Problem 40 - Calculus Early Transcendentals

Question

\(39-44\) Find equations of (a) the tangent plane and (b) the normal line to the given surface at the specified point. $$y=x^{2}-z^{2}, \quad(4,7,3)$$

Step-by-Step Solution

Answer

Tangent plane: \(8x + y - 6z = 23\); Normal line: \(x = 4 + 8t, y = 7 + t, z = 3 - 6t\).

Step 1: Understand the Problem

We need to find the equations for both the tangent plane and the normal line of the surface given by the equation \( y = x^2 - z^2 \) at the point \((4, 7, 3)\).

Step 2: Calculate Partial Derivatives

The first step in finding the tangent plane is to calculate the partial derivatives of the surface equation with respect to \(x\), \(y\), and \(z\). The given surface equation is \( y = x^2 - z^2 \). Differentiate with respect to \(x\): \( \frac{\partial y}{\partial x} = 2x \). Similarly, differentiate with respect to \(z\): \( \frac{\partial y}{\partial z} = -2z \). The partial derivative with respect to \(y\) is \( \frac{\partial y}{\partial y} = 1 \).

Step 3: Evaluate Partial Derivatives at Given Point

Substitute the point \((4, 7, 3)\) into the partial derivatives. So, \( \frac{\partial y}{\partial x} \) evaluated at \((4, 7, 3)\) is \( 2 \times 4 = 8 \), and \( \frac{\partial y}{\partial z} \) evaluated at \((4, 7, 3)\) is \( -2 \times 3 = -6 \). \( \frac{\partial y}{\partial y} \) remains 1.

Step 4: Equation of the Tangent Plane

The equation of the tangent plane to a surface \( f(x, y, z) = 0 \) at the point \((x_0, y_0, z_0)\) is given by \( \frac{\partial f}{\partial x}(x_0, y_0, z_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0, y_0, z_0)(y-y_0) + \frac{\partial f}{\partial z}(x_0, y_0, z_0)(z-z_0) = 0 \). Substitute the derivatives and point values: \( 8(x-4) + 1(y-7) - 6(z-3) = 0 \). Simplify to get the tangent plane equation: \( 8x + y - 6z = 23 \).

Step 5: Equation of the Normal Line

The direction vector of the normal line is given by the coefficients of \( x, y, z \) in the tangent plane equation: \((8, 1, -6)\). The parametric equations of the normal line are \( x = 4 + 8t \), \( y = 7 + t \), \( z = 3 - 6t \) where \( t \) parameterizes the line.

Key Concepts

Normal Line

When we talk about the normal line in the context of a surface, we refer to a line that is perpendicular to the tangent plane at a particular point on the surface. Think of it as the upright edge that sticks out from the face of the plane.The normal line is important because it helps us understand the orientation of the surface at a point more completely than just the tangent plane. The direction of the normal line is determined by the gradient vector of the surface at that point.In our example, for the surface equation \( y = x^2 - z^2 \), the tangent plane at point \((4, 7, 3)\) was found using partial derivatives. Once we had these derivatives, we could find the direction vector for the normal line.

• The gradient vector or the coefficients from the tangent plane equation \( 8x + y - 6z = 23 \) becomes the direction \([8, 1, -6]\).
• Using this direction vector, the normal line is described with parametric equations.
• These equations are \( x = 4 + 8t \), \( y = 7 + t \), \( z = 3 - 6t \), where \( t \) is a parameter.

This line shows how the surface extends perpendicularly to the surrounding space at that precise point.

Partial Derivatives

Partial derivatives are like the focus lenses of calculus that allow us to examine how a function changes as each variable changes, one at a time.For functions with more than one variable, such as surfaces, partial derivatives give us the rate of change with respect to each variable independently. This is essential for understanding the behavior and shape of the surface.

• In our exercise, the equation \( y = x^2 - z^2 \) requires us to take partial derivatives concerning \(x\), \(y\), and \(z\).
• We found this to be \( \frac{\partial y}{\partial x} = 2x \), \( \frac{\partial y}{\partial z} = -2z \), and \(\frac{\partial y}{\partial y} = 1\).

Calculating these values at the given point \((4, 7, 3)\) helps us to describe how the surface tilts or leans at this specific place.By using the partial derivatives, we've constructed a deeper understanding of the surface's texture and movement.

Surface Equation

A surface equation is a mathematical representation that describes a two-dimensional surface within a three-dimensional space. It's like a blueprint for the plane or shape, indicating how it behaves and appears.In our case, the surface equation we are working with is \( y = x^2 - z^2 \). This equation characterizes the relationship between the variables \(x\), \(y\), and \(z\) across a surface.

• The equation itself tells us how to calculate points on this surface.
• The role of surface equations in problems like these is foundational, providing a starting point for determining things like tangent planes or normal lines.

By exploring this equation with calculus tools such as partial derivatives, we can extract detailed information about the slope, curvature, and orientation of the surface at any specified point.Understanding this equation is crucial as it lays the groundwork for visualizing how the surface would look in three dimensions.

Parametric Equations

Parametric equations are like a set of instructions that tell us how to trace a path in space, representing complex curves and lines more dynamically than by a standard function.Instead of dealing directly with \(x, y, z\) coordinates, parametric equations introduce a new variable, often called a parameter (like \(t\)), which lets us express \(x\), \(y\), and \(z\) as functions of this new variable.

• In our example, the normal line is expressed in parametric form using \(t\):
• \(x = 4 + 8t\), \(y = 7 + t\), \(z = 3 - 6t\).

This use of parameter \(t\) allows us to describe each position along the normal line's direction seamlessly.The key advantage of using parametric equations is their flexibility. We can explore motion or paths that might be awkward or impossible to describe with simple equations. They are handy for defining paths in computer graphics and animations.