Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cross Products

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle is moving along the path c(t) = <cos(t),sin(t),2t>. Suppose its acceleration a(t) is given by a(t) = v(t) x B where v(t) is the velocity and B(x,y,z) is a constant vector (i.e. independent of (x,y,z)).

    Find v(t) and a(t)
    Find B?

    2. Relevant equations

    3. The attempt at a solution
    Finding v(t) is relatively simple: v(t) = <-sint(t),cos(t),2>. Finding a(t) is more tricky and I don't know how to begin. I would be appreciative of at least the first step.
  2. jcsd
  3. Oct 17, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper

    You found the velocity v(t) by taking the first derivative of c(t), v(t)=c'(t) right? The acceleration is the derivative of the velocity. Why is that trickier?
  4. Oct 17, 2011 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Well, how did you find v(t) ?

    a(t) has the same relationship to v(t) as v(t) has with c(t).

    So, if finding v(t) from c(t) was easy, finding a(t) from v(t) should be equally easy, maybe easier.
  5. Oct 17, 2011 #4
    I don't understand a(t) as simply being the derivative of v(t). That would be easy. Instead, I see a(t) as being defined as the cross product of v(t) with some vector B. Given this definition, how can I assume that a(t) = v'(t)?
  6. Oct 17, 2011 #5


    User Avatar
    Science Advisor
    Homework Helper

    v(t) and a(t) you find the usual way. a(t)=v(t) x B isn't a definition of a(t), it's a definition of B. Try and find B from that.
  7. Oct 17, 2011 #6
    Okay, that makes sense. Thank you!

    But another question: Suppose we have the surface, S, which is defined by x^2 + y^2 - z^2 + 1 = 0, and the plane, H, x - 3z = 0. I want to find all the points on S where the tangent plane is parallel to H. Here's how I tried to approach this:

    I know the vector normal to H is <1,0,-3>. Therefore, are points on S which have gradients that point in the same direction as <1,0,-3> should be parallel. So, finding the gradient of S gives

    graf f = <2x,2y,-2z> = <1,0,-3>, so,

    x = 1/2
    y = 0
    z = 3/2, so, the point (.5,0,1.5) should satisfy the condition, right? Except then I realize that this point doesn't lie on the surface S. So I don't know how to proceed.
  8. Oct 17, 2011 #7


    User Avatar
    Science Advisor
    Homework Helper

    Two vectors can point in the same direction without having the same components. c*<1,0,-3> for any constant c is also parallel to <1,0,-3>, right?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook