class 051101

1.  List the laws of exponents.

If you didn't get this, look it up and make sure you understand these laws and why they are so, and never forget them again.

2.  Plug the coordinates of the two points (3, 7) and (5, 9) into the form y = A * 2^(kx) and solve for A and k.  Use trial and error if necessary to solve for k.

We get

7 = A * 2^(k * 3) and

9 = A * 2 ^ ( k * 5).

We divide one equation by the other and solve for k, obtaining k = .18 approx.

We then substitute this value into the first equation and solve for A.  We get A = 4.8, approx..

So our y = A * 2^(kx) model becomes

y = 4.8 * 2^(.18 x).

3.  At annual interest rate 10%, compounded twice annually, how much would you end up with after 6 months, and after 3 years if you started with $1000?

Compounding 10% annual interest twice annually, you get 5% at the end of 6 months, and another 5% (on the amount you have at the end of 6 months) and the end of the year.

You keep compounding in this manner, adding 5% every 6 months.

After 6 months you have 1.05 * $1000 = $1050.

After another 6 months you have $1050 * 1.05 = $1000 * 1.05^2.

After 3 years you will have done this 6 times and you will have $1000 * 1.05^6.

After t years you will have added 5% how many times?

Every year we add 5% twice, so in t years we'll add 5% a total of 2 * t times.

So how much will you end up with?

$1000 * 1.05^(2t).

If you compounded daily, how much would you end up with after 3 years?

Conventionally we assume 360 days per year in this situation.  360 is nicer to work with than 365.25.

How much interest do we get per day?

We get 1/360 of 10%, or .1 / 360 = .000277.

Growth rate for each day is .1 / 360.

Our growth factor for each day 1 + .1 / 360, approximately 1.000277.

In 3 years you do this 3 * 360 times, so you get

$1000 ( 1 + .1 / 360) ^ (3 * 360) = $1349.80.

Note that if you compounded only once a year you would end up with $1331.  So you get extra money by compounding.

If you compounded continuously, how much would you end up with after 3 years?

meaning of the number e

Mth163

When I was looking at the tests I came across Explain what the number e represents?.......What do you mean...I thought that was just a variable....but I think that there is more to it?

e is the limiting value of (1 + 1/n) ^ n, as n gets larger and larger without bound.

It is the factor by which you would multiply your initial principle to get the final principle if you compounded 100% annual interest continuously for a year.

That process is described in detail in the first worksheet on exponential functions, and we will also be discussing it in class tomorrow.

 

mass function for tropical plant ___ growth rate and growth factor

Mth163

A certian tropical plant, after it reaches mass 8000 grams, grows annually at the rate of 6.3 percent per year.How much mass will there be 1, 2, 3, and 10 years after first reaching mass 8000 grams?I know that I think that I can find the answer by .....8000(.063) which = 504 then I can + 504 to 8000....8504...but I think that there is an easier way but I cant find it. Thanks

That will work, and the easier way is this:

8000 + 8000 ( .053 ) = 8000 ( 1 + .053) = 8000 * 1.053.

Adding .053 of a number to that number is the same as multiplying that number by 1.053.

That's why we call 1.053 the growth factor.

You multiply by 1.053 for each year.

So for 2 years you get 8000 * 1.053 ^ 2. For 3 years you get 8000 * 1.053 ^ 3. for 10 years you get 8000 * 1.053 ^ 10.

For t years you get 8000 * 1.053 ^ t.

That's why we say that the mass function is

m(t) = 8000 * 1.053 ^ t.

A certian tropical plant, after it reaches mass 8000 grams, grows annually at the rate of 6.3 percent per year. How much mass will there be 1, 2, 3, and 10 years after first reaching mass 8000 grams? .....I have another question about this same problem....What function m(t) gives mass as a function of number t of years?....I have know idea what to write down here...I think that it is that formula that I ask you about in my previous question.....also the next one....What are the growth rate and growth factor of this function?....I know that we worked on this in class but to be honest I really didnt understand were everything was coming from....I think that the growth rate in .063 but I am not sure....

This question is answered in the above.

spring length as linear function of suspended weight

Mth163

If a door spring has a lenght that changes lineary with the weight supported as long as the sping's length is at least 49cm, what is the linear function weight(load) that models the lenght, provided that the length is 76 cm when the load is 1.53 pounds, and that the lenght changes at a rate of .049 cm/pound?......I really dont even understand were to start...I cant think of a formula or anything to say....I might be able to say Lenght over Lenght...but I dont know ....Thanks.

Any linear function is of the form y = m x + b.

If the length changes linearly with weight, then we have y = m x + b with y = length and x = weight.

To help keep track of the meanings of our variables, let's write this as

length = m * weight + b.

The slope of a graph is the rate at which the vertical coordinate changes with respect to the horizontal coordinate. So the slope m is the rate at which length changes with respect to weight. That would be the .049 cm / pound.

So the function is

length = .049 cm/pound * weight + b.

Now you know that length is 76 cm when the weigth is 1.53 pounds so

76 cm = .049 cm/pound * 1.53 pounds + b.

We solve this equation for b and obtain b = 75.924 (check my mental arithmetic on this, I could be wrong).

This would give us the equation

length = .049 cm/pound * weight + 75.924.