Tuesday

September 23, 2014

September 23, 2014

Posted by **Chopsticks** on Sunday, March 1, 2009 at 10:59pm.

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 12:44amWow, 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 12:46amYeah, I'll try to finish it off.

- Exp. Growth + Decay (HELP ME PLEASE!) -
**Chopsticks**, Monday, March 2, 2009 at 1:10amI 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 1:21amOh 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 5:46amOn #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

**Answer this Question**

**Related Questions**

Finances and Math - Can someone please help me solve these problems or help me ...

Accounting - Calculate the future value of the following: o $5,000 compounded ...

Business Algebra - Invest $23,000 in a savings account at 4.25% interest ...

PLEASE HELP!!1 - Does $1000 at a rate of growth of 12% compounded continuously ...

math - Find the future value of an annuity if you invest $1,550 annually for 5 ...

Math - Use the present value formula to compute the amount that should be set ...

Finance - You are currently investing your money in a bank account which has a ...

Mathematics - In the problems 1 to 3, find the present value of the given (...

finances - Calculate the future value of the following: o $5,000 compounded ...

Algebra - How much money should I invest at an annually compounded interest rate...