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Derivative question.

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the rate of change of depth D with respect to time t
    Find the rate of change of depth when the depth is 22.5 m


    2. Relevant equations

    D=10(2-(t/14))^2


    3. The attempt at a solution
    I know I need to determine the derivative for both.
    dy/dx =10(2-(t/14))^2
    = 10(4+(t^2/196)-2t/14)
    = 40 - 10t^2/196 - 40t/14
    =-20t/196-40/14
    It just doesn't look right to me :/
     
  2. jcsd
  3. Sep 20, 2009 #2

    CompuChip

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    Homework Helper

    That looks fine, except that you accidentally turned a plus sign into a minus sign when you opened the brackets: 10(4 + t^2) is 40 + 10t^2, not 40 - 10t^2 :)
    That also affects the derivative.

    Now find out for which time, the depth D is equal to 22.5 m, and plug that number into the derivative.

    (Also, your notation could use some work, but I will leave that until after you have solved the question).
     
  4. Sep 20, 2009 #3
    Oh, wait. But then the derivative is positive.

    I should say the question: In order to make repairs to a retaining wall that is 40 m high, the water in a conservation reservoir is slowly drained out over a 4-week period. Starting at t=0, the water is released that after t days the water depth D (in metres) is approximated by: (the equation I gave).

    Did I do something wrong?
     
  5. Sep 20, 2009 #4
    Is this the equation you have?

    [tex]D=10(2-\frac{t}{14})^2[/tex]

    If so, then wouldn't you use the chain rule to get:

    [tex]
    \frac{dD}{dt} = 10(2)[2-\frac{t}{14}]*(-\frac{1}{14})
    [/tex]

    If you were avoiding the chain rule, and expanding instead, then you should have a -4 instead of a -2 here:

    And you also shouldn't say that it's equal to [tex]\frac{dy}{dx}[/tex], when you haven't differentiated yet. Then below, I don't think it should be a negative ten, but rather positive ten here:

    Once that's fixed it should be equal to the same derivative I got with the chain rule and that will be negative if you fill in 22.5.
     
  6. Sep 20, 2009 #5
    Ahhh, thank you! I see where I made the mistake. We were supposed to use the power rule on this question.

    I got my derivative as: 5t/49 -20/7

    is this correct? It looks different than what I got before but I think it's 'cause I simplified it.
     
  7. Sep 20, 2009 #6
    Hmm.
    Find the rate of change of depth when the depth is 22.5 m

    My answer for this is -15/7. I don't think this is right though. I equated D= 25 and solved for t which gave me 7 days. So does this mean in a week the rate of change, when the water is 22.5 m, is -15/7?
     
  8. Sep 20, 2009 #7
    When I did it, I got:

    [tex]

    \frac{dD}{dt} = 10(2)[2-\frac{t}{14}]*(-\frac{1}{14}) = -\frac{40}{14} + \frac{20t}{196} = \frac{5t}{49} - \frac{20}{7}

    [/tex]

    which is exactly what you got. So I think it's likely right. For future reference, if you put the numbers between tags like these: [tex ][/tex ] (without the spaces), you can normally make it a lot nicer looking. Between these tags, fractions are written like \frac{numerator}{denominator}, which looks like [tex]\frac{numerator}{denominator}[/tex].

    Did you get -10/7 or -15/7? Not sure either is right...
     
    Last edited: Sep 20, 2009
  9. Sep 20, 2009 #8
    Oh, thank you.

    I just solved it like this: [tex]22.5 = 10(2-(t/14))^2[/tex]

    [tex]2.25=(2-(t/14))^2[/tex]
    took the square root of both sides

    [tex]1.5 = 2 -t/14 [/tex]
    [tex]t = (1/2)(14)[/tex]
    t= 7.

    Do I sub this into the derivative or the function?
     
  10. Sep 20, 2009 #9
    That looks good. Since you're looking for the rate of change, you sub it into the derivative. And yup, -15/7 looks right.

    Sorry about doubting you before - I must have been having a weak moment! :wink:
     
  11. Sep 20, 2009 #10
    Thank you! I finally understand this much better :)!
     
  12. Sep 20, 2009 #11
    No problem. Good luck with the course!
     
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