Doubling time, half-life, and interest problems?

[dirtydarsh]

Can someone give me tips on how to start these problems?

1. If a population of bees double every hour, and there are 175 present initially, how many will remain after 6.5 hours?

2. The monthly rent in an apartment is $875, and will raise 3.25% each year. What will monthly rent be in 15 years?

3. The population of a city is 5000 and decreasing by 3.5% every 4 years. To the nearest year, how many years will it be before the population reaches 4000.

4. The half-life of rubidium is 12.5 seconds. With a starting sample of rubidium, what percentage of it will remain in one minute?

 

[Deleted member 4993]

Can someone give me tips on how to start these problems?

1. If a population of bees double every hour, and there are 175 present initially, how many will remain after 6.5 hours?

2. The monthly rent in an apartment is $875, and will raise 3.25% each year. What will monthly rent be in 15 years?

3. The population of a city is 5000 and decreasing by 3.5% every 4 years. To the nearest year, how many years will it be before the population reaches 4000.

4. The half-life of rubidium is 12.5 seconds. With a starting sample of rubidium, what percentage of it will remain in one minute?

What are your thoughts?

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[stapel]

Can someone give me tips on how to start these problems?

Tip: Use what you learned back in algebra! These all involve growth or decay ("negative growth") with fixed-period intervals (rather than "continuous" time), so they'll all fit into the compound-interest formula. (You can refresh here.)

1. If a population of bees double every hour, and there are 175 present initially, how many will remain after 6.5 hours?

Starting with the basic compound-interest formula:

. . . . .\(\displaystyle A\, =\, P\left(1\, +\, \dfrac{r}{n}\right)^{nt}\)

...they've given you that P = 175. They've given the growth in terms of hours, so that's your time interval. They've told you that you're looking for A at t = 6.5. Since the growth is in hours and the time-interval is hours, then n = 1. However, this leaves you still needing to figure out r. But if the population doubles every time interval, then the percentage growth, over the initial amount, must be... what percentage, expressed in decimal form?

Plug all of that information into the formula, just like you did back in algebra, and simplify to find the value of A.

2. The monthly rent in an apartment is $875, and will raise 3.25% each year. What will monthly rent be in 15 years?

P = 875, r = 0.0325, n = 1, and t = 15. Find A.

3. The population of a city is 5000 and decreasing by 3.5% every 4 years. To the nearest year, how many years will it be before the population reaches 4000.

In this case, you'll need to remember that the time interval is four years. They've given you P = 5,000, A = 4,000, and r = -0.035. Let's use n = 1, and then remember to make adjustments when we interpret the meaning of the solution value for t.

4. The half-life of rubidium is 12.5 seconds. With a starting sample of rubidium, what percentage of it will remain in one minute?

This is a continuous-growth exercise, so use that formulation. (Here.)

If you get stuck on any of these (after having studied lessons for each topic), please reply showing your thoughts and efforts so far. Thank you!

 

[dirtydarsh]

Tip: Use what you learned back in algebra! These all involve growth or decay ("negative growth") with fixed-period intervals (rather than "continuous" time), so they'll all fit into the compound-interest formula. (You can refresh here.)


Starting with the basic compound-interest formula:

. . . . .\(\displaystyle A\, =\, P\left(1\, +\, \dfrac{r}{n}\right)^{nt}\)

...they've given you that P = 175. They've given the growth in terms of hours, so that's your time interval. They've told you that you're looking for A at t = 6.5. Since the growth is in hours and the time-interval is hours, then n = 1. However, this leaves you still needing to figure out r. But if the population doubles every time interval, then the percentage growth, over the initial amount, must be... what percentage, expressed in decimal form?

Plug all of that information into the formula, just like you did back in algebra, and simplify to find the value of A.


P = 875, r = 0.0325, n = 1, and t = 15. Find A.


In this case, you'll need to remember that the time interval is four years. They've given you P = 5,000, A = 4,000, and r = -0.035. Let's use n = 1, and then remember to make adjustments when we interpret the meaning of the solution value for t.


This is a continuous-growth exercise, so use that formulation. (Here.)

If you get stuck on any of these (after having studied lessons for each topic), please reply showing your thoughts and efforts so far. Thank you!

Thanks for the guidance!

I think I'm on the right path.

for number 1, would r be 1? So the equation would be A=175(2)^6.5, and A would equal around 15839?
Im fairly certain this is right.

Number 2, A=875(1.0325)^15, and A comes out to 1414, rounded to the nearest dollar.

Number 3: I started with 4000=5000(0.965)^t (0.965)^t=0.8 Log base 0.965(0.8)=t. Finally I had t come out to 6.2632. Would i multiply this by 4 to get around 25 or closer to 26?

for number 4, why couldn't I use a half life decay formula or something like that? But I'll go through the way your link showed me. So A=Pe^(rt). I should convert 1 minute to 60 sec and solve for A? Should I use a starting sample of 100 and see what percentage remains? A=100e^(72.5), but I get some insane answer.

 

[Deleted member 4993]

Thanks for the guidance!

I think I'm on the right path.

for number 1, would r be 1? So the equation would be A=175(2)^6.5, and A would equal around 15839?
Im fairly certain this is right.

Number 2, A=875(1.0325)^15, and A comes out to 1414, rounded to the nearest dollar.

Number 3: I started with 4000=5000(0.965)^t (0.965)^t=0.8 Log base 0.965(0.8)=t. Finally I had t come out to 6.2632. Would i multiply this by 4 to get around 25 or closer to 26?

for number 4, why couldn't I use a half life decay formula or something like that? But I'll go through the way your link showed me. So A=Pe^(rt). I should convert 1 minute to 60 sec and solve for A? Should I use a starting sample of 100 and see what percentage remains? A=100e^(72.5), but I get some insane answer.

4000 = 5000(0.965)^t (0.965)^t = 0.8 Log base 0.965(0.8) = t.

That cannot be correct. Don't know what you are doing there!

Do it step-by-step and write one-line for each step (not string of = signs).

 

[dirtydarsh]

4000 = 5000(0.965)^t (0.965)^t = 0.8 Log base 0.965(0.8) = t.

That cannot be correct. Don't know what you are doing there!

Do it step-by-step and write one-line for each step (not string of = signs).

I'm sorry, I'm knew to this website so my bad haha.

well is this better for number 3?

Step 1: 4000=5000(0.965)^t

step 2: 0.8=0.965^t

Step 3: Log base0.965 (0.8)= t

Step 4 solve for t, t=6.263295778

Step 5 Multiply it by 4, because the decrease is happening every 4 years, to get around 25 years.



Can you tell me if I'm on the right track for number 4?

 

[stapel]

for number 1, would r be 1?

Since the growth is by 100% over every time interval, then, yes, r = 1.00 = 1 in decimal form.

So the equation would be A=175(2)^6.5, and A would equal around 15839?
Im fairly certain this is right.

Check to be sure. You started with 175, and that value doubles every time-interval, over 6.5 time-intervals. So:

. . . . .t = 0: 175
. . . . .t = 1: 2*175 = 350
. . . . .t = 2: 2*350 = 700
. . . . .t = 3: 2*700 = 1,400
. . . . .t = 4: 2*1,400 = 2,800
. . . . .t = 5: 2*2,800 = 5,600
. . . . .t = 6: 2*5,600 = 11,200
. . . . .t = 7: 2*11,200 = 22,400

Since 6.5 is between 6 and 7, then a value between 11,200 and 22,400 should be expected.

Number 2, A=875(1.0325)^15, and A comes out to 1414, rounded to the nearest dollar.

Yes.

Number 3:

Step 1: 4000=5000(0.965)^t

step 2: 0.8=0.965^t

Step 3: Log base0.965 (0.8)= t

Step 4 solve for t, t=6.263295778

Your calculator has no button for a log with base 0.965. How did you get from Step 3 to Step 4? (I agree with your value, but how did you obtain it?)

Step 5 Multiply it by 4, because the decrease is happening every 4 years, to get around 25 years.

Yes.

for number 4, why couldn't I use a half life decay formula or something like that?

You would. That's what the lesson at the link explains: you have to use the half-life info to obtain the decay constant "k".

But I'll go through the way your link showed me. So A=Pe^(rt). I should convert 1 minute to 60 sec and solve for A?

I'm not sure what you mean by this...? The time-interval is given in terms of seconds, so let t have a unit of "seconds". Then the time period for 100% of the initial sample to be reduced to 50% is t = 12.5:

. . . . .\(\displaystyle A\, =\, Pe^{kt}\)

. . . . .\(\displaystyle 0.5\, =\, 1\cdot e^{12.5k}\)

Solve for the decay constant "k".

A=100e^(72.5), but I get some insane answer.

How did you get that 60k equalled 72.5? What value did you get for the decay constant?