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Math Help - need to differentiate?

  1. #1
    Junior Member
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    need to differentiate?

    I'm not really sure what to do here:

    The area covered by a species of ground-cover plant has a rate of increase of ycm^2/week, given by:

    y = -t^3 + 5x^2 + 6t

    where t is the number of weeks after planting. According to the model, if the plant covered 100cm^2 when planted, what area will it cover after 4 weeks?

    Thanks!
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, freswood!

    The area covered by a species of ground-cover plant has a rate of increase of y cm²/week,
    given by: . y \:= \:-t^3 + 5t^2 + 6t where t is the number of weeks after planting.

    According to the model, if the plant covered 100 cm² when planted.

    What area will it cover after 4 weeks?

    Since y \:=\:-t^3 + 5t^2 + 6t is the rate of increase of ground cover,

    the amount (area) of ground cover at time t is given by:
    . . A(t) \;=\;\int(-t^3 + 5t^2 + 6t)\,dt \;= \;-\frac{1}{4}t^4 + \frac{5}{3}t^3 + 3t^2 + C


    We are told that, when t = 0,\;A = 100.

    So we have: . -\frac{1}{4}\cdot0^4 + \frac{5}{3}\cdot0^3 + 3\cdot0^2 + C \:= \:100\quad\Rightarrow\quad C = 100

    Hence, the area function is: . A(t) \:= \:-\frac{1}{4}t^4 + \frac{5}{3}t^3 + 3t^2 + 100


    When t = 4:\;\;A(4) \;=\;-\frac{1}{4}\cdot4^4 + \frac{5}{3}\cdot4^3 + 3\cdot4^2 + 100 \;= \; \frac{572}{3}\;=\;190\frac{2}{3} cm².

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