1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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
    It just doesn't look right to me :/
  2. jcsd
  3. Sep 20, 2009 #2


    User Avatar
    Science Advisor
    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?


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

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

    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
    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:


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


    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]

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook