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.
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?
Yeah, I'll try to finish it off.
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?
Oh for number 4
Did you do 2(1 + .1) to get 1.1? because of the 10%?
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
Sure! Let's break down each problem and find a solution:
1. For the first problem, calculating the value of the investment after a certain period of time can be done using the formula for continuous compound interest:
A = P * e^(rt)
where A is the final amount, P is the principal (initial investment), e is the base of the natural logarithm, r is the interest rate, and t is the time in years.
In this case, the principal (P) is $1,000, the interest rate (r) is 5% (or 0.05 as a decimal), and we need to find the value after one year, two years, and five years.
Plugging these values into the formula:
A = 1000 * e^(0.05 * 1) ≈ $1,051.27 (after one year)
A = 1000 * e^(0.05 * 2) ≈ $1,105.17 (after two years)
A = 1000 * e^(0.05 * 5) ≈ $1,284.03 (after five years)
So, based on the calculations you provided, the value after one year seems to be correct, but the values after two years and five years need to be adjusted slightly.
2. For the second problem, we need to approximate the population of the bacteria colony after one day, given that it grows at a rate of 50% per hour and the initial population is 500.
Since the growth rate is given per hour, we need to convert one day into hours. There are 24 hours in a day.
Now, we can use the formula for exponential growth:
A = P * (1 + r)^t
where A is the final population, P is the initial population, r is the growth rate, and t is the time in hours.
Plugging in the values:
A = 500 * (1 + 0.5)^24 ≈ 12,091
So, the approximate population of the bacteria colony after one day would be around 12,091.
3. For the third problem, we need to determine the percent of the glacier that remains after ten years if it is melting proportionately to its volume at a rate of 15% per year. The initial volume is one million cubic meters.
Since the glacier is melting proportionately to its volume, we can use the formula for exponential decay:
A = P * (1 - r)^t
where A is the final amount, P is the initial volume, r is the decay rate, and t is the time in years.
Plugging in the values:
A = 1,000,000 * (1 - 0.15)^10 ≈ 131,501 cubic meters
So, approximately 13.15% of the glacier would remain after ten years.
4. For the fourth problem, we need to determine the weight of the snowball after a certain number of seconds if the weight increases at a rate of 10% per second. The initial weight is 2 pounds.
Similar to the previous problems, we can use the formula for exponential growth:
A = P * (1 + r)^t
where A is the final weight, P is the initial weight, r is the growth rate, and t is the time in seconds.
Plugging in the values:
A = 2 * (1 + 0.1)^10 ≈ 6.1917 pounds (after 10 seconds)
A = 2 * (1 + 0.1)^20 ≈ 17.4915 pounds (after 20 seconds)
A = 2 * (1 + 0.1)^45 ≈ 270.4264 pounds (after 45 seconds)
A = 2 * (1 + 0.1)^60 ≈ 1,048.576 pounds (after 1 minute)
In terms of limitations, these calculations assume continuous exponential growth without any external factors affecting the results. Real-life scenarios might involve additional complexities that would need to be considered.
I hope this explanation helps you understand how to approach and solve these problems. Let me know if you have any further questions!