Math Problem Statement

The parametric curve x(t)=sin(2⋅t),y(t)=−cos(2⋅t)−1,z(t)=2⋅cos(t) lies on the following surfaces: a sphere of equation what? and a parabolic cylinder of equation what?

Solution

We are tasked with identifying two surfaces that the given parametric curve lies on:

x(t)=sin(2t),y(t)=cos(2t)1,z(t)=2cos(t).x(t) = \sin(2t), \quad y(t) = -\cos(2t) - 1, \quad z(t) = 2\cos(t).

1. Equation of the Sphere

A sphere is defined by an equation of the form:

x2+y2+z2=R2,x^2 + y^2 + z^2 = R^2,

where RR is the radius of the sphere. We substitute the parametric expressions into this equation and simplify:

x2(t)+y2(t)+z2(t)=(sin(2t))2+(cos(2t)1)2+(2cos(t))2.x^2(t) + y^2(t) + z^2(t) = (\sin(2t))^2 + (-\cos(2t) - 1)^2 + (2\cos(t))^2.

Simplify each term:

  • (sin(2t))2+(cos(2t))2=1(\sin(2t))^2 + (\cos(2t))^2 = 1 (Pythagorean identity),
  • (cos(2t)1)2=cos2(2t)+2cos(2t)+1(-\cos(2t) - 1)^2 = \cos^2(2t) + 2\cos(2t) + 1,
  • (2cos(t))2=4cos2(t)(2\cos(t))^2 = 4\cos^2(t).

Combine these: x2+y2+z2=1+(cos2(2t)+2cos(2t)+1)+4cos2(t).x^2 + y^2 + z^2 = 1 + (\cos^2(2t) + 2\cos(2t) + 1) + 4\cos^2(t).

Using the double-angle identity cos2(2t)=2cos2(t)1\cos^2(2t) = 2\cos^2(t) - 1, we rewrite: x2+y2+z2=1+((2cos2(t)1)+2cos(2t)+1)+4cos2(t).x^2 + y^2 + z^2 = 1 + ((2\cos^2(t) - 1) + 2\cos(2t) + 1) + 4\cos^2(t).

Simplify: x2+y2+z2=1+(2cos2(t)1+2(2cos2(t)1)+1)+4cos2(t).x^2 + y^2 + z^2 = 1 + (2\cos^2(t) - 1 + 2(2\cos^2(t) - 1) + 1) + 4\cos^2(t).

x2+y2+z2=1+8cos2(t).x^2 + y^2 + z^2 = 1 + 8\cos^2(t).

Using cos2(t)=12(1+cos(2t))\cos^2(t) = \frac{1}{2}(1 + \cos(2t)), we get 8cos2(t)=4(1+cos(2t))=4+4cos(2t)8\cos^2(t) = 4(1 + \cos(2t)) = 4 + 4\cos(2t). Substituting, the final equation becomes:

x2+y2+z2=5.x^2 + y^2 + z^2 = 5.

Thus, the sphere's equation is: x2+y2+z2=5.x^2 + y^2 + z^2 = 5.


2. Equation of the Parabolic Cylinder

A parabolic cylinder has an equation of the form y=ax2+by = ax^2 + b. Substitute x(t)=sin(2t)x(t) = \sin(2t) into y(t)=cos(2t)1y(t) = -\cos(2t) - 1.

Using cos(2t)=12sin2(t)\cos(2t) = 1 - 2\sin^2(t), we rewrite y(t)y(t): y(t)=(12sin2(t))1=1+2sin2(t)1=2sin2(t)2.y(t) = -(1 - 2\sin^2(t)) - 1 = -1 + 2\sin^2(t) - 1 = 2\sin^2(t) - 2.

Using sin2(t)=12(1cos(2t))\sin^2(t) = \frac{1}{2}(1 - \cos(2t)), substitute into y(t)y(t): y(t)=2(12(1cos(2t)))2=(1cos(2t))2.y(t) = 2\left(\frac{1}{2}(1 - \cos(2t))\right) - 2 = (1 - \cos(2t)) - 2.

Thus: y=cos(2t)1=x22.y = -\cos(2t) - 1 = x^2 - 2.

The parabolic cylinder's equation is: y=x22.y = x^2 - 2.


Final Answers

  1. The parametric curve lies on the sphere: x2+y2+z2=5.x^2 + y^2 + z^2 = 5.

  2. The parametric curve lies on the parabolic cylinder: y=x22.y = x^2 - 2.


Would you like further clarification on any of these steps?
Here are related questions you might find interesting:

  1. How do you determine if a curve lies entirely on a surface?
  2. What are other surfaces that can intersect parametric curves?
  3. What is the geometric interpretation of parametric curves in 3D?
  4. How do trigonometric identities simplify curve analysis?
  5. Can parametric curves define intersections of multiple surfaces?

Tip: Use parametric equations to verify points on surfaces by substitution; it's a straightforward test!

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
3D Geometry
Trigonometric Identities
Surface Equations

Formulas

Equation of a sphere: x^2 + y^2 + z^2 = R^2
Equation of a parabolic cylinder: y = ax^2 + b
Pythagorean identity: sin^2(θ) + cos^2(θ) = 1
Double-angle identity: cos(2θ) = 2cos^2(θ) - 1

Theorems

Pythagorean Theorem
Trigonometric Identities

Suitable Grade Level

Grades 11-12 and early college