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Math Help - differencial equation

  1. #1
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    differencial equation

    The rate, dy/dt, at which ice is forming on a pond is proportional to the square root of t. Let y be the thickness of the ice in inches at time t measured in hours since the ice started forming.

    (a) Write the differential equation. Use k > 0 as the proportionality constant.
    dy/dt =

    (b) Solve the differential equation.
    y(t) =

    (c) If the thickness of the ice is 3 inches after 9 hours, what is the value of k?
    k =

    a little help with these bad frickin johnnys...thanks

    mathlete

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  2. #2
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    Hello, mathlete!

    The rate, \frac{dy}{dt}, at which ice is forming on a pond
    is proportional to the square root of t.
    Let y be the thickness of the ice in inches at time t
    measured in hours since the ice started forming.

    (a) Write the differential equation. Use k > 0 as the proportionality constant.
    {\color{blue}\frac{dy}{dt} \:=\:k\sqrt{t}}


    (b) Solve the differential equation.
    We have: . \frac{dy}{dt}\;=\;k\!\cdot\!t^{\frac{1}{2}}

    Separate variables: . dy \;=\;k\!\cdot\!t^{\frac{1}{2}}dt

    Integrate: . y \;=\;\frac{2}{3}k\!\cdot\!t^{\frac{3}{2}} + C

    Assuming there was no ice at t = 0, we have: . C \:=\:0

    The ice-thickness function is: . {\color{blue}y \;=\;\frac{2}{3}k\!\cdot\!t^{\frac{3}{2}}}



    (c) If the thickness of the ice is 3 inches after 9 hours,
    what is the value of k?
    We are told: .when t = 9,\:y = 3

    So we have: . 3 \;=\;\frac{2}{3}k\!\cdot\!9^{\frac{3}{2}} \;=\;\frac{2}{3}k\cdot27\quad\Rightarrow\quad18k \:=\:3

    . . Therefore: . {\color{blue}k \:=\:\frac{1}{6}}

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  3. #3
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    The rate, dy/dt, at which ice is forming on a pond is proportional to the square root of t. Let y be the thickness of the ice in inches at time t measured in hours since the ice started forming.

    (a) Write the differential equation. Use k > 0 as the proportionality constant.
    dy/dt = k*sqrt(t)

    (b) Solve the differential equation.
    y(t) =
    dy = k(t)^(1/2) dt
    y = (1/2)k[t^(-1/2) +C
    y = (k/4)sqrt(t) +C

    When t=0, y=0, so,
    0 = (k/4)sqrt(0) +C
    C = 0

    Therefore, y(t) = (k/4)sqrt(t)
    (c) If the thickness of the ice is 3 inches after 9 hours, what is the value of k?
    k =
    3 = (k/4)sqrt(9)
    3 = (k/4)*3
    k = 4

    EDIT: Uh-oh! After seeing Soroban's solution, I admit mine is wrong. Wrong Integration of sqrt(t). :-)
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