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Find the growth factor problem

  1. Jul 8, 2015 #1
    1. The problem statement, all variables and given/known data
    These questions require you to see a graph - which can be accessed from the link below

    43.
    a). Find the growth factor between the numbers of stick insects in consecutive time intervals of two weeks. Hence find a relationship between ##x## (time in days) and ##y## (number of insects) that can be used as an approximate model for the data given in the graph.

    b). Use the model to estimate
    i) the number of insects 45 days after the first recorded entry on the graph
    ii) how many days after the first recorded entry the number of insects had grown to 100.
    Explain why these results can only be estimates.

    c). Find the rate at which the numbers are growing 10 weeks after the first entry and comment on the accuracy of the result.

    2. Relevant equations
    http://ibin.co/27ygaCJx9Hyx - link to the image of the graph

    3. The attempt at a solution

    Okay so I'm stuck on the part 'a' of this problem - I need to get that sorted before I move on to part b and part c.

    I figured out that the two week growth factor would be approximately ##1.5## because from the graph the number of stick insects grows from ##10## to ##15## in a two week period.

    I then presume that the question wants me to find an equation in the form ##y=ab^x## that will describe what's happening in the graph but in terms of days instead of weeks. From the points on the graph, I've tried setting up a system of equations to solve for ##a## and ##b## but I reckon I'm barking up the wrong tree as the textbook gives the answer as ##y=6.7(15^\frac{x}{14})## and I'm not sure how to get that answer. Would be very pleased if anyone can give me some pointers in the right direction :)

    Thanking you
     
  2. jcsd
  3. Jul 8, 2015 #2

    SteamKing

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    This estimate is OK for the first two-week growth period, looking at the graph, but what about some of the other periods, using data from the graph? By the time you get to week 14, the growth of the population apparently has accelerated.

    Clearly, you are not looking for a linear relationship here. How can you modify the assumed growth equation y = A * Bx to make it simpler to analyze? (Hint: don't think the usual cartesian coordinates.)
     
  4. Jul 10, 2015 #3
    Okay after thinking a bit I reckon I got part b) of the question sorted now.

    So breaking down ##y=ab^x## to make it easier to understand:
    ##y=## number of stick insects
    ##a=## initial number of stick insects (amount that I start with)
    ##b=## rate of growth (the ##1.5## growth factor I found in part a)
    ##x=## time

    To find the initial number of stick insects, I'm going backwards in time so I just divide the first recorded entry by my growth factor to get the initial number of stick insects that I'm starting with. So ##10/1.5 \approx 6.7##

    So plugging in the values in ##y=ab^x## I get ##6.7(1.5)^x## but the question wants it as a model for daily growth except my current one is doing it for two weekly periods. So dividing my time by ##14## I will get a model which will approximate daily growth.

    So I now have ##y=6.7(1.5)^\frac{x}{14}##

    So for question b).
    i) I need to find number of insects 45 days after first recorded entry which would be on day 59. Plugging in the values
    ##y=6.7(1.5)^\frac{59}{14}\approx 37##
    ii) To find when the number of insects has grown to 100, I test out different values in my model and find that it happens some time during day ##93##. And this is ##79## days after the first recorded entry.


    Now I'm a bit stuck on this last bit of the question.

    c) Find the rate at which the numbers are growing 10 weeks after the first entry and comment on the accuracy of the result

    10 weeks after the first entry would be at week 12. My problem is I'm not quite sure how to find the new growth rate now. I've tried a few things like dividing the numbers after week twelve and trying to create a new model for after week 12 etc.

    But my textbook gives the answer as a rate of increase ##\approx 5.5## per day. I can't see how it arrived at this answer. Any hints would be much appreciated.

    Thanking you
     
  5. Jul 10, 2015 #4

    HallsofIvy

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    The crucial point is that the given graph is NOT a pure exponential so does not have a constant "growth factor". Knowing that you are to find a constant growth factor you have to pick those times that will give one.
     
  6. Jul 15, 2015 #5
    Hi - I think I get at what you're saying. The graph isn't a pure exponential so an exponential model would only be approximate and get less accurate as the time increases. However I'm still quite lost with respect to the part c).

    c) Find the rate at which the numbers are growing 10 weeks after the first entry and comment on the accuracy of the result

    The textbook gives the answer as$$\approx5.5$$Would you mind showing me how it got there?

    P.S Sorry to ask such a trivial question it's just that I've been thinking about this problem a while now but can't seem to see how it got this result. I'm a self-learner so don't got no teacher to ask.

    Thank you very much :)
     
  7. Jul 15, 2015 #6

    HallsofIvy

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    I do not get 5.5. I get 6.5 but perhaps I am reading the graph wrong. Your graph seems to show that, at 8 weeks, there were about 32 insects, at 10 weeks 50, and at 12 weeks, 58. Between the 8th and 10th week the growth rate was (50- 32)/2= 9. Between the 10th and 12th weeks (58- 50)/2= 4. Between the 8th week and the 12th week, (58- 32)/4= 13/2= 6.5 which is, of course, the same as the average of 9 and 4.
     
  8. Jul 15, 2015 #7
    I'm incredibly sorry but I forgot to add that the textbook reckons the answer is a rate of increase of 5.5 per day. So we've got the 6.5 - but isn't this a weekly rate of increase? Sorry I think I'm a bit confused here. Anyhow, if I understand the question correctly, it wants me to find the rate of increase 10 weeks after the first entry - so that would be at week 12 as the first entry was on week 2. So I presume that it wants me to find the instantaneous rate of increase at week 12. Is there any conceivable way it could 5.5 per day?

    Thank-you very much for taking a look at this I really appreciate it. :)


    P.S Here are some more bits of information to give some context on the question:

    • The earlier parts of the question (part a and part b see above) were relatively straightforward and had me devising an approximate model of the form $$y=ab^x$$ to get an estimate of the number of insects at different points in time. So the model I used to answer those parts was $$y=6.7(1.5)^\frac{x}{14}$$ Is there any way that model could be used for this question??
    • This question was also the last question in a chapter on differentiation (introducing the quotient and product rule) so I'm not sure if it wants me to use calculus but I've assumed that it doesn't and that it's just an anomaly being in that chapter.
    • Here is the full answer at the back of my book: approx 5.5 per day; larger than the actual increase
     
  9. Jul 16, 2015 #8

    SammyS

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    Looking at the two week growth factors, as suggested in the problem instructions:

    The number of insects is multiplied by a factors of between 1.4 and about 1.56, except when going from the week 10 to the week 12 counts. The latter being about 1.15 .
    It's as if some calamity happened in the insect colony during that time frame.

    Below is a graph with the data points (as well as I could estimate them) along with graphs of two functions.

    PF trendline Prob_OP_funcs.png

    The graph in blue is your function: ##\displaystyle \ f(x)=(6.7)(1.5)^x\,,\ ## where ##\ x\ ## is in units of two weeks.

    The graph in red, has the multiplier, ##\ a\ ## adjusted to be 4.9, so that it fits the final 3 points quite well, still with an exponential base of 1.5.
    The function is ##\displaystyle \ g(x)=(4.9)(1.5)^x \ .##

    A trend-line through the first 5 points gives a result extremely close to your function: ##\ a\approx 6.64\ ## and ##\ b\approx 1.49\ .##
     
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