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Math Help - Calc- Word Problem

  1. #1
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    Calc- Word Problem

    Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by P(t) = 2t + \frac {1}{162t + 1}, 0 \leq t \leq 1, where t is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

    P'(t) = 2 - \frac {162}{(162t + 1)^2}

    P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}

    0 = 52 488t^2 - 160

    160 = 52 488t^2

    \pm 12.64911064 = 52488t

    0.000240991 = t

    P(0.000240991) = 0.963
    P(0) = 0
    P(1) = 2.006

    Therefore, the pollution in 0.000240991 year was at its lowest level of 0.963

    -----

    Textbook answer: 0.049 years

    Did I take the first derivative wrong? I checked it and I don't see anything wrong with it. Please show me my mistake and fix it for me?
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    Hello,

    Quote Originally Posted by Macleef View Post
    Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by P(t) = 2t + \frac {1}{162t + 1}, 0 \leq t \leq 1, where t is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

    P'(t) = 2 - \frac {162}{(162t + 1)^2}

    P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}

    [...]
    You've got a problem here :

    (162t+1)^2=162^2t^2+{\color{red}2 \cdot 162t}+1

    Remember :

    (a+b)^2=a^2+2ab+b^2 \neq a^2+b^2

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  3. #3
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    Quote Originally Posted by Macleef View Post
    Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by P(t) = 2t + \frac {1}{162t + 1}, 0 \leq t \leq 1, where t is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

    P'(t) = 2 - \frac {162}{(162t + 1)^2}

    P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}

    0 = 52 488t^2 - 160

    160 = 52 488t^2

    \pm 12.64911064 = 52488t

    0.000240991 = t

    P(0.000240991) = 0.963
    P(0) = 0
    P(1) = 2.006

    Therefore, the pollution in 0.000240991 year was at its lowest level of 0.963

    -----

    Textbook answer: 0.049 years

    Did I take the first derivative wrong? I checked it and I don't see anything wrong with it. Please show me my mistake and fix it for me?
    P(t) = 2t + 1/(162t + 1)

    *use quotient rule on second term*

    P'(t) = 2 - 162/(162t + 1)^2

    *minimum occurs when P'(t) = 0*

    2 - 162/(162t + 1)^2 = 0

    2 = 162/(162t + 1)^2

    2(162t + 1)^2 = 162

    2(26244t^2 + 324t + 1) = 162

    52488t^2 + 648t -160 = 0

    t = 0.04938 or t = -0.06173

    Cannot be negative, therefore t = 0.04938
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