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Vector v = a i +bxj

  1. May 26, 2008 #1
    I have a vector v = a i +bxj

    Now I want to find a.So I apply chain rule and differentiate dv/dx * dx/dt = bj * Vx = bj*ai ??

    Now what does this mean? Is this allright? Should I not care about the vector directions when differentiating and just use the magnitudes a = ab , if yes then why? And what will be the direction of the acceleration then?
  2. jcsd
  3. May 26, 2008 #2

    [tex]d/dt = \mathbf{v}\cdot \nabla[/tex]

    \frac{d\mathbf{v}}{dt} = \frac{\partial v_x}{\partial x}\frac{dx}{dt} + \frac{\partial v_y}{\partial y}\frac{dy}{dt}
    and we know that

    [tex]\frac{dx}{dt} = a, \frac{dy}{dt} = bx, \frac{\partial v_x}{\partial x} = 0, \frac{\partial v_y}{\partial y} = 0[/tex]

    and then you get...
  4. May 27, 2008 #3
    I dont get what you have written. Please tell me the physics and not just the math. How can a be zero ?
    Last edited: May 27, 2008
  5. May 27, 2008 #4
    Hold on a second I did it wrong, didn't I? No one corrected me yet...

    [tex]\frac{dv_x}{dt} = \frac{\partial v_x}{\partial x}\frac{dx}{dt} + \frac{\partial v_x}{\partial y}\frac{dy}{dt}[/tex]

    [tex]\frac{dv_y}{dt} = \frac{\partial v_y}{\partial x}\frac{dx}{dt} + \frac{\partial v_y}{\partial y}\frac{dy}{dt}[/tex]

    Which is just applying the chain rule to the components of v separately. Now that should make sense, it didn't before because it was wrong.
  6. May 27, 2008 #5
    [tex]a_x = 0[/tex]

    [tex]a_y = ab[/tex]

    I'm sorry about my stupidity.
  7. May 27, 2008 #6
    just differentiate the magnitudes. derivatives does not effect the basis vectors. so you write the related basis vector after the derivation. and it will give you the direction of the acceleration.
  8. May 27, 2008 #7
    That's wrong. Consider the counterexample-- centripetal motion. The speed is constant in time, so by your method you would conclude that acceleration is 0, but it's not.

    And in this problem, differentiating the magnitude would also give you the wrong answer.
  9. May 28, 2008 #8
    that's not totally wrong. if you should use "the polar coordinate" system and then you can find the direction.
  10. May 28, 2008 #9
    No, the polar coordinate basis vectors are not constant.

    Face it, once you leave 1 dimensional motion, you can not characterize position, velocity and acceleration by their magnitudes.
  11. May 28, 2008 #10
    I'm not sure I understand, is a a constant? what do you understand by the vector a ?
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