \) back into the original surface equation, yielding the point \( (3, -1, -14) \) where the tangent plane is horizontal.