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Vector Derivatives

  1. Jul 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the equation [tex]\vec{P} = t^{3}\hat{x} + 5t^{2}\hat{y} + 10t\hat{z}[/tex]

    The tangent to the curve is [tex]3\hat{x} + 10\hat{y} + 10\hat{z}[/tex]
    When evaluated at t = 1, we get [tex]3t^2 \hat{x} + 10t \hat{y} + 10\hat{z}[/tex]

    If we take the dot product of the equation "tangent to the curve" with the same equation evaluated at t = 1 and set it to zero, we get value(s) of t where they are perpendicular.
    [tex]9t^2 +100t +100 = 0[/tex],
    which would give us two roots t = -10, and t = -10/9.


    Could someone explain to me why "we get value(s) of t where they are perpendicular", and how they are perpendicular?

    What if we had a cubed root [or higher]. If we followed the same steps, would we get values that would be perpendicular? What exactly is perpendicular?
     
  2. jcsd
  3. Jul 11, 2009 #2
    (Your "tangent to the curve" and "when evaluated at t=1" equations need to be switched).

    Two vectors are perpendicular if and only if their dot product is zero. This exercise seems to be just getting you acquainted with taking derivatives of curves and using the dot product. You took the dot product of an equation of vectors with a vector and set it equal to zero. By solving for your parameter t, you found all the vectors in your "equation of vectors" that are perpendicular to the other vector. The two vectors you found are perpendicular to 3x+10y+10z and you can check that they are by taking their dot product. Just take t=-10 for example. Now take the dot product of 3x+10y+10z and 3(-10)2x+10(-10)y+10z and you should get zero, if you did your work right.

    The geometric definition of perpendicular is: two lines are perpendicular if their intersection forms right angles. For example, if you have the unit circle, the tangent to the circle at the point (0,1) is perpendicular to the tangent at the point (-1,0). Just draw it out.
     
  4. Jul 11, 2009 #3
    thanks so much,


    Jeff
     
  5. Jul 11, 2009 #4
    No problem!
     
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