new here, not sure how this works. Calculus problem, a tree grows at a rate inversely proportional to its height, i.e. 1/h. Initial height is 2 feet. After 1 month it's 3 feet. what will the height be at 12 months. I know the integral of 1/x is ln|x|, but not sure where to go from there. Pointers to worked example(s) would be great.

Question

jpsmarinho · Answer

I'm not seeing what in this problem has relation with integral...

anonymous · Answer

OK, that's probably where I'm getting stuck. How would you solve it?

jpsmarinho · Answer

Is something about using geometric progression?

anonymous · Answer

these can be viewed as a sequance but the sequence will be made by the growth
so
when height is $$h=2,h=3...$$ the growth will be inverse ie $$g=1/2,1/3...$$

anonymous · Answer

I was thinking along the line - I can make a formula for the rate of change of height, and if I take the antiderivative of that it will give me formula for height

jpsmarinho · Answer

A growth vs. time graph?

jpsmarinho · Answer

Sorry, height vs. time

anonymous · Answer

yes, so height = h, rate of growth at time t = k/h. 
need a formula for h at time t.

anonymous · Answer

If the instantaneous rate of change of height is 1/h, then the function is ln h + C, find a C such that the two points given, (0,2) and (1,3), satisfy the function, then find h when t=12.

anonymous · Answer

yes, that's the problem, how to get the answer? can you point me to a worked example

anonymous · Answer

and there should be a k in there, the instantaneous rate of change of height is proportional to 1/h, i.e. = k/h then the function is k.ln h + C, i get that bit, its the next step find h when t =12

anonymous · Answer

The k you're looking for is probably related to e. You need to find the proper constants such that you can satisfy the two points that are given to you.

anonymous · Answer

yes, thanks,but how do i find them?

anonymous · Answer

That is the problem. I don't have the answer. Look at a graph of ln x and play around with some ideas.

anonymous · Answer

We are trying to set it up so that there is an interval with a distance of 1 between a and b, and also a distance of 1 between log(a) and log(b). For ln x, we have this condition satisfied for a=1/(e-1), b=e/(e-1).

anonymous · Answer

Now you have to modify the equation so that those points correspond to the points you're looking for, by translating the graph