Scribe post: Day 7

Hey everyone my name is Ashlynn and I will be your scribbler for today =D. Sorry if I was late on my scribe, I lost my first version and I had to start all over, and it frustrated me a bit. But it's okay now =) . Today's class began with a talk about which scribes belong in the Hall of Fame. Well class began with these questions:

1) If the graph of a function of the form y = Ce^rt passes through the points (0,3) and (2,7), determine the constants C and r.

I started by plugging the value (0,3) into the equation y = Ce^rt;
* The input value is 0, which is t, and the output value is 3, which is y. So I just substituted them into the equation*
3 = Ce^r0

Anything raised to a 0 equals 1 so e^r0 = 1.

3 = C(1)

therefore;

C=3

To solve for r, we start by using C = 3 and the points (2,7) into the equation.

7 = 3e^r2

7/3 = e^r2

Take the natural log of both sides

ln (7/3) = 2r lne

We know that ln e = 1. ln e = 1 because ln e=loge e ( I don't know how to make subscripts so bare with me =) ) Base e raised to the power of e is one. This is a rule that applies all the time. I understand what this is about, but I am having difficulty explaining it, so please be free to leave a comment.
ex) log5 5=X
5^x = 5
5^x = 5¹

Now back to our question;

ln(7/3) = 2r
(1/2) ln(7/3) = r

so, r = .4236489302 , but we can round it off to four decimal places

r = .4236

2) Six months after some laboratory rats were delivered to a new research lab, there were 786 rats present. Nine months after the original delivery there were 1720 rats. Assuming exponential growth for the rat population, how many were in the original delivery?

The equation we use is P=P _o(model)^t

We know,
P₆ = 786 t = 6
P₉ = 1720 t = 9
P₀ = ?

change in t = 9-6
= 3

We let Po = 786 then t=3. Plug these values into the equation P=P_o(model)^t

1720 = 786 (model)³

1720/786 = (model)³

Now we use natural log because it deals with population. Populations grow continuously, and that is why we use base e.

ln(1720/786) = 3 ln(model)
(1/3) ln(1720/786) = ln(model)

* *remember when using your calculator to find the answer remember to store your answer*
Sto --> , Aplha A

ln(model) = 0.2610409258 , we can round it to 4 decimal places.

model = e^.2610

Our equation is P = 786e^.2610t

Okay now we found our model now we have to find the original value. There are 2 methods.

Method 1:

P = 786e^.2610t , but we are looking for six months before so t = -6

P_-6 = 786 e^.2610(-6)

P_-6 = 164

Method 2:

Let P₆ = 786

786 = P_oe^.2610(6)

786/e^.2610(6) = P_o

164 = P_o

Homework for tonight is 1.5 in the textbook, odd numbered questions, and to sign up for an account on the websites Mr. K provided in the previous post.

This was pretty much what we did today. I hope my work was clear and easy to understand. Please be free to make any corrections if something is unclear. I'll be happy ot fix it. The next person I pick for Scribe is Loso.