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Math Help - derivative graph

  1. #1
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    derivative graph

    You are given the following graph.
    Use the figure above and the fact that P = 30 when t = 0 to find the values below.
    P(1) =
    P(2) =
    P(3) =
    P(4) =
    P(5) =
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  2. #2
    Senior Member Peritus's Avatar
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    dP/dt = -1_____0 <= t <=2
    x-3____2<= t <= 4
    1____4 <= t <= 5

    start from integrating the function in the first interval:

    P = -t+C_____0<=t<=2

    now we know that P(0) = 30 thus C = 30 ----> P = -t+30_____0<=t<=2
    no because the derivative is continues we can use P(2) = 28 as the initial condition for the function in the next interval
    P = (t^2)/6 - 3t + C____2<= t <=4, P(2) = 28----> P(t) = (t^2)/6 - 3t +100/3

    continue in the same way and solve for the last interval...
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  3. #3
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    Hello, mathaction!

    This is a long and messy one . . .


    You are given the following graph.

    Use the figure above and the fact that P(0) = 30
    . . to find: . P(1),\;P(2),\;P(3),\;P(4),\;P(5)

    On [0,2],\;P'(t) \,=\,-1

    Integrate: . P(t) \:=\:-x +C_1

    Since P(0) = 30\!:\;\;-0 + C_1 \:=\:30\quad\Rightarrow\quad C_1 \:=\:30

    Hence, on [0,2],\;P(t) \:=\:-x + 30

    . Therefore: . {\color{blue}P(1) =29,\;\;P(2) =28}



    On [2,4],\;P'(t) \:=\:x - 3

    Integrate: . P(t) \:=\:\frac{1}{2}x^2-3x+C_2
    Since P(2) =28,\;\;\frac{1}{2}(2^2) - 3(2) + C_2 \:=\:28\quad\Rightarrow\quad C_2 = 32
    Hence, on [2,4],\;P(t) \:=\:\frac{1}{2}x^2 - 3x + 32
    . Therefore: . {\color{blue}P(3) = \frac{55}{2},\;\;P(4) = 24}



    On [4,5],\;P'(t) \:=\:1

    Integrate: . P(t) \:=\:x + C_3

    Since P(4) = 24,\;\;4 + C_3 \:=\:24\quad\Rightarrow\quad C_3 = 20

    Hence, on [4,5], \;P(t) \:=\:x + 20

    . Therefore: . {\color{blue}P(5) = 25}

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  4. #4
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    thank you for the help, however i think you miss calculated p(4) and p(5)...for p(4), the equation...1/2x^2-3x+32...plug in 4 for x

    16/2 - 12 +32= 8+20=28 p(4)=28

    P(t)=x-C3...4-C3=28=24
    P(t)=x+24...p(5)=5+24 = 29

    p(5)=29


    thanks for the other ones tho...
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