How to calculate the time it takes for a tree to reach a certain height?

[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 modeled 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?

 

[Integral]

Are you sure that you have the correct equation?

What have you done so far?

 

[sutupidmath]

Are you sure that you have the correct equation?

What have you done so far?

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...

 

[Paulo2014]

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

 

[yourdadonapogostick]

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?

 

[Paulo2014]

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

 

[Snazzy]

Looks right to me.

 

[sutupidmath]

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

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.

 

[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.

 

[Snazzy]

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.

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.

 

[sutupidmath]

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.

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.

 

[Integral]

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.

 

[sutupidmath]

Sutupidmath, please restrain yourself.

Ok, sorry, i'll try!

 

[Paulo2014]

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

 

[sutupidmath]

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

What is happening with this attachment??

 

[Paulo2014]

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

 

[sutupidmath]

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

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...

 

[Paulo2014]

How do you take the ln of both sides?

 

[sutupidmath]

well, first divide through 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??

 

[Paulo2014]

nope...

 

[sutupidmath]

nope...

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.

 

[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.)

 

[sutupidmath]

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.

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.

(Not to mention that this is the pre-calculus section. Last I checked, diff eq's weren't a pre-requisite for pre-calc.)

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!

 

[Paulo2014]

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

And what is the difference between log and ln?

 

[sutupidmath]

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

And what is the difference between log and ln?

YOu forgot a 1 in your answer, check it out again.

Well, it depends some mathematicians use log in general, Ln is the logarigthm with base e. While log is used with a base 10. unless differently stated. But like i said, some people use log to mean ln, it depends how u like it.

 

[Paulo2014]

oops i mean:

ln8=t*ln1+0.08

 

[Dr Zoidburg]

oops i mean:

ln8=t*ln1+0.08

minor quibble, but you should really write that as ln(1.08), not ln1+0.08, as it does make it look like you're wanting to multiply t by the ln of 1 and then add 0.08.
If you did this, you're answer would end up being out by a little over 8%.

 

[Paulo2014]

What would I do next?

At a guess i would say ln(1.08)/ln8=t. is that right?

 

[sutupidmath]

What would I do next?

At a guess i would say ln(1.08)/ln8=t. is that right?

NO.

How do you find t from

ln8=t *ln(1.08) YOu would divide by ln(1.08) right? What do you get?