Scribe Post: Day 6

Okay, I'm the scribe for today, Manny. So for today's class we started by touching up on our homework (homework lab1). It was nothing too difficult (things we did in pre-cal), but Mr.K wanted us to be absolutely correct and KNOW IT IS! He told us a story about an honour role engineer scoring above 90's in school, and getting a job to give a report on soil where a building will be built. Well to make a short story shorter... he gets the report done, and gives it to the boss in charge and the engineer asks "is it right?" ... Uh, thats why YOU were hired! And that was the story.

After that we did these two following word problems, with the solutions:

1) The half-life of a toxic substance is 11 250 years.

a) If 153 g of the substance is present now, write a formula that gives the amount as a function of time, t years from now A(t).
In this problem, lots of information is given. We have the intial value, the multiplication factor, and the period it takes to get to the multiplication factor. So we would look at this equation

A = A_o (M) ^{t / p}where...
A is the final value
A_o is the intial or original value
M is the multiplication factor
t is the time
p is the period

And now we can just plug in the values. Leaving you with the equation...

A(t) = 153(1/2) t /11 250

b) When will only 0.1 g of the substance remain?
We use the equation found in a) above and just plug in the values to look like...

0.1 = 153(1/2) ^{t /11 250}

and now we solve for t. To do this, we first need to isolate t as much as possible (it looks better and makes it easier to handle), making it look like this...

0.1/153 = (1/2) ^{t /11 250}

so to make it easier to deal with the exponent we would use logarithms or natural logs (base "e"). In this example I'll be using the natural logs just because it's one less letter to type, just like the words of Mr.K. So therefore it'd look like this..

ln 0.1/153 = ln (1/2) ^{t /11 250}
ln 0.1/153 = t/11 250 ln(1/2)
(ln 0.1/153)/(ln (1/2)) = t /11 250
11 250 [(ln 0.1/153)/(ln (1/2))] = t

then you're left with this as your exact answer...

t = 11 250 [(ln 0.1/153)/(ln(1/2))] years

REMEMBER: A logarithm is an exponent!

2) In 1951 the population of India was 357 million people. By 1981 it had grown to 684 million. If the population is growing exponentially, what is the population of India today? - i.e. this month

In the problem, it only states two pieces of information. The population in two different years. When little information is given in continuous growth problems, we would use this equation..

P_(t) = P_o(model)^t

From the data, we can let 1951 be 0, so let t=0 in 1951
and since 1981 is 30 years later than 1951, let t=30 in 1981
so now we can say P_o = 357 million and P₃₀ = 684 million.

The only piece of information we're missing is the model in which it grows at. To solve for it, plug in the values, making our equation look like this..
683 = 357 (model)³⁰ isolate,
683/357 = (model)³⁰

then now we have a choice of using the logs or natural logs, or we can just raise both sides to the exponent of (1/30), like this
[683/357] ^1/30 = [(model)³⁰] ^1/30 and so now our

(model) = [683/357] ^1/30 and you can just plug in this value into the equation.

There's all sorts of ways of tackling a problem, you just gotta find the one that's more suitable for you. For me its the natural log because, on the exponent on base "e", it gives you the growth rate as a percent already. So we would use the "pert" formula. For example... (continuing from the isolation above)
683/357 = (model)³⁰
ln 683/357 = ln (model)³⁰
ln 683/357 = 30 ln (model)
[ln 683/357] / 30 = ln (model)
(model) = 0.0217 (rounded to 4 decimal places)
0.0217 happens to be the rate of growth, so in percent it would be 2.17%.

So now using base "e" as our (model), we plug in the values. And since it asked for the population in India today, that will be our time. [ 2006 - 1951 = 55 years + .75 because of september is 3/4 of the year. ]

P = 357(e ^{r t} )
P = 357(e ^{(0.0217) (55.75)} )

P = 1196.9247 million people (rounded to 4 decimal places)

And that was the end of our day! My brain is fried right now, it's like late! I got home from work past 12 =(. My sleeping habits always get ruined over the weekends, sigh...

Homework in our textbook on 1.4 all odd questions + 6, 12, and 14.

So anyway! I hope I was clear and concise about everything. I tried adding in some useful annotations for each step. If things are unclear, comment me and i'll fix it asap. Now, it's bedtime for me... the next scribe will be ashlynn just because you're first alphabetically =).

Good Stuff.