Question

A person deposits $25,000 in a bank that pays 5% per year interest, compounded continuously. The person continuously withdraws from the account at the rate of $750 per year. Find V(t), the value of the account at time t after the initial deposit.

Answer 1

@ganeshie8 @paki @dan815 @Abhisar @TuringTest @Compassionate

Answer 2

@Hero @abb0t @cwrw238

Answer 3

@ganeshie8 @iambatman @dan815

Answer 4

Let \(V(t)\) is the amount of money in the account (with time \(t\) in years), then \(\dfrac{dV}{dt}\) is the rate at which this amount changes per year. The rate is proportional to the amount of money minus the yearly withdrawals, so you have
\[\frac{dV}{dt}=0.05V(t)-750t\]
Solve this linear ODE with the initial condition, \(V(0)=25,000\).

Answer 5

@SithsAndGiggles
are you sure it is
dV/dt = .05*V - 750*t
instead of
dV/dt = .05*V - 750

Answer 6

Ah yes that's a typo, thanks for pointing out @perl

Answer 7

:)

Answer 8

i was a bit confused by the problem since it says compounded continuously, but the rate is a discrete rate given per year

Answer 9

and you are subtracting 750 at the end of the year

Answer 10

oh it says 'continually withdraws at the rate of 750 per year"

Answer 11

The interest rate for these sorts of problems is almost always annual. It comes from the derivation of the compound interest formula, \(y=Ce^{rt}\). If it weren't annual we would only need to make a slight adjustment to the exponent.

Answer 12

And yeah I assumed the way the question was written suggested $750 was withdrawn at the end of each month.

Answer 13

if we approach this problem using a limit
lim (n->oo) 25,000 ( 1 + .05/n)^(n*t) - 750*t , where t is in years

we get
25,000 e^(.05t ) - 750t

but this is different than the solution of the differential equation
dV/dt = .05*V - 750
the solution is
10000 e^(.05t) + 15000

Answer 14

i think the problem is that we have an instantaneous rate of change (the compounding interest) , but the yearly withdrawal is discrete.
somehow we can incorporate the 750 continual withdrawal.

Answer 15

here is the differential equation solution
http://www.wolframalpha.com/input/?i=dy%2Fdx+%3D+.05*y+-+750%2C+y%280%29%3D25000

i'm not sure if the logic makes sense here:
lim (n->oo) 25,000 ( 1 + .05/n)^(n*t) - 750*t , where t is in years

Answer 16

I think the differential equation does work then :)

Answer 17

Alternatively we can try the future value formula for this scenario: http://en.wikipedia.org/wiki/Future_value#Compound_interest

Answer 18

the problem is that we are taking compound interest of what is left in the balance after the withdrawal, but the withdrawal is continually being withdrawn at a rate of 750 dollars per year. That means over small dt , he has withdrawn 750*dt

Answer 19

in addition to the interest on the interest gained from compounding

Answer 20

going back to your formula, you wrote
"Let V(t) is the amount of money in the account (with time t in years), then dV/dt is the rate at which this amount changes per year."
isnt dV/dt the rate at which the money changes in an instant of time ?

Answer 21

also a fun exercise, for what annual interest rate will the balance in the account stay fixed.
for what annual interest rate will the balance decay (so you can no longer withdraw 750 dollars)

Answer 22

the balance will stay fixed when k = 0.03
http://www.wolframalpha.com/input/?i=dy%2Fdx+%3D+.03*y+-+750%2C+y%280%29%3D25000

The balance will grow when k > 0.03
and decay when k < 0.03

I was just fiddling with different k .
I don't know how to find this generally.
ie
find
dy/dt = ky - 750 , y(0) = 25,000 , such that the solution will be constant

Answer 23

The general solution for
dy/dt = ky - 750 , y(0) = 25000 is :

y(t) = (250 ((100 k-3) e^(k t)+3))/k

http://www.wolframalpha.com/input/?i=dy%2Fdt+%3D+k*y+-+750%2C+y%280%29%3D25000

Answer 24

ok i think i got it.
The general problem :
A person deposits p dollars in a bank that pays 100k % per year interest, compounded continuously. The person continuously withdraws from the account at the rate of m dollars per year. Find y(t), the value of the account at time t after the initial deposit.
The differential equation is:
dy/dt = k*y - m , y(0) = p

The solution is :
y(t) = (( kp - m)e^(-kt) + m ) /k

now we see that if (kp -m) = 0 , then the whole expression becomes constant.
or more specifically if k = m/p , then y(t) = (0 + m ) / ( m/p) = p

if k > m/p then y(t) is increasing
if k < m/p then y(t) is decreasing.

Answer 25

here is the diff. equation solution
http://www.wolframalpha.com/input/?i=dy%2Fdt+%3D+k*y+-+m%2C+y%280%29%3Dp

Answer 26

Yeah that makes sense to me. Including the factor of \(t\) would have more to do with taking out arithmetically increasing quantities of money (i.e. 750 the first month, then 1500 then next, then 2250, and so on).

Answer 27

oh , withdrawing increasing amounts (linearly increasing)

Answer 28

that could be a nice variation on the problem , more challenging perhaps.
But the unit of time is year here, not months

Answer 29

i mean , in the given problem above :)

Answer 30

I was a bit surprised the original solution does not decay but increases
but the threshold is pretty small. changing k from .05 to less than .03 will cause it to decay

Answer 31

Well, in the first year the account grows from \($25,000\) to about \($26,281.80\), and the yearly withdrawal is less than the growth.

Answer 32

good point

Answer 33

but what kind of real word problem would satisfy this diff. equation

dy/dt = .05*y - 750*t

you were saying if the withdrawal is increasing per year?

Answer 34

how can we state this exactly. something about the instantaneous change i find troubling though

Answer 35

the account is compounding continuously at 5% annual interest and withdrawing continuously* at a rate of 750 dollars per year, but the rate of 750 per year is increasing linearly?

Answer 36

we might be pushing the problem to unrealistic edge

Answer 37

Yeah this might be a bit more than @Idealist10 bargained for :)

Suppose \(t=\dfrac{1}{12}\) (i.e. we consider the first month after the initial deposit). The ODE suggests we continuously withdraw $750 each year \(t\), which would mean in the first month we've accrued a total withdrawal of \(\dfrac{$750}{12}=$62.50\). I think the setup still works and makes sense, though in reality how do you withdraw an infinitesimally small amount of money?

I think if we wanted to explicitly state that $750 is withdrawn at the end of the year, we could make use of some special function, or perhaps define our own, like
\[f(t)=\begin{cases}750t&\text{for }t\in\mathbb{N}\cup\{0\}\\0&\text{elsewhere}\end{cases}\]
so we have
\[\frac{dV}{dt}=rV(t)-f(t)\]

Answer 38

I was reading this pdf, which says you can withdraw continuously

Answer 39

can't attach file, brb refresh

Answer 40

That \(f(t)\) might not be right... If we tried using the Laplace transform, we get
\[sF(s)-25,000=0.05F(s)~~\iff~~F(s)=\frac{25,000}{s-0.05}~~\implies~~V(t)=25,000e^{0.05t}\]
but this shouldn't be the solution.

Answer 41

i think on second page

Answer 42

Oh nice, so the setup of the ODE was right (aside from the \(t\) typo in my first comment).

Answer 43

the problem given by the OP is equivalent to
"A similar situation is paying off a loan. Suppose that you took out college loans totaling
$60; 000 with interest of 7.5%. You have an online payment plan which continuously deducts money from your bank account at a rate which comes out to $15; 000 per year. How long will it take you to pay off the loan?"

Answer 44

well not exactly equivalent, but similar in setup

Answer 45

if you turned the original problem into a problem of paying off a college loan,
"A person takes a loan of $25,000 from a bank that charges 5% per year interest, compounded continuously. The bank continuously withdraws from the account at the rate of $750 per year."
under these conditions you will never be able to pay back the loan

Answer 46

if we wait till the end of the year to withdraw the 750, yes that would produce a different function
It does seem odd in practice, how do you withdraw 750 continuously (infinitesimal amounts of money)

Answer 47

i dont think banks actually practice continuous withdrawal, they probably withdraw at the end of a day or hour , or something . (or month)

Answer 48

Thank you guys so much!