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Why are these perpendicular?

  1. Sep 27, 2014 #1
    Hi all,

    When you have a surface defined by [itex] F(x, y, z) = 0 [/itex] where [itex]x = f(t)[/itex], [itex]y= g(t)[/itex] and [itex]z= h(t)[/itex] and a point on this surface [itex] P_0 = (x_0, y_0, z_0) [/itex], could someone explain to me why a line through [itex] P_0 [/itex] with direction numbers [itex] [\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}] [/itex] is perpendicular to a line through [itex] P_0 [/itex] with direction numbers [itex] [\frac{\partial F}{dx}, \frac{\partial F}{dy}, \frac{\partial F}{dz}] [/itex]?

    I'm having real trouble picturing it in my head, which means I'm struggling to understand why it is so.

    Thanks!
     
  2. jcsd
  3. Sep 27, 2014 #2

    Erland

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    Hint: Use the Chain rule to calculate rhe derivative w.r.t t of the composite function F(f(t),g(t),h(t)). What is this derivative? How can it be interpreted in terms of scalar product (inner product)?
    It must be assumed here that the curve (x,y,z)=(f(t),g(t),h(t)) lies on the surface F(x,y,z)=0.
     
  4. Sep 27, 2014 #3
    Well, the derivative wrt [itex] t [/itex] is: [tex] \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} + \frac{\partial F}{\partial x}\frac{dz}{dt} [/tex] Setting this to zero and comparing it to the scalar product of the two vectors comprising the direction numbers would mean that the angle between them had to be 90 degrees, hence they are perpendicular... I guess it's ok to set the derivative to zero, since the original surface [itex] F(x,y,z) [/itex] is zero, right? So that all kinda makes sense to me. Yeah... Thanks! I think I've got it!
     
  5. Sep 27, 2014 #4

    Erland

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    You got it :)
     
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