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Derivative and water flow

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 6.00t 0.8 - 3.35t + 23.00, with t 0, m in grams, and t in seconds.
    (a) At what time is the water mass greatest?

    (b) What is that greatest mass?

    (c) In kilograms per minute, what is the rate of mass change at t = 2.00 s?

    (d) In kilograms per minute, what is the rate of mass change at t = 5.00 s?
    2. Relevant equations
    The first derivative would be 4.8t^-.2-335


    3. The attempt at a solution

    a)I set the derivative equal to zero and figured t to be 2.3375, but it says thats wrong.
    b)I assume I'd plug a into the original and try tht, but I can't get a.
    c and d)I tried to put 2 and 5 into t as the origional, but they are not right.
     
  2. jcsd
  3. Jan 31, 2008 #2
    a) is, as you said, the solution for t when

    [tex]\frac{dm}{dt}=0.[/tex]

    However, the solution to

    [tex]t^{0.2}=\left[\frac{4.8}{3.35}\right][/tex]

    is not 2.3375, check your algebra.

    b), as you said is [tex]m(t)[/tex] for the solution above
    c) and d) can both be found by substituting the times into the expression for [tex]\frac{dm}{dt}.[/tex]
     
  4. Jan 31, 2008 #3
    Thanks. Where did the 3.35 come from? How would I solve t^-.2?
     
  5. Feb 1, 2008 #4
    The 3.35 is from the original expression for m, unless you've mistyped it. I used logarithms and the change of base rule to solve for t.
     
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