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Related Rates problem

  1. Dec 11, 2017 #1
    1. The problem statement, all variables and given/known data

    ((I cannot, for the love of life, understand related rates, so please bear with me. Thank you! ))

    A cylindrical tank with radius 5m is being filled with water at a rate of 3m3/min. How fast is the height of the water increasing?

    I'm having trouble interpreting this question - actually, most related rates questions, having trouble differentiating between what I would call the volume and what not.

    2. Relevant equations
    v= (pi)r2h


    3. The attempt at a solution
    So, I'm gonna say that I know the rate of change of the volume - which means I need to get the height in terms of the volume?

    since v= (pi)r2h
    h= v/((pi)r2)

    h' = (dy/dx [v]((pi)r2) - dy/dx[(pi)r2](v))/((pi)2r4)
    h' = (3(2pir^2) - 2pirv)/ ((pi)2r4)


    I'm pretty sure this is wrong since I ended up with a "v" in my numerator which I don't know the value of.
    How would I go about solving this?
     
  2. jcsd
  3. Dec 11, 2017 #2

    Orodruin

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    What is your dy/dx? You have no variable called x nor one called y in your expressions.
     
  4. Dec 11, 2017 #3

    I probably shouldn't have written that, my bad - I just meant to use it as a way to say derivative of.

    so, derivative of volume multiplied by the denominator minus the derivative of the denominator multiplied by the volume, over the denominator squared.
     
  5. Dec 11, 2017 #4

    Orodruin

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    I am sorry, but that is a really bad notation. What you want to do is to take the derivative of the expression with respect to time ##t##. That derivative operator should be written ##d/dt##, not anything else. You need to specify what you are taking the derivative with respect to. There is no general "derivative of".

    Also, think about what variables in your expression actually depend on ##t##. The derivatives of the other variables with respect to ##t## are going to be zero.
     
  6. Dec 11, 2017 #5

    so would it be d/dt[v](piR^2)/(pi^2R^4) ?
    Does this mean my reasoning was correct? I messed up with the actual derivative part, but I've been having much more trouble understanding what to do or what reasoning is correct
     
  7. Dec 11, 2017 #6

    Orodruin

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    Your idea of taking the derivative of the expression for h is correct, yes. I am still not completely sure what you mean by your expression. Can you insert the numbers to get the final result so that we can be sure?
     
  8. Dec 11, 2017 #7
    I am still not completely sure what you mean by your expression. Can you insert the numbers to get the final result so that we can be sure?[/QUOTE]

    I got the same answer as the textbook after plugging in my answers but if I take the expression I came up with:
    d/dt[v](piR^2)/(pi^2R^4)

    d/dt[v] = 3
    R = 5
    so: [3(pi)(25)]/[(pi)2(625)]
    = 3/25(pi)
     
  9. Dec 11, 2017 #8

    PeroK

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    Note that ##r## is a constant in this problem. So is ##\pi##!

    This is the first time I've ever seen anyone differentiate ##\pi##.

    You don't differentiate constants.
     
  10. Dec 11, 2017 #9
    I didn't differentiate ##\pi##. It was being multiplied so I kept it as is but I did differentiate r which I wasn't supposed to.
     
  11. Dec 11, 2017 #10

    Orodruin

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    I would suggest the following:
    • You are overcomplicating your answer by not cancelling the ##\pi R^2## from your numerator with the corresponding terms in the denominator.
    • You should always use units. This will help you along by giving you a consistency check in your answer.
    In this case, the units would give you dv/dt = 3 m^3/min (you really should write it like this rather than d/dt[v]), R = 5 m, R^2 = 25 m^2. Hence
    $$
    \frac{dv}{dt}\frac{1}{\pi R^2} = \frac{3\ \mbox{m}^3/\mbox{min}}{\pi (25 \ \mbox{m}^2)} = \frac{3}{25\pi} \ \mbox{m/min}.
    $$
     
  12. Dec 11, 2017 #11

    PeroK

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    Okay, I see what you did now.
     
  13. Dec 11, 2017 #12

    Orodruin

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    Also, note that you can also differentiate before you solve for ##h##, it will give you the same result. (Writing ##v'(t)## and ##h'(t)## rather than ##dv/dt## and ##dh/dt##)
    $$
    v(t) = \pi R^2 h(t) \quad \Longrightarrow \quad v'(t) = \pi R^2 h'(t).
    $$
    Then ##h'(t)## is just the rate you are after and so you can solve for it in terms of ##v'(t)##, which you have.
     
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