Help me solve this math problem

[Devlyn]

I guess they figure the mathematical incantations were camouflage enough! Truthfully, I think the construction crews in the area relocated it here.



Next math problem. If Devlyn uses a $600 GPS to find soggy paper in empty pill bottles, how many rocks does she have in her head?



It has begun!



Mucho thanks for the help with this!    The Geohound and I had fun finding it and acting nonchalant while people zipped by on their bikes. I filled the jar with toys and trinkets for the next finders.

Hugs, Devlyn

[KathyLauren]

Kewl!  Glad to help.  Being a math nerd is good for something after all. 

[Newfie]

So what did you use? Did the dandelions only reproduce once? If so, I think the formula should have been:

N=sum (200^t) for t={0,1,2, ... n}

If not, it should have been :

N=1 for t=0
N=(1+200)×200^(t-1) for t>0

A more elegant and commonly used version of the second equation is:

N=201*e^(ln(200)*(t-t0)) so that it follows the form
b*e^r*(t-t0)

[Laura_7]

So what did you use? Did the dandelions only reproduce once? If so, I think the formula should have been:

N=sum (200^t) for t={0,1,2, ... n}

If not, it should have been :

N=1 for t=0
N=(1+200)×200^(t-1) for t>0

I tried the formula for reproducing once, coordinates made no sense.

The author of the puzze used this solution :

https://www.susans.org/forums/index.php/topic,208870.msg1851080.html#msg1851080

(and suggested to do it without calculator )


hugs

[Newfie]

Haha ok, that's not actually the answer but I'm glad you got the goods in the end! That equation would have each plant producing 201 seeds every year instead of 200.

The reason I'm invested in this particular question is that I do population dynamics. This is a well- known equation in my field for exponential growth.

[Devlyn]

So what did you use? Did the dandelions only reproduce once? If so, I think the formula should have been:

N=sum (200^t) for t={0,1,2, ... n}

If not, it should have been :

N=1 for t=0
N=(1+200)×200^(t-1) for t>0

A more elegant and commonly used version of the second equation is:

N=201*e^(ln(200)*(t-t0)) so that it follows the form
b*e^r*(t-t0)

Kathy's solution brought me right to it. Some of the other geocachers said it turned out to be easier than they thought, they just needed to find big calculators to do it.

Each year, you have 201 times as many plants as the year before: the 200 new ones for each original one, plus the original ones that are still there.  So, in year n, you have 201^(n-1) plants.

In year 100, you have 201^99 plants.  That's a really big number.  I can't even count the digits because, on my screen, it is wider than the full screen, but it's around 250 digits long.  You need to calculate at least the first 26 digits to get the full alphabetic translation required for the coordinates.  You don't actually use all 26 letters, but you have to get as far as the letter 'T', which means 20 digits.  I tried the calculation in Excel, but it pooped out after 15 digits.

So, no, I can't get you the coordinates.

No calculator, my ***!

[KathyLauren]

That equation would have each plant producing 201 seeds every year instead of 200.

The formula was 201^(n-1).  The 201 comes from the 200 child plants plus the parent plant.

In the first year, you have 1 plant.

In the second year, you have the 200 offspring plus the parent ( = 200+1 = 201)

In the third year, each of the 201 plants from the second year has produced 200 more plants ( = 201 * 200) plus the 201 plants from the second year are still there.  So the total in the third year is 201 * 200 + 201 = 201*201.

In the fourth year, each of the 201*201 plants from the third year has produced 200 new plants ( = 201*201*200), plus the 201*201 plants from the third year are still there.  The total in the fourth year is therefore 201*201*200 + 201*201 = 201*201*201.

The general formula is therefore 201^(n-1). 

[Newfie]

Oops, you're right. And I made an additional mistake.

Let me amend my first case to be:

N = e^ln(200)(t-t_0)

Proof:
N_t=e^ln(200)(t-t_0)=(e^ln(200)t)*(e^-ln(200)t_0)
e^-ln(200)t_0 = 1 because t_0 = 0, so in this case the t_0 is really quite redundant. I'll eliminate it for now to make the math more legible. Unfortunately the sub and sup options don't seem to be working for me.

N_(t+1) = e^(ln(200)(t+1)) = (e^ln(200)t)*(e^ln(200)) = 200*e^ln(200) = 200*N_t

This follows the be^rt with b = 1 and r = ln(200). r is the instantaneous rate of change, given by (dN/dt)/y(t)

For the other case, you're absolutely right that I failed to take into account that it was cumulative past the first year. As you showed, the effect of it being cumulative is to increase r by 1, because at any time t the number which we multiply by 200 remains, and therefore the cumulative at time t+1 is Nt * 201.

The end result would be:

Nt = e^ln(201)(t-t_0)

[Devlyn]

So my boss proposed a fun math problem. Using the information below,

The total land surface area of Earth is about 57,308,738 square miles, of which about 33% is desert and about 24% is mountainous. Subtracting this uninhabitable 57% (32,665,981 mi2) from the total land area leaves 24,642,757 square miles or 15.77 billion acres of habitable land.

and the same dandelion rules as the original equation, with each dandelion covering one square inch, how long until the dandelions populate all the habitable land on earth?

Go! Tick, tock, tick, tock..........

[Newfie]

Do you want us to do this without a calculator as well?

Let's call this population maximum N_inf, and the time at which we reach N_inf to be t_a.

Then, N_inf = e^(ln(201)t_a
ln(N_inf) = ln(201)t_a
ln(N_inf) / ln(201) = t_a

N_inf = 24642757 sq mi * (5280^2 sq ft / 1 sq mi) * (144 sq in / 1 sq ft) is approximately 9.8928092 * 10^16
=> ln(N_inf) / ln(201) is approximately 7.379. Assuming that the dandelions spawn only once a year, that would mean you need 8 years to cover the total land surface of the earth.

Enjoy!

[KathyLauren]

To be consistent with the first problem, we have to count the year with the single parent dandelion as year 1.  The dendelions would cover every square inch in year 9.

Newfie is correct that the elapsed time would be 8 years.