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# P3 May 2002 help watch

1. 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?

2. (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?

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
3. 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
4. (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...
5. (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.
6. (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?
7. (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'
8. (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.
9. because if u consider the definition of an integral,it would'nt make sense.
Attached Images

10. 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:
Attached Images

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