1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show that the vector value function is on the surface of a sphere

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data

    r(t) = [cos(t^3)sin(t), sin(t)sin(t^3), cos(t)], where t is an element of (0,2pi).

    Show that the graph of this function is on the surface of a sphere. Then find it's radius.

    2. Relevant equations

    T(t) = r'(t)/norm[r'(t)]

    equation for a sphere in 3-space: r^2=x^2+y^2+z^2

    3. The attempt at a solution

    My thought is, a tangent vector must lie on the surface of this function (which is a sphere), and therefore the tangent vector at some point (t) would be sufficient proof. But that seems too easy.

    Secondly, to find the radius of a sphere, set up the parametric equations in the form of x= cos(t^3)sin(t), y = sin(t^3)sin(t), z = cos(t), in the form of the equation for a sphere:

    (cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2 = r^2

    then substitute one of the elements to find the radius squared?
     
  2. jcsd
  3. Nov 11, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Too easy and wrong too. The tangent vector is always tangent to any surface.
    You need to show that is an identity. Just simplify it.
     
  4. Nov 11, 2011 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Just expand your expression a little bit and use trig. Like sin(x)^2+cos(x)^2=1.
     
  5. Nov 11, 2011 #4
    r^2 = x^2 + y^2 + z^2

    r^2 = (cos(t^3)sin(t))^2 + (sin(t^3)sin(t))^2 + (cos(t))^2

    r^2 = [cos^2(t^3) sin^2(t)] + [sin^2(t^3)sin^2(t)] + [cos^2(t)]

    since r^2 = 1, and the other side simplifies to 1, the graph of the function lies on the surface of a sphere?

    Since it simplifies to 1, and the root of 1 is one, the radius is 1.

    Am i on the right track?
     
  6. Nov 11, 2011 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. If you show |r(t)| = 1 for all t that does it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Show that the vector value function is on the surface of a sphere
  1. Vector-Valued Function (Replies: 1)

Loading...