September 15, 2006

Scribe Post: Day 6

ScribeBadge11Okay, 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 = Ao (M) t / p
where...
A is the final value
Ao 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) = Po(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 Po = 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.

11 comments:

Lani said...

Hi Manny,

I'm thinking it's good you've shared Mr. K's story so that later in the year it will be here if you you want to refer back to it. Sometimes good stories get lost when they're not recorded.

What a good technique!--your use of frames to draw your readers to the answers following your explanations--

Best,
lani

charlotte said...

HI Manny,

You are brilliant on using color, like light orange on Questions, green on Solution and red on answer. I really like that becuase its easier to go back and find either the Question, solution or the answer! I just wanna say Godd job, because i understand everyting u wrote and it helps me alot!

jann said...

Hey Manny,
Well done on the post. Your detailed scribe helped me understand exponential functions. Since i have no idea what happened in grade 12, Im doing my best to understand these concepts. Anyways, good job again with the scribe.

ashlynn said...
This comment has been removed by the author.
Suzanne said...

This post is very detailed, and the step-by-step explanations make it easy to understand even if you got confused part way through the in-class explanation. Not to mention the excellent technical aspects with frames and colour, as have already been mentioned in previous comments. I think this post deserves to be in the hall of fame.

M-A-R-K said...

Wow, nice scribe Manny. it's really simple and there is not much diagrams, but the use of your color, bold and italics is excellent. Your scribe post is an excellent example of how a scribe post can still stand out, without the use of big flashy diagrams. The transitions from the colors helped. Your explanations were also clear and easy to understand =), which has set me for tomorows quiz. Thanx Manny for a great SP. oh yeah, Mr. k i'd like to see Manny's SP in the HOF.

danny said...

I like way you explained everything in basically almost step by step, the colour was great and the frames for the final answers were a smart thing to do. Everything is well organized and I think this should be in the HOF good job Manny, you cleared up whatever things that may have confused me a bit during the class.

danny said...

I like way you explained everything in basically almost step by step, the colour was great and the frames for the final answers were a smart thing to do. Everything is well organized and I think this should be in the HOF good job Manny, you cleared up whatever things that may have confused me a bit during the class.

linger said...

Holy moly Manny, lol. I agree with the previous comments. Your scribe is simple yet it is very easy to understand and helpful too! You make it look so easy to make a post when we all know how time consuming it is. I definitely think your post HOF worthy... keep up the good work =)

P.S. We have a quiz tomorrow? dun DUN DUNN.

lindsay said...

manny, your post was easy to understand. i don't quite follow things very well in class but your scribe made me feel less left behind. =). i like how you presented your work and i think it's HOF worthy, too!.

be happy.

Stephanie said...

Hi manny,

I dont no who u r but i luv the way its detailed in every way tht u put it and the words and ummm the thing ppl need to under stand u highlighted them with a different colour you explain them really well so ppl will understand!!!

good job,

Stephanie:)