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calculus

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A quantity has the value P at time t seconds and is decreasing at a rate proportional to sqrt(P).

a) By forming and solving a suitable differential equation, show that P= (a - bt)^2 , where a and b are constants.

Given that when t= 0, P = 400,
b) find the value of a.

Given also that when t= 30, P = 100,
c) find the value of P when t = 50.

  • calculus -

    decreasing rate dP/dt= k sqrt (P)

    dP/(sqrtP)= k dt
    integrate both sides.

    -1/2 sqrtP=kT+ C
    square both sides
    1/4 P= (C+kT)^2
    P= 4 (C+kT)^2 and by choosing the constansts C, k
    P= (a-bt)^2

    400=(a-b*o)^2
    a= 20

    100=(20-b30)^2
    10=20-30b
    b=1/3

    P=(20-1/3*50)^2=(20-17.7)^2=...

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