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Really understanding how to apply calculus.

  1. Jan 17, 2013 #1
    Aside from learning all the tools to solving differential equations and taking derivatives, I've been trying to practice actually applying this knowledge to the creation of my own equations to find out certain things. Where I'm uncertain is, have I achieved the actual expression?

    Let's start with the simple, Newton's Second Law
    ---
    Assume mass is constant with time.
    ---
    (1) [itex]F=ma[/itex]
    (2) [itex]F=m\frac{dv}{dt}[/itex] ; F(t,v)
    (3) [itex]F=m\frac{d^{2}u}{dt^{2}}[/itex] ; F(t,u,[itex]\frac{du}{dt}[/itex])

    Rate at which v changes with time (acceleration)
    (4) [itex]\frac{dv}{dt}=\frac{F}{m}[/itex]

    Rate at which F changes with time
    (5) [itex]\frac{dF}{dt}=m\frac{d^{3}u}{dt^{3}}[/itex]

    Rate at which u changes with time
    (6) [itex]\frac{du}{dt}=\frac{1}{m}∫Fdt[/itex]

    ---
    Assume m(t), no longer constant.
    ---
    Rate at which v changes with time
    (7) Same as (4)

    Rate at which F changes with time
    (8) [itex]\frac{dF}{dt}=m\frac{d^{3}u}{dt^{3}}+\frac{d^{2}u}{dt^{2}}\frac{dm}{dt}[/itex]

    Rate at which u changes with time
    (9)Absolutely no clue, still working on it.


    I'll leave it there.Is there anything wrong with my notation? Does each expression capture the descriptions?
     
  2. jcsd
  3. Jan 17, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    (6) needs an additional integration constant, which corresponds to the initial velocity.

    "u" for a position can be a bit confusing. In (6), I would use a different variable in the integral (like t'). Apart from that, it looks fine.

    Hint for (9):
    ##\frac{du}{dt}=c+\int_0^{t} \frac{dv}{dt} dt'##
     
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