Posted by Chopsticks on Sunday, March 1, 2009 at 10:59pm.
1. Suppose you invest $1,000 in a CD that is compounded continuosly at the rate of 5% annually. What is the value of this investment after one year? two years? five years?
A=1000(e^(.5x1)) = 1648.72
A=1000(e^(.5x2)) = 2718.28
A=1000(e^(.5x5)) = 12182.49
2. A colony bacteria is growing at ar ate of 50% per hour. What is the approximate population of the colony after one day if the initial population was 500
A=500(1+.5)^??
I am confused on this one.
3. Suppose a glacier is melting proportionately to its volume at the rate of 15% per year. Approximately what percent opf the glacier is left after ten years if the initial volume is one million cubic meters?
I have no idea how to do this one either. I know this is expo decay though.
4. A snowball is rolling downa snow covered hill. Suppose that at any time while it is rolling down the hill, its weight is increasing proportionately to its weight at a rate of 10% per second. What is its weight after 10 seconds if its weight initailly was 2 pounds? after 20seconds? after 45seconds? after 1 minute? What limitations might exist on this problem?
I have no clue on this one at all. Please help me start this out.
Can you please check if I did my first problem right aswell? Thanks.

Exp. Growth + Decay (HELP ME PLEASE!)  Reiny, Monday, March 2, 2009 at 12:44am
Wow, that's a big post
#1 you want 1000(e^.05t) not ^.5t
so 1000(e^.5) = 1051.24  one year
1000(e^(.10) = 1105.17
etc
#2. your equation is
number = 500(1.5)^t, where t is in hours
so one day is 24 hours
number  500(1.5)^24 = 8 417 056
#3 amount = 1mill(1  .15)^t , where t is in years.
so after 10 years
amount = 1mill(.85)^10
= 1mill(.1968)
so 19.7% of the glacier is left.
#4 looks like weight = 2(1.1)^t , where t is in seconds.
Can you take it from here? 
Exp. Growth + Decay (HELP ME PLEASE!)  Chopsticks, Monday, March 2, 2009 at 12:46am
Yeah, I'll try to finish it off.

Exp. Growth + Decay (HELP ME PLEASE!)  Chopsticks, Monday, March 2, 2009 at 1:10am
I have a problem on number 1 and number 4.
For number 1, I can't get the same answer your getting for some reason
I punch in the same numbers and I keep getting the same answers
A=1000(e^(.5x1)) = 1648.72
A=1000(e^(.5x2)) = 2718.28
A=1000(e^(.5x5)) = 12182.49
I read your directions on that, but I still do not understand.
For number 4.
Where did you get 1.1 from? 
Exp. Growth + Decay (HELP ME PLEASE!)  Chopsticks, Monday, March 2, 2009 at 1:21am
Oh for number 4
Did you do 2(1 + .1) to get 1.1? because of the 10%? 
Exp. Growth + Decay (HELP ME PLEASE!)  drwls, Monday, March 2, 2009 at 5:46am
On #1, you are not following Reiny's direction to multiply by e^0.05t, rather than e^0.5t
On #4, there is a difference between increasing 10% every 1 second and increasing at a rate of 10% per second, because the rate is continuously increasing, and will be higher at the end of each second.
In the former case, the growth is given by weight = 2*(1.1)^t
After 1 second, the weight is 2.2 lb in this case.
In the latter case,
dW/dt = 0.1W
log W = 0.1 t + log Wo
W = Wo*e^0.1t (where Wo = 2 lb)
W(t=1 s) = 2*e^0.1 = 2.2103 lb