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Proof of distributive and product rule

  1. Aug 31, 2014 #1
    1. Prove a) r=(u*v)=r*u+r*v and b) d/dt(r*s)=r*ds/st+dr/dt*s



    2. Relevant equations: b) dr/dt=lim t->0=Δr/Δt and Δr=r(t+Δt)-r(t)



    3. Attempt at the solution:
    Okay, so I was able to work out part a but I'm not quite sure how to start part b. Could anyone point me toward a useful resource to explain how to approach this problem? I'm not looking for an answer, just a means to an answer.

    (* is dot product)
     
  2. jcsd
  3. Aug 31, 2014 #2
    Are you familiar with a proof of the standard product rule; i.e. the version that says (fg)'=f'g+fg'? The proof of this version shouldn't look significantly different.
     
  4. Aug 31, 2014 #3
    Thank you! I looked it up and you're right, I see where they start with the limit and expand from there. Much appreciated!
     
  5. Aug 31, 2014 #4
    Alright then. The key things you'll want to verify (if you haven't already) are that the algebra $$\mathbf{u}\cdot(\mathbf{v}+\mathbf{w})=\mathbf u\cdot\mathbf v+\mathbf{u}\cdot\mathbf{w},\ \ c(\mathbf{u}\cdot\mathbf{v})=(c\mathbf{u})\cdot\mathbf{v}$$ and calculus
    $$\lim_{t\rightarrow a}\left(\mathbf u(t)+\mathbf v(t)\right)=\lim_{t\rightarrow a}\mathbf u(t)+\lim_{t\rightarrow a}\mathbf v(t), \ \ \lim_{t\rightarrow a}\left(\mathbf u(t)\cdot\mathbf v(t)\right)=\lim_{t\rightarrow a}\mathbf u(t)\cdot\lim_{t\rightarrow a}\mathbf v(t)$$
    work the same with vectors/vector-valued functions as they do with numbers/real-valued functions.
     
  6. Aug 31, 2014 #5

    HallsofIvy

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    This certainly isn't true! You mean "r*(u+ v)= r*u+ r*v" don't you?

     
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