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How do I work this out?

  1. Mar 29, 2008 #1
    Wiremu planted a tree that was 1.5 m high.
    He is told that the tree will increase in height at a rate of 8% a year.
    The height h metres of the tree can be modelled by the function
    h = 1.5(1 + 0.08)t
    where t is the time in years since the tree was planted.
    When will the height of the tree be 12 m?
  2. jcsd
  3. Mar 29, 2008 #2


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    Are you sure that you have the correct equation?

    What have you done so far?
  4. Mar 29, 2008 #3
    WOW, what happened to my answer??? It took me 5 min to write that down, don't tell me that i waisted those 5 min..lol...
    Last edited: Mar 29, 2008
  5. Mar 29, 2008 #4
    Lol I don't have a clue how to start this question
  6. Mar 29, 2008 #5
    You have an equation for height which is a function of time. You have the desired height and are looking for the required time. Can you rearrange the equation to find the time given the height?
  7. Mar 29, 2008 #6
    I got the equation to rearange to t=1.62/h. Is that right?
  8. Mar 29, 2008 #7
    Looks right to me.
  9. Mar 29, 2008 #8
    The deal is that i do not think you have set right the very first equation at the first place. I wrote a detailed answer and posted it here, but i guess a moderator deleted it, due to forum rules, that we aren't supposed to give explicit answers.
  10. Mar 29, 2008 #9
    Because i think that the differential equation that describes that situation, given that the height increases at a rate of 8% of the previous height is:

    [tex]\frac{dh}{dt}=.08h[/tex] Now if you solve this diff. eq. you will end up with the solution function:


    So, i think that this equation instead describes the height of the tree at any point in time.
  11. Mar 29, 2008 #10
    It doesn't have anything with e. Now that I look carefully at the derivation of the thread author's formula, it's wrong. It does, however, have to deal with compounding.

    The formula for compounding is F = I(1 + r)^t, not F = I(1 + r)t.
  12. Mar 29, 2008 #11

    Well, the deal is that this is an inappropriate way of describing the growth of a tree, since it implies that actually the tree grows only once a year in an abprupt way at a specific point in time, which is not true. The tree grows contnuously that is at every point in time. So, if we wanted to use the formula for compounding the appropriate one would be

    [tex] F=I(1+\frac{r}{n})^{nt}[/tex] where I would be the initial value, n the number of compoundings per year, and t the number of years.

    If we take [tex]n\rightarrow \infty[/tex] We actually get the result

    [tex] F=Ae^{rt}[/tex] which is actually the solution to the differential equation


    So, in other words, the diff eq. that i used is the proper way to describe this situation.
    Last edited: Mar 29, 2008
  13. Mar 30, 2008 #12


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    Stupidmath, please restrain yourself.

    I deleted your original post, because it was a complete solution.

    Next, I saw no evidence in the origianal post that the Differential equation for a constant precentage growth rate was part of the problem. He was GIVEN an equation for the growth rate. The equation he posted does not meet the requirement of the given problem, therefore he has been asked to double check what he posted.

    OP, please check what you have posted, I do not think your equation is correct, something is missing.
  14. Mar 30, 2008 #13

    Ok, sorry, i'll try!
  15. Mar 30, 2008 #14
    Here is an attachment to the exact question I was given... it is question 6

    Attached Files:

  16. Mar 30, 2008 #15
    What is happening with this attachment??
  17. Mar 30, 2008 #16
    The equation is h = 1.5(1 + 0.08)^t, not, h = 1.5(1 + 0.08)t
  18. Mar 30, 2008 #17
    Well, first you have to notice that this is a hypothetical situation, since like i explained before the tree grows continuously rather than just once a year, like it is claimed in this problem. So, lucky you, no need for diff. equations in this problem.

    You are asked to find t-? when h(t)=12. In other words

    [tex]12=1.5(1+0.08)^{t}[/tex] do you know how to isolate t here? jut take ln on both sides and you will be fine..lol...
  19. Mar 30, 2008 #18
    How do you take the ln of both sides?
  20. Mar 30, 2008 #19
    well, first devide throught my 1.5 and you'll get:
    [tex]8=(1+0.08)^{t}[/tex] now we take the ln of both sides so we get

    [tex]ln8=ln(1+0.08)^{t}[/tex], now do you know how to solve for t??
  21. Mar 30, 2008 #20
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