Question

Find an exponential model for the data.

Answer 1

The table gives the approximate number of degrees Fahrenheit above room temperature of a cup of tea as it cools.

Find an exponential model for the data.

Answer 2

Hello! There are various exponential models, but why not start with this one and determine whether it's useful or not here? y=ae^x.

Then, stealing data from that table,

If x=0 and y =132, and we substitute those values into the model above, we could find a value for a:

132=ae^0 = a(1) =a

so, perhaps y=132e^x, perhaps not.
the next point is (1,120). Subst. that info into our model and see whether the equation is true or not. 120=132e^1 ?? Nope.

Have you any idea of what to do next? Hint: We may have to modify the model.

Answer 3

No, my lesson plan really didn't give good detail. They kind of left me for dead on this one.

Answer 4

Glad I had the good luck to drive by and find you lying there. First, here's some pure water.

What do you know about exponential functions?

Answer 5

I know that if you have a data set like the one we have, we have to see if the two rows have constant differences. In order for it to be exponential, it has to be either divided or multiplied.

Say that x has a constant difference of 1

X: 1, 2, 3, 4

And we have a row "f(x)":

f(x): 5, 20, 80, 320

The model would be exponential.

But that's all I could really make out of the lesson, and that is this truth.

Answer 6

The ratio of that bottom line would be (multiply by)4

Answer 7

Ric: Yes, you're given a table of x and y values, and you attempt to find an 'exponential model' that correctly predicts y for each given value of x.

The simplest and most common expo function is y=e^x; how much experience have you with that?

Answer 8

Not much, only, really from what you have told me.

Answer 9

Do you have access to a calculator that has "e" on it?

Do you have a function key marked e^x?

Answer 10

\[e ^{x}\]

Answer 11

I think I have one. What would I type down?

Answer 12

Try this: e^1. You should get 2.7182818... something like that.

Answer 13

Yes, that is exactly what I got.

Answer 14

Cool. You're on the right track.

Now it's a pretty big leap from that to this:

Assume that your exponential model looks like

\[y=ae ^{bx}\]

Answer 15

It'd be our job to find values of a and b (both constant) that correctly predict the y values in your table when the x values are given. Does this make any sense to you?

Answer 16

Okay, so, how would I start that? Can you do the first just as an example so I can write it down and try the others myself?

Answer 17

sure!

Your first data point, from the table, is (0,132). In other words, when x = 0, y = 132.
Agree with that?

Answer 18

Now we're going to plug both of those coordinate values into our model.
Our model is \[y=ae ^{bx}\]
and we're letting x =0 and y=132.
Do you happen to know the value of e^0?
If not, try it on your calculator.

Answer 19

e^0 = 1
I did learn that.

Answer 20

In case there's any confusion here, the aim of doing this is to determine a value for a.

If your x=0 and your y=132, then you have the following:\[y=132=ae ^{0}=a. \] What's the value of a?

Answer 21

Hint: you don't need the calculator for that.

Answer 22

would a=132? because 132*1=132?

Answer 23

That's right! therefore, our model becomes\[y=132e ^{bx}\]

Answer 24

OK with that?

Answer 25

Yes, that seems good to me. Then what from there?

Answer 26

Please take another look at the table and obtain from it your second point.
Write it as (x,y).

Answer 27

The first point was (0,132). The second is ( ? , ? ).

Answer 28

(1,120) would it become 120=ae^1 ?

Answer 29

You've identified the 2nd point correctly and also correctly substituted the x and y values into our model. Great. That reduces to 120=ae^(b*1). Be sure to keep that b! Our job is to find b.

Also, remember that we've already found that a=132.

So our model now says 120=132e^(b*1). Agreed or not?

Answer 30

How did 132=a? That slipped me

Answer 31

Please go back to your third or fourth response before your last response. You said there, "Would a=132?"

Answer 32

How are you doing?

Answer 33

Oh yes! Okay I see now.

Answer 34

OK: So then you have the new equation 120=132e^(b).

How would you go about determining the value of b? I'd use logarithms in your shoes.

Answer 35

would we try to find a common difference of the line and then substitute that difference for b?

Just shooting a guess, I think that is what it had me do earlier, but I could very well be mistaken. I've been working on algebra all day.

Answer 36

Let's modify that statement a bit. Your data doesn't have a common difference, but it does have a common multiple.

Here's how you could verify that: If you multiply 132 by 10/11, you'll get 120. Please try that.

Answer 37

it would be 120, that is what I got.

Answer 38

Great. Now would you multiply that 120 by the same fraction, (10/11)? compare your result to the y values in your table.

Answer 39

i got 109.09090909.....

Answer 40

Which is not exactly 110, for x=3, but is not all that far off.
Does this convince you that your data has a common ratio, 10/11?
And not a common difference?

Answer 41

Yes! so now I have to make a model for this data. How would I do that?

T(t) = ?

That's how I have to put it.

Answer 42

Right.

Answer 43

We still have one last thing to do: We need to determine the value of b.

alternatively, since e is a constant base, we need to determine the value of e^b without necessarily solving for b. Care to choose which approach you'd prefer? Either is fine.

Answer 44

I am sorry but I am still a tad bit confused. We found the values of x and y when we needed them, then we found that 120=132e^b

So now I have to find b. Would b = 10/11?

Answer 45

no, but e^b would be 10/11.
Does that make sense to you? i asked y ou again and again to multiply by 10/11, which you did, and we got the expected results. If we were to find the correct value of b, then e^b would = 10/11.

Answer 46

Yes. So e^b = 10/11

Answer 47

That's right. so our model becomes y=132e^(bx) OR y=132(10/11)^x.
You can verify this by letting x=0; you'll get 132, as expected. Next, you could let x=1. What do you get for y in this case? Does your y value agree with the table y value for x=1?

Answer 48

AHHHHH! Yes! okay! I see it!

So y=132(10/11)^x

So if x=1 we would get 120=132(10/11)^1

Answer 49

yes, and that would come to 132(10/11) = ?

Answer 50

120?

Answer 51

Is that (or is is it not) what you'd expected?

Answer 52

Honestly, you asked me and I thought to myself, "Isn't that what I just answered?"

Answer 53

I've beeen asking you to verify that your formula "works."
to answer my own question, yes, 120 is what we expected for x=1. LIkewise, 110 is what we expected for x=2 (except that we get 109 instead of 110).

So, we can conclude that our formula t=132(10/11)^x is a valid model.
What do you think?
You don't have to go througha ll this testing, but i do recommend it.

Answer 54

Yes, I understand a lot more clearly. And yes, I do think that the model would fit! Now it is time for me to go and take what you learned and do it with my other questions.

Answer 55

sorry this took so long, but I do think the testing is imiportant, as is the process of finding appropriate values for a and b (or e^b) based upon the data.

Answer 56

You've done very well and I appreciate your sticking with me through all that detail.
Hope you can now solve more problems of the same kind effectively.
I won't necessarily be on OpenStudy all night, but will look forward to your next questions.
All the best to you.