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Second derivatives when pouring juice into a cup

  1. Oct 31, 2017 #1
    Its a question about volume increase (in units cm^3) and height increase (cm) when pouring juice into a cup. Its stated that the volume of the juice in the cup increases at a constant rate, so I know the volume derivatives are zero. But the shape of the cup is inconsistent and there is alot of variance in it, so the height increase is not constant. I know the first derivative of the height increase is positive... my professor hinted that the second derivative of height increase would be negative... how is that so?
     
    Last edited by a moderator: Oct 31, 2017
  2. jcsd
  3. Oct 31, 2017 #2

    Charles Link

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    The derivatives of interest are ## \frac{dh}{dt} ## and ## \frac{d^2 h }{dt^2} ##. You are given ## \frac{dV}{dt}=constant ##. Suggestion is to write ## V=V(h) ##, and compare derivatives ## \frac{dV}{dt} ## and ## \frac{d^2 V}{dt^2} ## using the chain rule. Also ## \frac{d^2 V}{dt^2}=0 ##. ## \\ ## Additional hint: What can you say about ## \frac{d^2 V}{dh^2} ## if the volume is written as ## V(h)=\int\limits_{0}^{h} A(h') \, dh' ##, and ## \frac{d A(h)}{dh} >0 ## ?
     
    Last edited: Oct 31, 2017
  4. Oct 31, 2017 #3

    fresh_42

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    The first derivative measures the rate of change, so it's not zero for the volume. It's a positive constant. The second derivative of the volume is zero, because it measures the rate to which the rate of change changes. The same is true for the height. As it doesn't change constantly, it's first derivative actually changes with time. And as this change isn't constant, the first derivative of this change isn't zero, which is the second derivative of the height function. Whether this is positive or negative depends on whether the rate of change is sublinear or hyperlinear.
     
  5. Oct 31, 2017 #4
    What does the A stand for in your last equation?
     
  6. Oct 31, 2017 #5

    fresh_42

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    The surface area at height ##h##.
     
  7. Oct 31, 2017 #6

    Charles Link

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    I had a slight error. I corrected it. (I originally wrote ## V(h)=A(h) h ##. The correct expression is ## V(h)=\int\limits_{0}^{h} A(h') \, dh' \, ##).
     
  8. Oct 31, 2017 #7
    your answers have been helpful
    i know that the height increase and volume increase first derivatives are positive. thats helpful
    the units of the first volume derivative is cm^3/s (seconds)
    second derivative is cm^3/s^2
    height first derivative is cm/s
    second derivative of height is cm/s^2
     
  9. Oct 31, 2017 #8
    not sure why the photo of the problem didn't go through last time. here it is
    calculus.jpg
     

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    Last edited by a moderator: Oct 31, 2017
  10. Oct 31, 2017 #9
    I guess I was confused because I thought the derivative of a constant was always zero
     
  11. Oct 31, 2017 #10

    Charles Link

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    The trickiest part of this for someone just learning calculus is applying the chain rule a couple of times. Suggestion: ## dV/dt =(dV/dh)(dh/dt)=constant ##. Finding ## d^2V/dt^2 =0=... ## using the chain rule takes a little more work, but it should give you the result you need. ## \\ ## Additional hint: By the chain rule, ## d^2 V/dt^2=(d(dV/dh)/dt)(dh/dt)+(dV/dh)(d^2 h/dt^2)=0 ##. The first term there needs to be evaluated with the chain rule one more time..
     
    Last edited: Oct 31, 2017
  12. Oct 31, 2017 #11

    FactChecker

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    You will get answers that are more specific to your questions if you show us the Figure 7 of the cup and the rest of the problem statement.
     
  13. Oct 31, 2017 #12
    thanks everyone
    thanks charles =)
     
  14. Oct 31, 2017 #13

    Charles Link

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    See my edited post of post 10. That is very close to the result you need.
     
  15. Oct 31, 2017 #14
    here it is, for those curious
     

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  16. Oct 31, 2017 #15
    nevermind. my phones too blurry. i give up
     
  17. Oct 31, 2017 #16

    Charles Link

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    I was able to read it. They want to know the sign of the terms ## (dh/dt) ## and ## (d^2 h/dt^2) ##, etc. If you look at my hint in post 10, you need to compute ## \frac{d (\frac{dV}{dh})}{dt } ## using the chain rule. Once you have that, it will allow you to compare ## d^2 h/dt^2 ## with ## d^2 V/dh^2 ##. (See also my hint for ## d^2V/dh^2 ## in the last part of post 2).
     
    Last edited: Oct 31, 2017
  18. Oct 31, 2017 #17

    FactChecker

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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. [CORRECTION: Of course it is always positive. The second dirivative is not always positive.]
    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.
     
    Last edited: Oct 31, 2017
  19. Oct 31, 2017 #18

    Charles Link

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    @FactChecker input can be quantified by computing ## d^2 V/dh^2 =dA(h)/dh ## as my hint in the last part of post 2 is suggesting. Using that, along with the equation in post 10, and the problem is really very straightforward.
     
  20. Oct 31, 2017 #19

    FactChecker

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    Yes, I know. I can do it easily. But I think that the original poster is less familiar with the application and meaning of derivatives in real problems and I'm not sure that the problem was asking for actual formulas as opposed to a more qualitative characterization.
     
  21. Oct 31, 2017 #20
    you're right factchecker, I am new to all this. I took a 6 year break from pre calc to this class (calc 1) and prior to that, it'd been like 6 years since I took algebra or trig. so I am very behind. I own calculus for dummies, and a complete idiots guide to calculus, and I watch alot of kahn academy and youtube videos. but i am struggling and am very behind academically
     
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