Question

Help me make a function for this data table!

Answer 1

Morning, Ric! Wheres that data table?

Answer 2

This table shows the mass in grams m of the radioactive substance iodine-131 remaining in a container t days after the beginning of an experiment.

Write a function that models the data.

(I know that is an exponential function)

Answer 3

How do you know it's an expo fn?
Have you fitted functions to data before?

Answer 4

Yes, I took the data set and found the first differences and second differences and the ratios, then plotted it on a graph and it looks like an exponential function.

Answer 5

OK, we'll ASSUME that this is indeed an expo function and proceed with an expo model. If that doesn't work, we cld always try fitting a polynomial to the data.

Answer 6

The simplest expo function is y=e^x. Can y ou picture what that looks like, in your mind's eye?

Answer 7

Yes!

Answer 8

Oh, wait, then mine is not an exponential function..

Answer 9

What kind of function does that look like to you?

Answer 10

This is the picture I had to go by in the lesson, it looks like none of these..

Answer 11

I do think that the straight line graph you've shared with me represents a linear function with a negative slope. What do you think?

Answer 12

I don't know, I thought that too, but it seems like the line has a bit of curve into it. It could jus be me.

Answer 13

Remember, when you're curve fitting, you're generally not aiming for perfection.
We could start with either a linear model y=mx+b or an exponential model with a negative exponent (decaying exponential). Probably neither will give us a "perfect" result.

Answer 14

Which model would you like to start with?

Answer 15

Again, remember: this is not an earth-shaking decision. I would definitely try the linear model because the given graph looks so straight (even if it's not perfectly so).

Answer 16

I think it would be a linear function.

Answer 17

OK, then all we have to do is to determine the slope of the given line and determine the y-intercept of that line. Please take a stab at determining the y-intercept.

Answer 18

isn't the y-intercept 1000?

Answer 19

and the x-intercept would be 4?

Answer 20

yes. You could write that as a point: (0,1000), or as b in y=mx+b: b=1000.

Answer 21

Yes, otherwise known as (4,0).

Answer 22

So now you have your "rise" and your "run".;

Please calculate the slope of the line shown.

Answer 23

Hmm, how would I determine the slope? I forgot how to do that.

Answer 24

rise
slope = ------- = ?
run

Answer 25

so

Slope = 1000/4 ?

Answer 26

Except that it should be -1000/4, since the "rise" is actuallyl a "drop".
You are going from (0,1000) to (4,0).

Answer 27

Fill in the blanks:

m = ??

b = ??

y = ( )x + ( )

Answer 28

so

y = 4x + 1000 ?

Answer 29

or would it be

y = -250x + 1000 ?

Answer 30

Yes, y = -250x + 1000.

Be sure you are comfortable with this ... if you are not, then we need to practice more.

Answer 31

I understand it more clearly now. So that would be my answer?

Answer 32

Before I say "yes," I'd like for y ou to check this work by going to the table you gave me e3arlier. choose an x value between 0 and 4 and insert it into the equation. Does the result agree with the table's corresponding y-value?

Answer 33

Or was that table data really not related to the problem we're currently working on?

Answer 34

No, the data table is, I have to find an equation that best fits that data table.

Answer 35

We started by looking at that data table. But later on you introduced the graph. I'm curious: why the graph?

Answer 36

because part of what it told me to do in the lesson was to see what type of line it was (linear, exponential, etc.). And I had to make an equation that fits the data set, on way of finding that equation is to see what kind of line the data table made.

Answer 37

From looking at the points in the data table plotted on the graph we found what kind of line it made, and i turn; found the type of equation that best fits the model.

Answer 38

Oh, is that really all we have to do? Just look at the table, do a few simple calculations, and determine just the category of function we're dealing with?
Please go back to that table. Does y seem to be increasing or decreasing as x increases?

Answer 39

x increases by one, that is our dependent variable.
y increases by a variety of numbers, it is not constant (i.e. our independent variable)

Answer 40

To save time, I'll share my views on this matter:
Looking at the table, I see that if we increase x, y decreases. Every new y value is smaller than the one before it, if we let x increase.

It's x that is the independent variable, and y the dependent one.

Please respond in terms of how clear (or unclear) this is to you and with any questions that come to your mind.

Answer 41

You are manipulating x, choosing values for it; therefore x is independent.
You will calculate y, either calculating its values or obtaining them from the table. Therefore, y is the dependent variable; it depends on x.

Answer 42

I thought

Time t (days) = x
Mass m (g) = y

Answer 43

So yes, y does depend on x

Answer 44

If you do that, Ric, you are implying that there is a relationship between mass and time. One example of that would be a material which decays over time.

Anyway, I'd like to return to the homework problem at hand.
You were apparently just to identify which type of graph best (if not perfectly) represents the behavior of the given data, right?

Answer 45

Yes, I have to find out which type of line the table makes and then
"Write a function that models the data."

Answer 46

OK.
My first impressions are that the data could be approximated by either a straight line with negative slope or a decaying exponential function.
Could you quickly sketch a line with a negative slope and a decaying exponential, just to get a picture of the behavior of each?

Answer 47

so would y=-x work?

Answer 48

as a line with a negative slope?

Answer 49

Sure, that's a decreasing linear function. All I ask is that you take a pencil and in one quick stroke draw a line that slopes downward, as the graph of y=-x does.
Then

Answer 50

quickly do the same for a decaying exponential.

Answer 51

greaet graph. now, on to the graph of a decaying exponential.

Answer 52

so would it be y=-e^x ?

Answer 53

While that is an exponential function, the graph would start in negative y territory and continue to decrease from there.
A better model would be e^(-x). Please draw that with your graphing utility and show me the result. Does this graph start in positive y territory as we hope it will?

Answer 54

great graphs!
Now we must determine which model would best represent the data in your table.
Care to take a stab at that? i just want to learn what you're thinking. I have a clear preference myself for which model but want to know y ours first.

Answer 55

I'd say it would be more of a linear with a negative slope

Answer 56

OK. Before I respond to that, please take another look at the table data.
Look at the first y value, and then at the 2nd. By how much does y decrease, going from the first y value to the 2nd?

Answer 57

Just subtract the 2nd y value from the first.

Answer 58

oh so,

1000 - 917.40 = 82.6
or
917.40 - 1000 = -82.6

Answer 59

right. the drop is about 83.
Now do exactly the same thing for the 2nd and 3rd y values.

Answer 60

917.40 - 841.62 = 75.78
or
841.62 - 917.40 = -75.78

Answer 61

Now do exactly the same thing for the 3rd and 4th y values. just do one calculation; we don't need both.

Answer 62

841.62 - 772.10 = 69.52

Answer 63

OK, I should ask you now whether the drops from one y value to the next are uniform or different. You've given me 83, 76 and 70.

Answer 64

different, first by 7 and then by 6

Answer 65

Here's my point (you must be thinking by now, "Finally!!!"):

If y does not decrease at a constant rate, then a linear approx. is not as appropriate as would be an exponential approx. Please respond as you see fit; does this make sense?

Answer 66

Yes, I thought it would have been exponential because the numbers are all close to the same

Answer 67

Actually, the following statement would be a bit more fitting:

"I'd thought the appropriate model would be that of a decaying exponential because the decreases in y with uniform increases in x are not the same."

Answer 68

That does seem more detailing

Answer 69

Anyway, I recommend using a decaying expo model.

Let this model be y=a*e^(bx), where a repres. the initial value of the function at t=0 and b is the decay constant.

Answer 70

and that is all I need, I need that right there.

Answer 71

y=a*e^(bx) is what I needed,

Answer 72

So all you have to do is to take a couple of x-y pairs and substitute the x and y values into the model to determine a and b. Want to try that? Please note: I'm helping another student at the same time, but will be back to you for certain.

Answer 73

Yes, i think I can get it on my own from here, but I do have a hair appointment with my barber in about 45 minuets. so I am going to get ready. thank you for all of your help @mathmale ! It is greatly appreciated

Answer 74

Very happy to work with you! I look forward to next time.