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Chapter 14: Problem 60

(a) Where does the surface \(x^{2}+y^{2}-(z-1)^{2}=0\) cut the \(x y\) -plane? What is the shape of the curve? (b) At the points where the surface cuts the \(x y\) -plane, do vectors perpendicular to the surface lie in the \(x y\) -plane?

Short Answer

Expert verified
The surface intersects the \(xy\)-plane as a circle. Perpendicular vectors at intersections do not lie in the \(xy\)-plane.

Step by step solution

01

Introduction to the Surface Equation

The given equation is \(x^2+y^2-(z-1)^2=0\). We can rewrite it as \(x^2 + y^2 = (z - 1)^2\). This represents a set of surfaces where, for every \(z\), \(x^2 + y^2\) equals \((z - 1)^2\).
02

Intersection with the xy-plane

To find where the surface cuts the \(xy\)-plane, we set \(z = 0\). Substituting \(z = 0\) into the equation gives \(x^2 + y^2 = (0 - 1)^2 = 1\). Thus, the curve is \(x^2 + y^2 = 1\), a circle with radius 1 centered at the origin.
03

Shape of the Curve in xy-plane

The curve \(x^2 + y^2 = 1\) is a circle centered at the origin, with a radius of 1. This is a standard equation for a circle in the \(xy\)-plane.
04

Checking Perpendicular Vectors at Intersection

To check if the vectors perpendicular to the surface lie in the \(xy\)-plane, we find the gradient \(abla F\) of the surface equation \(F(x, y, z) = x^2 + y^2 - (z - 1)^2\). The gradient is \(abla F = (2x, 2y, -2(z - 1))\).
05

Evaluating the Gradient at the Intersection

At the points where \(z = 0\) (intersection with the \(xy\)-plane), the gradient becomes \(abla F = (2x, 2y, 2)\). Since the \(z\)-component is 2, perpendicular vectors at these points are not confined to the \(xy\)-plane and have a component in the \(z\)-direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Intersection
A surface intersection occurs when a three-dimensional surface crosses a specified plane, such as the xy-plane. In this particular case, we're examining where the surface defined by the equation \(x^2+y^2-(z-1)^2=0\) intersects the xy-plane. By setting \(z = 0\), we substitute it into the equation, which simplifies to \(x^2 + y^2 = 1\). This describes a circle with a radius of 1, centered at the origin in the xy-plane. Understanding how surfaces and planes intersect helps visualize 3D structures and their corresponding cross-sections in 2D planes.
  • Set z to 0 for xy-plane intersection.
  • The result is a 2D equation in x and y.
  • The shape of the intersection is a circle.
Gradient Vector
The gradient vector is a crucial concept in multivariable calculus. It gives us a way to find a vector normal, or perpendicular, to a surface at any given point. For the surface described by \(F(x, y, z) = x^2 + y^2 - (z - 1)^2\), the gradient vector is calculated as \(abla F = (2x, 2y, -2(z - 1))\).
  • The components of the gradient vector: \(2x\), \(2y\), and \(-2(z - 1)\).
  • The gradient points in the direction of greatest increase of the function.
  • It is useful for finding normals to surfaces.
When calculated at a point, this vector indicates the direction of steepest ascent on the surface. Its components can reveal insights about the surface's geometry at that precise location.
Circle Equation
The equation of a circle in the xy-plane is one of the most fundamental forms in geometry and calculus. The standard equation is \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle. From the solution provided, at the intersection of the surface with the xy-plane, we find that the curve is \(x^2 + y^2 = 1\).
  • This is a circle centered at the origin \((0,0)\).
  • The radius of this circle is 1.
  • It confirms the intersection's geometry.
Circles have special properties, like constant radius, which make them simple yet powerful in understanding intersections within the plane.
Perpendicular Vectors
Perpendicular vectors play a vital role in analyzing surfaces and their orientations. When examining the intersection of a surface with the xy-plane, it鈥檚 important to consider whether vectors normal to the surface lie wholly within the xy-plane. To assess this, we can use the gradient vector, which serves as a normal to the surface.
  • Calculated perpendicular vector: \(abla F = (2x, 2y, 2)\) at \(z = 0\).
  • A '2' in the z-component shows presence above the plane.
  • Confirming not confined to xy-plane entirely.
Since the z-component of the gradient is 2, it indicates that at the points where the surface intersects the xy-plane, the perpendicular vectors indeed extend out of the xy-plane, affirming a three-dimensional aspect to their orientation.

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