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Second derivatives when pouring juice into a cup

  1. Oct 31, 2017 #21
    "Today at 10:46 AM#17
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    First you should consider how the shape of the cup effects the rate of height increase and the derivative of that. I think that your statement in the original post about the first derivative of height increase being positive is not always true as the cup fills up.
    Consider the derivatives where the cup diameter is expanding, where it is contracting, the sudden change in the middle, and the constant curvature of the top and bottom halves."

    hey factchecker. even if the height derivative fluctuates at the shape of the glass (I know itll slow down drastically at the wide parts, and increase in height quicker at skinnier parts of the cup) wouldn't it still be a positive derivative? i guess i'm trying to imagine this on a graph. I've attached the graph. lets say that at the skinny parts of the cup, it increase 10cm/s. and then at the widest parts of the cup, it decreases to an increase of 4cm/s. isn't it still positive because it's still an increase? or am i just not understanding

    thankyou in advance for all your help =)

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  2. Oct 31, 2017 #22
    i forgot to include my other increases. i'm not sure if my image came up clearly. so if its 10cm/s at the skinniest part, and decreases down to 4cm/s at the widest part.... the first second would be 10cm increase, the second second would be 8cm increase, the third second would be 6 cm increase, the fourth second 4cm increase, and then progressively back up to 10cm increase (I'm just using these numbers as an example based loosely upon the shape)... isn't it all still increases? even if the increases fluctuate? or would the derivative not be a positive nor a negative nor a zero, would it be a fluctuating derivative (if that makes sense?) would it bounce between being a positive and a negative?
  3. Oct 31, 2017 #23

    Charles Link

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    In this complicated orange juice cup, if you worked with the formulas that you can derive from calculus, you could treat every different part of the cup very quickly. Otherwise, it is tricky trying to do a qualitative explanation for the different parts of the cup. ## \\ ## The formulas will tell you that if ## dA/dh >0## , that ## d^2 h/dt^2 <0 ##, once you compute the various derivatives, (from the chain rule), and equate them to each other. Alternatively, you have parts of the cup that have ## dA/dh<0 ##, and for those parts, ## d^2 h/dt^2>0 ##.And if the cup has a part that is like a cylinder, then ## dA/dh=0 ##, so that ## d^2 h/dt^2=0 ##.
    Last edited: Oct 31, 2017
  4. Oct 31, 2017 #24


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    Note: I corrected my earlier post. The rate of increase (first derivative) is always positive as you originally stated. The second derivative is not always positive.
  5. Nov 1, 2017 #25
    thankyou guys. handing this in today =)
  6. Nov 3, 2017 #26


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    You didn't provide the most important thing, i.e., Fig. 7!
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