Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent plane to the given surface at the indicated point. \(z=x e^{-2 y} ;(1,0,1)\)

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = x - 2y \).

Step 1: Identify the Surface Equation and the Point

The given surface is described by the equation \( z = x e^{-2y} \), and we need to find the equation of the tangent plane at the point \((1,0,1)\). We will denote \(x_0 = 1\), \(y_0 = 0\), and \(z_0 = 1\).

Step 2: Find the Partial Derivatives

Calculate the partial derivatives of \( z = x e^{-2y} \) with respect to \(x\) and \(y\).- The partial derivative with respect to \(x\) is \( \frac{\partial z}{\partial x} = e^{-2y} \).- The partial derivative with respect to \(y\) is \( \frac{\partial z}{\partial y} = -2x e^{-2y} \).

Step 3: Evaluate the Partial Derivatives at the Given Point

Substitute \((x_0, y_0) = (1,0)\) into the partial derivatives to get:- \( \frac{\partial z}{\partial x}(1,0) = e^{0} = 1 \)- \( \frac{\partial z}{\partial y}(1,0) = -2 \cdot 1 \cdot e^{0} = -2 \)

Step 4: Use the Tangent Plane Formula

The equation of the tangent plane is given by:\[ z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0) \cdot (x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0) \cdot (y - y_0) \]Substitute the known values:\[ z - 1 = 1(x - 1) + (-2)(y - 0) \]

Step 5: Simplify the Equation

Simplify the equation to find:\[ z - 1 = x - 1 - 2y \]Finally, rearrange to give:\[ z = x - 2y \]

Key Concepts

Partial Derivatives

Partial derivatives are an essential part of multivariable calculus. They help us understand how a function changes when only one of its variables is slightly varied, while the others remain constant. In this exercise, we have the function \( z = x e^{-2y} \), which involves two variables: \( x \) and \( y \). We calculate the partial derivative with respect to \( x \) by treating \( y \) as a constant. This results in \( \frac{\partial z}{\partial x} = e^{-2y} \). For the partial derivative with respect to \( y \), we treat \( x \) as a constant. We then find \( \frac{\partial z}{\partial y} = -2x e^{-2y} \).

• \( \frac{\partial z}{\partial x} = e^{-2y} \)
• \( \frac{\partial z}{\partial y} = -2x e^{-2y} \)

By understanding partial derivatives, we can analyze how the surface behaves near a certain point, providing a basis for finding tangent planes.

Surface Equation

The surface equation \( z = x e^{-2y} \) represents a three-dimensional surface in space. This equation tells us the height \( z \) of the surface at any point \( (x, y) \). Surfaces like these can be imagined as sheets or shapes that curve and vary over a plane. They are common in calculus problems related to rates of change, slopes, or quantities that depend on multiple factors.To understand this better, consider performing a simple evaluation: - Substitute values for \( x \) and \( y \) to see changes in \( z \). - For instance, \( z \) at \( (1,0) \) is \( 1 \times e^{-2 \times 0} = 1 \).This awareness of how \( z \) changes over \( x \) and \( y \) is vital in finding tangent planes and analyzing more complex surfaces.

Calculus Problem Solving

In calculus, solving problems usually involves finding relations between variables, rates of change, or tangent lines/planes. This exercise requires finding a tangent plane, which is a flat surface that "just touches" a curved surface at a particular point.The process involves:

• Identifying the surface equation and the given point.
• Finding partial derivatives to understand rate changes along each axis.
• Using the tangent plane formula: \[ z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0)(y - y_0) \]This guides us through forming the equation of the tangent plane. It's most useful for visualizing surfaces and solving engineering problems where flat approximations simplify complex calculations.

Evaluation at a Point

Evaluating at a point is a crucial step in determining the tangent plane. It involves substituting the specific coordinates of the given point into the partial derivatives to find the exact slopes. In this exercise, the point \( (1,0,1) \) on the surface assures us the partial derivatives give the correct inclination of the tangent.For the function \( z = x e^{-2y} \), at \( (x_0, y_0) = (1,0) \):

• \( \frac{\partial z}{\partial x}(1,0) = 1 \)
• \( \frac{\partial z}{\partial y}(1,0) = -2 \)

These evaluated values give us direct slopes in the \( x \) and \( y \) directions at the point. This ensures the tangent plane formula accurately reflects the nearby behavior of the surface. It simplifies the original complex surface into a planar equation, \( z = x - 2y \), near that point.