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Calc 2: what is the point of

  1. Apr 26, 2008 #1
    ok we're learning the part of calc 2 where you have y = 3 + 4t; x = 4 + 3t

    then u multiply dy/dt * dy/dx / dx/dt

    but then u just plug in the value of t into dy/dt.... so what was the point of finding the whole second part? lol it seems random and pointless to me. my friend says when u do this t "sometimes" cancells out. seems like a lot of work for nothing
  2. jcsd
  3. Apr 27, 2008 #2


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    what are you trying to do?
    dy/dt * dy/dx / dx/dt=(dy/dx)^2

    maybe you intended
    in which you see how y changes with x
    since you know how x and y change with t and maybe that is not interesting
    maybe y is how happy you are at time t
    and x is how many potatoes you have eaten at time t
    when what you would really like to know is how happy does eating potatoes make you
    dy/dt is increase in happiness [happy]/[hr]
    dx/dt is rate of potatoe ingestion [potatoe]/[hr]
    dy/dx is the happyness given by each potatoe [happy]/[potatoe]

    so maybe knowing you get 4 happier each hour and eat 3 potatoes is not as interesting as knowing each potatoe makes you 4/3 happier.
  4. Apr 27, 2008 #3
    The textbook our school uses talks about those problems dealing with the motion of particles. Overall, like many things, its just preparing you for what lies beyond... :)
  5. Apr 28, 2008 #4
    Are you trying to find the derivative of y with respect to x where y and x are defined parametrically? In this case, the "point" is to avoid having to solve for y as a function of x and instead to use the provided parametric formulas.

    Notice that by the chain rule,

    [tex]\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} [/tex]

    [tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]

    This formula lets you find the derivative of y with respect to x directly from the parametric equations. The formula you gave was incorrect if this is the goal you wanted to accomplish.
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