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Time derivative of the vector

  1. Dec 5, 2014 #1
    1. The problem statement, all variables and given/known data
    I have somewhat general question about time derivative of a vector.
    If we have
    r=at2+b3
    it's easy to find instantaneous acceleration and velocity(derivative with respect to dt)
    v=2at+3bt2
    a=2a+6bt​
    But consider this position vector
    r=b(at-t2)​
    where b is constant vector and a is positive constant.

    2. Relevant equations
    I know that in first derivative, vector can change in both magnitude and direction, but I don't have an idea how to differentiate this thing.
    How would you find acceleration and velocity in latter case?
    What if b is vector but not constant?

    By the way,for my second question,I'm studying from University physics by Young and Friedman,and my physics course is kinda more advanced(also first year intro).Our official book is Fundamentals of Mechanics by Irodov(problem book also by Irodov),but it's so devoid of any explanations that it's hopeless to use it for anything else other than reference book.Could you recommend me some equivalent on this level but more pedagogically sound?

    3. The attempt at a solution
    I don't even know how to start.
     
  2. jcsd
  3. Dec 5, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

    You should note that [itex] \mathbf b=b_x \hat x+b_y \hat y+b_z \hat z [/itex] and [itex] f(t)\mathbf b=f(t)b_x\hat x+f(t)b_y\hat y+f(t)b_z\hat z [/itex]. Then you should know that the time derivative of a vector, is just another vector which its components are the time derivative of the components of the original vector.(Of course if you're in Cartesian coordinates, because in other coordinates, the unit vectors are changing too.)
     
  4. Dec 5, 2014 #3
    Suppose b is not given in component(cartesian) form,rather just vector form.Should I use the chain rule? If yes should i treat both b and t like independent variables and then apply chain rule?I'm still confused.
     
  5. Dec 5, 2014 #4

    ShayanJ

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    Gold Member

    [itex] \mathbf b [/itex] is a constant vector, right? So its derivative is zero and chain rule doesn't apply here!
    Also note that whether you like it or not, any vector has a component representation!
    I don't know why your complicating this simple issue for yourself. Just treat the vector as a constant and differentiate the rest w.r.t. time!
     
  6. Dec 5, 2014 #5
    If we represent a vector ##\vec{V}## in terms of its magnitude V and a unit vector ##\vec{i}_V## (oriented in the same direction as ##\vec{V}##), and, if the vector ##\vec{V}## is changing both magnitude and direction, then:

    [tex]\frac{d\vec{V}}{dt}=\frac{dV}{dt}\vec{i}_V+V\frac{d\vec{i}_V}{dt}=\frac{dV}{dt}\vec{i}_V+V\left[\frac{d(\vec{i}_V\centerdot \vec{i}_x)}{dx}\vec{i}_x\frac{dx}{dt}+
    \frac{d(\vec{i}_V\centerdot \vec{i}_y)}{dy}\vec{i}_y\frac{dy}{dt}+\frac{d(\vec{i}_V\centerdot \vec{i}_z)}{dz}\vec{i}_z\frac{dz}{dt}\right][/tex]
     
    Last edited: Dec 5, 2014
  7. Dec 5, 2014 #6
    Guess I didn't know how to differentiate unit vector properly.Thanks.
     
  8. Dec 5, 2014 #7
    My apologies. I guess I didn't either. I had to go back and correct it to this:

    [tex]\frac{d\vec{V}}{dt}=\frac{dV}{dt}\vec{i}_V+V\frac{d\vec{i}_V}{dt}=\frac{dV}{dt}\vec{i}_V+V\left[\frac{d(\vec{i}_V\centerdot \vec{i}_x)}{dx}\vec{i}_x\frac{dx}{dt}+
    \frac{d(\vec{i}_V\centerdot \vec{i}_y)}{dy}\vec{i}_y\frac{dy}{dt}+\frac{d(\vec{i}_V\centerdot \vec{i}_z)}{dz}\vec{i}_z\frac{dz}{dt}\right][/tex]

    Chet
     
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