Very Simple exponential growth problem

  • Thread starter lmlgrey
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  • #1
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1. It is estimated that x years from now the value of an acre of farmland will be increasing at the rate of [tex]\frac{0.4x^3}{sqrt(0.2x^4+8000)}[/tex] dollars per year. If the land is worth 500 per acre, how much will it be worth in 10 years?



2. Use integral



Since the the function of money of time is the integral of the rate given, i integrated the function 0.4x
^3/sqrt(0.2x^4+8000)....The answer therefore represents D(t) (Dollar Vs. Time)... Then I substitue 10 years into the function, I got how much is it worth in 10 years. but the question is, whats the use of the detail "now its worth 500 per acre"? should i add 500 to the answer i have now???? thanks!
 

Answers and Replies

  • #2
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You have to include the constant of integration because you're working with an indefinite integral.

For example,
[itex]\int x^2 dx[/itex] = 1/3 x^3 + C

Your function D(t) should be D(t) + C. D(0) + C should be equal to 500.
 
  • #3
HallsofIvy
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Since "now" equals "0 years from now", another way to do exactly what Mark44 said is to use a definite integral from 0 to 10- and then, since the integral from "0 to 0" is 0, add the $500. I suspect that is what Imlgrey meant originally.
 
  • #4
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ok... It seems that i completely forgot the contant part :P
now I integrated the function and got:
D(x)= sqrt(0.2x^4+8000)+C
D(0)= sqrt(8000)+C
500= sqrt(8000)+C
C= 410.557
Then D(10) = sqrt(0.2*10^4+8000)+410.557= 510.557 acre ---- does it look right?

...also, if I do what Hallsofivy said, I would have
sqrt(0.2x^4+8000)+C|100

and solving that, I got a totally different answer...Did I do anything wrong?
 

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