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Rate of Change Mass vs Time

  1. Jul 25, 2008 #1
    I have been reading alot about rockets and how the mass flow rate is calculated.

    If i had a graph of the change in mass with respect to time of a rocket motor throughout its burn. How would i calculate the mass flow rate (change in mass at any given time t) ?

    Besides using density, velocity and area.


    That is where ive been reading.
  2. jcsd
  3. Jul 26, 2008 #2


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    ??? You just said "If i had a graph of the change in mass with respect to time" and then you said " How would i calculate ... the change in mass at any given time t"!

    You would read it off the graph, of course! Or did you mean to ask some other question?
  4. Jul 27, 2008 #3
    Well I suppose I don't even understand what im asking. Ha. In calculus when you find the slope of one point on a line by taking the derivative i dont understand how that applies to rate of change.

    Im talking about finding the rate of change at a certain point on a non linear line.

    Say from beginning of the curve to the end of the curve is the total burn of the motor as mass decreases. I want to know mass/sec that the motor burned off at the end. Would this be an average rate of change? I would also want to know the mass/sec the motor burned off at any other time. Maybe 2/3 the way through the burn.

    Last edited: Jul 27, 2008
  5. Jul 28, 2008 #4


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    Oh, so you have a graph of mass with respect to time, not "change of mass". In that case, the derivative is the rate of change of mass with respect to time.

    It is not easy to get the derivative directly from a graph. Strictlh speaking, the derivative is the slope of the tangent line to the curve at that point. One method of approximating it is to draw a straight line between two close points on the graph and use that to get the slope of the straight line : (y2- y1)/(x2- x1) where the two points are (x1,y1) and (x2,y2). That will approximate the rate of change.

    When I was in highschool, we learned this method: place a small mirror on the curve at the point in question. Rotate the mirror until it looks like the curve is going "smoothly" into its own image. Use the mirror as a straight edge to draw a line perpendicular to the curve. Now do the same thing, turning the mirror so that straight line appears to go "smoothly" into its own image. Drawing the line of the mirror gives the tangent to the curve.-
  6. Jul 28, 2008 #5
    Cool, I'll have to try that method. I have actually been through Calculus I during my Freshmen year at college. But i never really understood the concept of it all. Teachers(in this area) spend not near enough time on application of Calculus and explaining concepts of it.

    It reached the point where I was just finding the derivative or finding a tangent line. I really didnt know what i was achieving. I ended up with an A in that class because i was good at Algebra and could work the problems. But, i didnt and still do not understand the concept behind several topics we covered.

    I dont really understand why the rate of change at just one point is important. When i think of "Rate of change" i think of change over a PERIOD of time. Like the change between 0 and 3 seconds. ( I suppose when you differentiate you are finding the change over a period of time. But that time is extremely small)

    But then I also realize the the rate of change isnt constant in that interval.

    Just really confused i suppose.

    Is there a write up somewhere that explains in detail several Calculus topics, that you might know of? (In particular rate of change. With applications) I want to have a excellent understanding of Calculus. I want to be able to apply it to my own situations and use it to solve my own problems rather than just solve a bunch of word problems about shadows moving.(To do this I feel I do need a better understanding of the concept) I cant seem to find many "motion" type applications in our textbooks. I remember working one problem with Position, Velocity and Acceleration functions. That was it.

    Thank you for your time!
    James K.
    Last edited: Jul 28, 2008
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