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    A Car has a Value £V at time t years. A model for V assumes that the rate of decrease of V at time t is proportional to V.
    a) By forming and solving an appropriate differential equation show that V= Ae ^-kt

    The value of a new car is £20 000 and when it is 3 years old its value is £11 000

    b) Find to the nearest £100 an estimate for the value of the Car wen its 10 years old

    A car is regarded as scrap when its value falls below £500

    c) Find an approxiamte age of the car wen it becomes scrap?

    Can any1 please help me. Im reallllly stuck just on this q!!
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    (Original post by ruth_lou)
    A Car has a Value £V at time t years. A model for V assumes that the rate of decrease of V at time t is proportional to V.
    a) By forming and solving an appropriate differential equation show that V= Ae ^-kt

    The value of a new car is £20 000 and when it is 3 years old its value is £11 000

    b) Find to the nearest £100 an estimate for the value of the Car wen its 10 years old

    A car is regarded as scrap when its value falls below £500

    c) Find an approxiamte age of the car wen it becomes scrap?

    Can any1 please help me. Im reallllly stuck just on this q!!
    so dV/dt is proportional to - V
    so therefore dV/dt=-kV, and (1/V)dV/dt=-k
    therefore ∫(1/V)(dV/dt)dt=-∫k dt => ln|V|=-kt+C =>V=e^(-kt+C)=e^(-kt)e^C
    therefore V=Ae^(-kt) (Let e^C =A)
    b)at t=0 V=20,000 =>20000=A(1)=>A=20000
    so V=20000e^(-kt)
    at t=3 V=11000, so 11000=20000e^(-3k)
    =>e^(-3k)=11/20 therefore k=-(1/3)ln(11/20)
    for part c just sub in t=10, and for the last part sub in V=500
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    a) dV/dt = -kV (you get this from the question)

    => 1/V dV/dt = -k

    integrate wrt t:

    ln|V| = -kt+C

    exponentiate:

    V = e^(-kt+C) = Ae^-kt


    b) use the initial conditions to get A = 20000, k = 0.2

    stick t=10 in to get V = £2700

    c) stick V = 500 in to get t=19 years
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    (Original post by elpaw)
    a) dV/dt = -kV (you get this from the question)

    => 1/V dV/dt = -k

    integrate wrt t:

    ln|V| = -kt+C

    exponentiate:

    V = e^(-kt+C) = Ae^-kt


    b) use the initial conditions to get A = 20000, k = 0.2

    stick t=10 in to get V = £2700

    c) stick V = 500 in to get t=19 years
    Hmm. People here seem to do differential equations slightly differently from how we do it in IB.

    Id go:

    dv/dt = -kv

    1/V dv = -kdt

    integral 1/v dv = integral -k dt

    It is of course teh same thing, just that the notation is slightly different...
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    (Original post by Jonatan)
    Hmm. People here seem to do differential equations slightly differently from how we do it in IB.

    Id go:

    dv/dt = -kv

    1/V dv = -kdt

    integral 1/v dv = integral -k dt

    It is of course teh same thing, just that the notation is slightly different...
    Thats more like the method I am used to seeign as well.
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    (Original post by Jonatan)
    Hmm. People here seem to do differential equations slightly differently from how we do it in IB.

    Id go:

    dv/dt = -kv

    1/V dv = -kdt

    integral 1/v dv = integral -k dt

    It is of course teh same thing, just that the notation is slightly different...
    yeah i've seen that before,treating the differentials like fractions.But the problem i see is that when you integrate don't u have to integrate with respect to something?
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    (Original post by IntegralAnomaly)
    yeah i've seen that before,treating the differentials like fractions.But the problem i see is that when you integrate don't u have to integrate with respect to something?
    he is isn't he?..... (1/v) wrt 'v' and (-k) wrt 't'
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    (Original post by mockel)
    he is isn't he?..... (1/v) wrt 'v' and (-k) wrt 't'
    is he? because dV and dt are in the expression before integrating.
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    because if u consider the definition of an integral,it would'nt make sense.
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    i just find the "pretend the differential is a fraction" argument is a pointless time-consuming step in the process. you can integrate both sides wrt t straight away, because:
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