# Exponential Word Problem

• Apr 18th 2011, 05:47 AM
BuhRock
Exponential Word Problem
I haven't the clue to where to start on this one.

Under ideal conditions, a population of rabbits has an exponential growth rate of 11.7% per day. Consider the initial population of 100 rabbits.

a)Find the exponential growth function
b)Graph it
c)What will the population be after 7 days, and then 14 days.
d)find the doubling time.

Isn't the function P(t)= P sub 0 e^kt , K>0
• Apr 18th 2011, 06:42 AM
e^(i*pi)
That is the formula for exponential growth but you're expected to simplify it. For example you're told P_0 (the initial population)

t can be measured in any unit of time but if t is measured in days then k is simply 11.7% expressed as a decimal so days is easiest.

edit:

b) The graph is standard exponential with point (0,P_0). If you have to do it by hand then an estimate will be fine as long as you get the shape and intercept right. With software it's easy enough

c) Find P(7) and P(14) respectively

d) P(t) = 2P_0 , solve for t
• Apr 18th 2011, 08:21 AM
HallsofIvy
Be careuful here. If the population one day is P, and it increases by 11.7% each day, on the next day, the population will be 1.117P. Now, if you are writing this as P= P_0e^{kt} then you are saying that 1.117P= Pe^{k} so that e^{k}= 1.117, NOT k itself.

In fact, I would have done this simply as P(t)= 100(1.117)^t. That is, after one day, the population has been multiplied by 1.13, after another day, again by 1.17, so the original population by (1.117)(1.117)= (1.117)^2, the third day, again 1.117, so the original population is multiplied by (1.117)^2(1.117)= (1.117)^3, etc.
• Apr 18th 2011, 05:01 PM
BuhRock
I thought it would be

P(7) = 100e(.117)(7) which = 226.823
P(14) = 100e(.117)(14) which = 514.487

Why do you said P_0 = 1.117