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The derivative without time.

  1. May 15, 2013 #1
    So I know that the derivative of the area of a circle is 2∏r or the circumference, and that the derivative of the area of a square is the perimeter or 2x. I don't get how these are the derivatives because there isn't a time involved. I thought that the derivative measured the rate of change and if so how do these derivatives without time measure this change?
     
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  3. May 15, 2013 #2

    Office_Shredder

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    Rate of change with respect to ________

    You can fill in the blank with anything. For example, as you go up in an airplane the temperature of the air around you gets colder. You can have a function T(h) which is the temperature at a height of h, and T'(h) is the rate of change of the temperature with respect to height - it tells you approximately how much the temperature changes if you go one meter higher.

    Similiarly, the derivative of the area of the circle with respect to the radius tells you how much the area of the circle increases by if you increase the radius of the circle
     
  4. May 15, 2013 #3
    Ok so how would you write down the rate of change of the area of a circle. for example it would be two miles per hour but what would it be for the circle. I hope this makes some sense. and thank you for replying. It seems nobody has replied on my other posts.
     
  5. May 15, 2013 #4

    Office_Shredder

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    Assuming that all the units are in meters for example, the derivative of the area with respect to radius would be meters squared per meter. The units of a derivative are always units of the value of the function divided by units of the input of the function
     
  6. May 16, 2013 #5

    Mark44

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    Isaac Newton considered derivatives to be with respect to time, only. His notation for the derivative time derivative of x looked like this: ## \dot{x}##, while Leibniz used dx/dt to represent the same thing. The "prime" notation that is used in calculus today derives from Newton's dot notation.

    For your example of the circle, if both A and r are differentiable functions of t (time), then dA/dt = ##\pi##r2(t), so dA/dt = dA/dr * dr/dt = 2##\pi##r dr/dt.

    Here the expression 2##\pi##r is dA/dr, the rate of change of area with respect to change in radius.
     
  7. May 16, 2013 #6
    Ok, to office shredder, the radiuses derivative would be meters squared per meters because the radius is the acceleration and acceleration is squared? This is how I came up with this. accelerationxtime=rate, accerlerationxtime=d/t, accerlerationxt^2=distance?
     
  8. May 16, 2013 #7

    Mark44

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    No. Assuming you mean the rate of change of the radius with respect to time, dr/dt, the units would be meters/seconds or meters/minutes, or whatever the units of time are.
    This is pretty mixed up. Assuming length units of meters, and time units of seconds,
    acceleration = meters2/seconds2
    velocity = meters/seconds
     
  9. May 16, 2013 #8
    Im not getting the example you had because I don't want it to have a function of time. But would this be true....
    A=∏r^2
    A primed= 2∏r
    So how would you put this into a function of time?
     
  10. May 16, 2013 #9

    SteamKing

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    The perimeter of a square with side x is 4*x, not 2*x.
     
  11. May 16, 2013 #10

    Office_Shredder

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    He was explaining how there is no time involved with taking that derivative, but if you have the radius as a function of time then you can alternatively find the derivative of area with respect to time
     
  12. May 16, 2013 #11

    arildno

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    Think of a circle that expands, or contracts, uniformly over time. Then, the rate of change of the area over time will be 2*pi*r*(dr/dt), so that if the the radial increase pr.time unit Equals one distance unit, you will have dA/dt=2*pi*r
     
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