noor264
Sep 28, 2011, 03:15 PM
The percent of carbon-14 that remains in a specimen after various numbers of years is shown below.
Year5730=50% year11460=25% year17190=12.5% year22920= 6.25% year28650=3.125% year 34380=1.5625%
How can you use the function P(t)=100(0.5)^t/5730 to model this situation and determine the age of a natural specimen?
A.What percent of carbon is remaining for t =(0)?
B.Draw a graph of the function P(t) = 100(0.5)^t/5730, using the given
Table of values.
C.What is a reasonable domain for P(t)? What is a reasonable range?
D.Determine the approximate age of a specimen, given that P(t)= 70.
E.Draw the graph of the inverse function.
F.What information does the inverse function provide?
G.What are the domain and range of the inverse function?

Unknown008
Sep 28, 2011, 11:32 PM
A. What do you think you should do to get the percent of carbon for t = 0? Did you answer the first question? This would tell you how to answer this question.

B. Have you drawn the graph?

C. The domain is the horizontal 'bracket' of the graph, where it starts and where it stops. It includes the word reasonable because this equation actually never stops. What happens is that you will have to choose a point where there is so little left that you can call it to end there.

Range is the vertical 'bracket' of the graph. Use the same concept as for the domain.

D. Put P(t) = 70 in the function and solve for t.

E. Do you know the relationship between a function and its inverse?

F. That is for you to infer. There is not absolute correct answer, as long as you support it with logical reasons.

G. If you did C well, you should be able to do that well as well.

noor264
Sep 29, 2011, 03:08 PM
I did A,B and C but I am stuck on D. Could you please help me ?

Unknown008
Sep 30, 2011, 01:10 AM
Okay, what did you get for A and C?

For D, you are asked to find the age of the specimen. The age of the specimen is given by t, right?

And you are given P(t) = 70

You have an equation equivalent to P(t), right? Use that, you will get an equation in terms of t.

noor264
Oct 2, 2011, 01:46 PM
A) is 100%
C) domain is greater than zero. The Range is greater than zero and smaller than or equal to 100.
D) age of specimen is 2948.5 is it correct?

Is this the inverse function -5730log^x/100 ?

Unknown008
Oct 2, 2011, 10:53 PM
A) Correct.

C) The keyword in the question is 'reasonable'. From a math perspective, your answer is correct, but from a practical objective, you have to arbitrarily set an 'end point'. From my side, I would say that the domain is between t = 0 and t = 40110, where the percentage of carbon remaining is less than 1%.

And the range would then be between P(t) = 100% and P(t) =0.78%

ie: 0 \geq t \geq 40110 and 0.78% \geq P(t) \geq 100%

D) Let's see.

P(t) = 70

P(t) = 100(0.5)^{t/5730}

70 = 100(0.5)^{t/5730}

5730\ \log_{0.5} (0.7) = t

t = 2948.5

Right! :)

Not quite yet, you missed the 0.5 part. Using those properties:

If a = x^y, then \log_x(a) = y

And \log_x(y) \equiv \frac{\log(x)}{\log(y)}

You should end up with:

P(t) = 5730\ \log_{0.5} $$\frac{t}{100}$$

You have to make t the subject of formula, then substitute the t by P(t) and vice versa.

I guess you can post the answer for the last part now? :)

noor264
Oct 3, 2011, 02:08 PM
For the inverse graph will the x-axis be the carbon% and the y axis the years?
The inverse function indicates the number of years that have passed on the specimen providing the percentage of carbon 14 is known. am I right?
G) Switch the domain and range together?

Thank you so much :)

Unknown008
Oct 4, 2011, 12:42 AM
F. You're right in some way, but I'm sure that the real answer that was being looked for has is worded another way:

The inverse function enables us to know the percentage of carbon 14 left, after a known number of years.

Always take the x-axis as the independent variable, that is the data that you know, the data that you have. Thus, using the years (that you have), you get the percentage.

The initial graph showed the opposite. Knowing the percentage of carbon 14 allowed us to know the age of the rock/substance.

:)

G. Exactly right! :D

Well done noor264 :)