Question

Use a computer algebra system to graph the curve formed by the intersection of the surface and the plane. Find the slope of the curve at the given point. $$ z=9 x^{2}-y^{2} \quad x=1 \quad (1,3,0) $$

Short answer

The slope of the curve at the point (1,3,0) is -6.

Step-by-step solution

Find the equation of the curve

Substitute \(x=1\) into the equation \(z=9x^{2}-y^{2}\) get the function describing the curve.

Simplify the equation

Simplify to get the equation \(z=9-y^{2}\). This is the equation of the curve formed at the intersection of the surface and the plane.

Find the slope of the curve

To find the slope of the curve at the given point (1,3,0), we need to find the derivative of \(z\) with respect to \(y\). The derivative of \(z=9-y^2\) with respect to \(y\) is \(dz/dy=-2y\).

Calculate the slope at the given point

Substitute \(y=3\) into \(dz/dy=-2y\) to find the slope at the given point. We get \(dz/dy=-2(3)=-6\).

Key concepts

Intersection of Surfaces

The intersection of surfaces occurs when two or more surfaces cross each other, creating a curve or line that represents all the points common to the surfaces involved. In the context of our exercise, a surface defined by the equation z=9x^2-y^2 is intersecting with the plane where x=1. By substituting the value of x into the surface equation, we effectively slice the surface along a plane parallel to the yz-axis, revealing the curve of intersection.

Graphing this intersection with a computer algebra system provides a visual representation of both the surface and the plane, and the resultant curve can be distinctly viewed where they meet. This visual aid is essential for students to fully grasp the concept and locate the point of intersection accurately within a three-dimensional coordinate system.

Plane and Surface Intersection

Understanding plane and surface intersections is crucial as it extends further than the simple line intersections in two-dimensional space. A plane is defined by a linear equation in three variables (typically x, y, and z), and it extends infinitely in two dimensions. When this plane intersects a surface, which is a two-dimensional form in three-dimensional space, the result is frequently a one-dimensional curve.

In our exercise, setting x=1 imposes a vertical plane that cuts through the surface. The equation of the curve produced by this section is simplified to z=9-y^2, which is far easier to work with and enables us to investigate properties of the curve, such as slope, at specific points.

Finding Slope of a Curve

The slope of a curve at a point is an essential concept in calculus analogous to the slope of a straight line but applied to curves which require consideration of instantaneous rates of change. To find the slope of the curve resulting from the intersection of the plane x=1 and the surface z=9x^2-y^2, we calculate the derivative of the curve's equation with respect to one of its variables.

In this case, since we are dealing with a vertical plane where x is constant, we focus on how z changes with respect to y. The curve's slope at a specific point provides us with the gradient or 'steepness' of the curve at that location, hence allowing us to visualize and understand the behavior of the curve in the vicinity of the point.

Derivative with Respect to Y

The concept of a derivative is fundamental in calculus, representing the rate at which a function is changing at any given point. For curves in three-dimensional space, you may be required to find the derivative with respect to variables other than x, such as y or z. This is especially true for curves represented in the yz-plane.

In our exercise, taking the derivative of z=9-y^2 with respect to y yields dz/dy=-2y. This derivative allows us to compute the slope of the curve at the given point (1,3,0), which is particularly useful for determining the direction and steepness of the curve at that specific location, affecting the interpretation of the graph's shape and aiding in the understanding of the curve's properties in its context.