Problem 6 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The equation of the tangent plane is \(z = 2x + 2y + 1\).

Step 1: Understand the Problem

We are given a surface defined by the equation \(z = e^{x^2 - y^2}\) and asked to find the equation of the tangent plane at the point \((1, -1, 1)\).

Step 2: Find the Partial Derivatives

To find the equation of the tangent plane, we first need to compute the partial derivatives of the function with respect to \(x\) and \(y\). These partial derivatives represent the slope of the tangent plane in the \(x\) and \(y\) directions, respectively. The partial derivative with respect to \(x\) is \(\frac{\partial z}{\partial x} = 2xe^{x^2 - y^2}\). The partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = -2ye^{x^2 - y^2}\).

Step 3: Evaluate the Partial Derivatives at the Given Point

Next, we'll evaluate these partial derivatives at the specified point \((1, -1, 1)\). For \(\frac{\partial z}{\partial x}\), substitute \(x = 1\) and \(y = -1\) to get \(2(1)e^{1^2 - (-1)^2} = 2e^{0} = 2\). For \(\frac{\partial z}{\partial y}\), substitute \(x = 1\) and \(y = -1\) to get \(-2(-1)e^{1^2 - (-1)^2} = 2e^{0} = 2\).

Step 4: Set Up the Equation of the Tangent Plane

The general formula for the tangent plane to the surface \(z = f(x,y)\) at point \((x_0, y_0, z_0)\) is: \(z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0)(y - y_0)\). Substituting, we have \(z - 1 = 2(x - 1) + 2(y + 1)\).

Step 5: Simplify the Equation

Simplify the equation from Step 4: \(z - 1 = 2x - 2 + 2y + 2\). Simplifying further, this gives \(z = 2x + 2y + 1\). Thus, the equation of the tangent plane is \(z = 2x + 2y + 1\).

Key Concepts

Understanding Partial Derivatives

Partial derivatives help us understand how a multivariable function changes with respect to one variable while keeping the others constant.
They are crucial when dealing with surfaces, as these derivatives show how the surface tilts in the direction of each axis.
For a surface defined by a function like \(z = e^{x^2 - y^2}\), we compute the partial derivative with respect to \(x\) as \(\frac{\partial z}{\partial x}\).
This helps us understand the slope along the line parallel to the \(x\)-axis.

• In our exercise, \(\frac{\partial z}{\partial x} = 2xe^{x^2 - y^2}\).

To find the slope along the \(y\)-axis direction, we compute \(\frac{\partial z}{\partial y}\):

• Here, \(\frac{\partial z}{\partial y} = -2ye^{x^2 - y^2}\).

Evaluating these derivatives at a specific point gives the actual slopes of the surface at that point.
This is necessary for finding the equation of the tangent plane, as these slopes determine how the plane sits against the surface.

Forming a Surface Equation

The surface equation represents a set of points in three-dimensional space, typically of the form \(z = f(x, y)\).
This equation describes how \(z\) values change based on varying \(x\) and \(y\) coordinates.
For the given exercise, we work with the surface \(z = e^{x^2 - y^2}\).
This exponential function forms a unique surface where the values of \(z\) depend on the difference \(x^2 - y^2\).
Understanding the shape of this surface helps in visualizing concepts like slopes and tangent planes.
At any point, we can "slice" this surface to find a tangent plane, allowing us to model how the function behaves locally.
The tangent plane equation will approximate the surface near a given point, showing how it changes in \(x\) and \(y\) directions.

Deriving the Tangent Line Equation

A tangent plane linearly approximates a surface at a specific point, much like a tangent line does for curves in two dimensions.
To derive the tangent plane equation at a point \((x_0, y_0, z_0)\), we use the equation:

• \(z - z_0 = \frac{\partial z}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial z}{\partial y}(x_0, y_0)(y - y_0)\)

This equation assesses the surface's tilt at a point with slopes given by the partial derivatives.
In the exercise, substituting the partial derivatives and the point \((1, -1, 1)\) gives \(z - 1 = 2(x - 1) + 2(y + 1)\).
Simplifying further, we obtain \(z = 2x + 2y + 1\), which describes the plane just touching the surface at the given point.
This expression helps us visualize how the plane aligns with the surface, offering a simple way to study the local behavior of complex shapes.