# Homework Help: How do I work this out?

1. Mar 29, 2008

### Paulo2014

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. Mar 29, 2008

### Integral

Staff Emeritus
Are you sure that you have the correct equation?

What have you done so far?

3. Mar 29, 2008

### sutupidmath

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
4. Mar 29, 2008

### Paulo2014

Lol I don't have a clue how to start this question

5. Mar 29, 2008

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?

6. Mar 29, 2008

### Paulo2014

I got the equation to rearange to t=1.62/h. Is that right?

7. Mar 29, 2008

### Snazzy

Looks right to me.

8. Mar 29, 2008

### sutupidmath

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.

9. Mar 29, 2008

### sutupidmath

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:

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

$$h(t)=1.5e^{.08t}$$

So, i think that this equation instead describes the height of the tree at any point in time.

10. Mar 29, 2008

### Snazzy

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.

11. Mar 29, 2008

### sutupidmath

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

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

If we take $$n\rightarrow \infty$$ We actually get the result

$$F=Ae^{rt}$$ which is actually the solution to the differential equation

$$\frac{dF}{dt}=rF$$

So, in other words, the diff eq. that i used is the proper way to describe this situation.

Last edited: Mar 29, 2008
12. Mar 30, 2008

### Integral

Staff Emeritus

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.

13. Mar 30, 2008

### sutupidmath

Ok, sorry, i'll try!

14. Mar 30, 2008

### Paulo2014

Here is an attachment to the exact question I was given... it is question 6

#### Attached Files:

• ###### exm-07.pdf
File size:
356.1 KB
Views:
106
15. Mar 30, 2008

### sutupidmath

What is happening with this attachment??

16. Mar 30, 2008

### Paulo2014

The equation is h = 1.5(1 + 0.08)^t, not, h = 1.5(1 + 0.08)t

17. Mar 30, 2008

### sutupidmath

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

$$12=1.5(1+0.08)^{t}$$ do you know how to isolate t here? jut take ln on both sides and you will be fine..lol...

18. Mar 30, 2008

### Paulo2014

How do you take the ln of both sides?

19. Mar 30, 2008

### sutupidmath

well, first devide throught my 1.5 and you'll get:
$$8=(1+0.08)^{t}$$ now we take the ln of both sides so we get

$$ln8=ln(1+0.08)^{t}$$, now do you know how to solve for t??

20. Mar 30, 2008

### Paulo2014

nope...

21. Mar 31, 2008

### sutupidmath

Well, we can see that the op has tried to solve this, but without sucess, so i think it is all right to go all the way through.

$$ln8=ln(1+0.08)^{t}$$

Or, better, i will just give some more hints, and try to put him/her on the right direction.

To the OP: Do you know the properties of logarithms? If not here are some:

$$log_a(xy)=log_ax+log_ay$$

$$log_ax^{n}=nlog_ax$$

$$log_a\frac{x}{y}=log_ax-log_ay$$ etc.

I hope this helps now. Does it? YOu need to figure it out which one of these rules to use on your problem. Remember that ln, is simply log base e. So, all these properties apply also at ln.

22. Mar 31, 2008

### drpizza

Not to bicker too much sutupidmath, but a continuous growth model is not correct. I'm looking out my window at this very moment. The view of all the snow and trees without leaves has served as a constant reminder for the last 6 months that tree growth isn't a continuous growth model. There's absolutely NOTHING wrong with the OP's equation, even in cases of a continuous growth model, providing t is measured in whole years. The original problem states that it's 8% "annual" growth. i.e. at the end of 1 year, it's 8% taller than at the beginning of the year. Thus, if you were going to switch to a continuous growth model, you would have to solve for r; it wouldn't be .08. And, while differential equations could be used, assuming continuous growth, they're not necessary to solve this problem. (Not to mention that this is the pre-calculus section. Last I checked, diff eq's weren't a pre-requisite for pre-calc.)

Last edited: Mar 31, 2008
23. Mar 31, 2008

### sutupidmath

Well, i don't want to blindly argue that a continuous model is actually correct, because since it is claimed that the tree was just planted, the fact that we cannot notice the growth of the tree at every moment in time, doesn't necessarly mean that the tree in fact does not grow continuously.

HOwever, since the question actually is estimating the tree growth only once a year, that formula that the op presented actually can be used, like you said.

LOL, what was the last time you checked? It has become a standard prerequisite of pre-calc nowa days!!

Sorry, this is a bad joke!

24. Mar 31, 2008

### Paulo2014

does the eqation go to: ln8=t*ln+0.08?

And what is the difference between log and ln?

25. Mar 31, 2008