Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(1-8\), 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 Function and Point

We are given the surface equation \( z = x e^{-2y} \) and asked to find the equation of the tangent plane at the point \((1, 0, 1)\). The point \((1, 0, 1)\) lies on the surface since substituting \(x=1\) and \(y=0\) into the equation yields \(z = 1 \cdot e^{0} = 1\).

Step 2: Compute Partial Derivatives

Find the partial derivative of \( z \) with respect to \( x \), holding \(y\) constant: \( \frac{\partial z}{\partial x} = e^{-2y} \). Find the partial derivative of \( z \) with respect to \( y \), holding \(x\) constant: \( \frac{\partial z}{\partial y} = -2x e^{-2y} \).

Step 3: Evaluate Partial Derivatives at the Given Point

Evaluate \( \frac{\partial z}{\partial x} \) at \((1,0)\): \( e^{0} = 1 \). Evaluate \( \frac{\partial z}{\partial y} \) at \((1,0)\): \(-2 \cdot 1 \cdot e^{0} = -2\).

Step 4: Write the Tangent Plane Equation

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 \((x_0, y_0, z_0) = (1, 0, 1)\), \( \frac{\partial z}{\partial x}|_{(1,0)} = 1 \), and \( \frac{\partial z}{\partial y}|_{(1,0)} = -2 \): \[ z - 1 = 1 \cdot (x - 1) - 2 \cdot (y - 0) \] Simplify to get: \[ z = x - 2y \]

Step 5: Verify the Equation

Check that the point \((1, 0, 1)\) satisfies the plane equation: Substituting \(x = 1\), \(y = 0\), \(z = 1\) into \( z = x - 2y \), we get \( 1 = 1 - 0 \), which is true. Therefore, the tangent plane equation is verified.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how a multivariable function changes when we vary one of the variables, keeping the others constant. For a function of two variables, such as our surface equation \( z = xe^{-2y} \), we need to explore its behavior in both the \( x \) and \( y \) directions.

• **Partial with respect to \( x \)**: We set \( y \) constant and differentiate \( z \) with respect to \( x \). For \( z = xe^{-2y} \), this yields \( \frac{\partial z}{\partial x} = e^{-2y} \). This tells us how \( z \) changes as \( x \) changes alone.
• **Partial with respect to \( y \)**: Similarly, keeping \( x \) constant, the differentiation with respect to \( y \) gives \( \frac{\partial z}{\partial y} = -2xe^{-2y} \). This demonstrates how \( z \) adjusts with changes in \( y \).

Understanding these partial derivatives is crucial for constructing the tangent plane, which approximates the surface near the point of interest.

Surfaces in Three Dimensions

Surfaces in three dimensions can be visualized as collections of points forming an infinite sheet across the 3D space. When defined by a function like \( z = f(x, y) \), it provides a height \( z \) at every point \((x, y)\) in the plane. For our specific case, \( z = xe^{-2y} \) describes a surface that stretches out as \( x \) increases and decays as \( y \) increases.

• **Visualizing the Surface**: Visualizing a surface involves imagining each point \( (x, y) \) having a corresponding height \( z \) which moves up or down. For our surface, it forms intricate exponential hills and valleys.
• **Impact of Components**: The term \( x \) impacts the overall elevation, while \( e^{-2y} \) exponentially shrinks the height with increasing \( y \). Hence, larger \( y \) values pull the surface closer to the \( xy \)-plane.

Understanding the graphical nature of surfaces helps to conceptualize the idea of tangent planes, which essentially "flatten" the surface at a specific point.

Differential Calculus

Differential calculus is a branch of mathematics that deals with how functions change when their inputs change. It's about delving into derivatives, which represent the rate of change, describing how one quantity varies in relation to another.

• **The Role in Tangent Planes**: In the context of finding tangent planes to surfaces, differential calculus aids by providing the slope of the surface across different directions. It allows us to compute partial derivatives, which are the backbone for the tangent plane equation.
• **Application in Problem**: For our problem, we've used differential calculus to derive the partial derivatives \( \frac{\partial z}{\partial x} = 1 \) and \( \frac{\partial z}{\partial y} = -2 \) at the point \((1,0,1)\), which form the coefficients in the tangent plane equation \( z = x - 2y \).

By understanding the principles of differential calculus, you gain the ability to tackle a range of problems involving changes, slopes, and instantaneous rates of change.