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Question 953769: Hi, I'd like some explanation this question please:
3. A population of rats grows at a rate which is proportional to the number of rats, y, existing at the end of month t. Initially there are 50 rats, and after 2 months there are 100 rats.
(i) Find a relationship between y and t. (given answer --> y = 50(sqrt(2))^t)
(ii) Find the number of rats present after 1 year. (given answer --> 3200)
My answer to (i) is different but still gives the correct values for 2 months and 12 months:
y= exp ( (ln2)/2 * t + ln50)
Obviously mine is much less elegant, but I can't see how the given answer is arrived at.
My equation is derived from:
dy/dt = ky
Thanks,
Robert
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
there are 2 basic equations that i know of that can be used.
the first is f = p * (1 + r)^t
the second is f = p * e^(rt)
using the first equation of f = p * (1 + r)^t .....
f = 100
p = 50
r = what you want to find.
t = 2 months.
equation becomes:
100 = 50 * (1 + r)^2
divide both sides of this equation by 50 and then take the square root of both sides of the equation to get:
100/50 = (1 + r)^2 which becomes:
2 = (1 + r)^2 which becomes:
sqrt(2) = 1 + r
your equation started off as:
100 = 50 * (1 + r)^2
and it ended off as:
100 = 50 * (sqrt)^2
the general equation becomes:
y = 50 * sqrt(2)^t
you keep the present value at 50.
you replace 100 with y.
you replace 2 with t.
that generalizes the equation for any values of t.
the value of t = 2 was just used to find the rate per time period.
t is the number of time periods.
in this case the number of time periods are expressed in months.
the generalized equation of y = sqrt(2)^t finds the value of y for any value of t.
the alternate method is to use f = p * e^(rt)
in your situation, when you make f = 100 and p = 50 and t = 2, this equation becomes:
100 = 50 * e^(2r)
divide both sides of the equation by 50 to get:
2 = e^(2r)
take the natural log of both sides of the equation to get:
ln(2) = ln(e^(2r))
since ln(e^(2r)) is equal to 2r * ln(e) and since ln(e) is equal to 1, your equation becomes:
ln(2) = 2r
solve for r to get r = ln(2)/2
replace r in your original equation with ln(2)/2 and you get:
100 = 50 * e^(ln(2)/2 * 2)
replace 100 with y and replace t = 2 with 5 and your equation becomes:
y = 50 * e^(ln(2)/2 * t)
well, that's not exactly the equation you got.
let's test to see if all 3 equations provide the same answer.
we'll let t = 15.
first equation of y = 50 * (sqrt(2)^t becomes y = 50 * (sqrt(2)^15 which becomes:
y = 9050.966799
second equation of y = 50 * e^(ln(2)/2 * t becomes y = 50 * e^(ln(2)/2 * 15) which becomes:
y = 9050.966799
third equation of y = e^(ln(2)/2 * t + ln(50)) becomes y = e^(ln(2)/2 * 15 + ln(50)) which becomes:
9050.966799
the third equation is your equation.
i've never seen it done that way, but it works.
let me see if i can understand why.
i got y = 50 * e^(ln(2)/2 * t)
you got y = e^(ln(2)/2 * t + ln(50))
I believe that e^(ln(50)) is equal to 50, so we'll replace 50 in my equation with e^(ln(50)) to get:
y = e^(ln(50)) * e^(ln(2)/2 * t) which results in:
y = e^(ln(50) + ln(2)/2 * t) which is identical to your equation.
i can see that they're equivalent but i don't see how you arrived at your equation.
if you get a chance, let me know.
in the meantime, i hope the explanation of how y = 50 * sqrt(2)^t is satisfactory for you.
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