• Support PF! Buy your school textbooks, materials and every day products via PF Here!

A tree grows

  • Thread starter noslen
  • Start date
26
0
Can't seem to set this problem up

Suppose a tree grows at a yearly rate equal to 1/10 of its height. If the tree is 10 ft tall now, how tall will it be in 5 years?

dy/dx=x/10

y=(x^2)/20+c
Im almost guessing this is right so far, if so will the 10ft be my constant?
 

mrjeffy321

Science Advisor
875
0
Since the tree is growing (not shrinking), then each year, the tree is 110% of what its height was the previous year.
if it starts out at 10 ft tall, then after the first year it would be 11 ft, since
10 * 110% = 11

If the tree continues at this rate each year,
y = 10*1.1^x
where y is the height of the tree and x is the number of years that have elapsed, and 10 is the original (constant) height of the tree.

Agree?
 
26
0
Yes but im trying to solve it using differentials which is what im struggling with
 

Curious3141

Homework Helper
2,830
86
This question deals with exponential growth. The whole issue is whether the growth is considered to be discrete, in which case, the approach outlined by mrjeffys is correct (this is how compound interest works), or the growth is continuous, in which case, a first order linear differential equation must be solved.

Since the tree grows continuously, the latter case is expected. The equation of this set-up is :

[tex]\frac{dh}{dt} = \frac{h}{10}[/tex]

where h is the height in feet and t is the time in years.

Solve it by separation of variables :

[tex]\frac{dh}{h} = \frac{dt}{10}[/tex]

[tex]\int_{h_0}^{h}\frac{1}{h}dh = \int_0^t\frac{1}{10}dt[/tex]

[tex]\ln{\frac{h}{h_0}} = \frac{t}{10}[/tex]

[tex]h = h_0e^{\frac{t}{10}}[/tex]

Hence the answer is [itex]10*e^{\frac{1}{2}} = 16.5[/itex] feet (approx)
 
Last edited:

HallsofIvy

Science Advisor
Homework Helper
41,709
876
mrjeffy321 said:
Since the tree is growing (not shrinking), then each year, the tree is 110% of what its height was the previous year.
if it starts out at 10 ft tall, then after the first year it would be 11 ft, since
10 * 110% = 11

If the tree continues at this rate each year,
y = 10*1.1^x
where y is the height of the tree and x is the number of years that have elapsed, and 10 is the original (constant) height of the tree.

Agree?
No, the problem said the rate of growth was 1/10 the height, not the added height.
 

HallsofIvy

Science Advisor
Homework Helper
41,709
876
noslen said:
Can't seem to set this problem up

Suppose a tree grows at a yearly rate equal to 1/10 of its height. If the tree is 10 ft tall now, how tall will it be in 5 years?

dy/dx=x/10

y=(x^2)/20+c
Im almost guessing this is right so far, if so will the 10ft be my constant?
This is an example of why it is a good idea to write out explicitely what your variables mean.

Since dy/dx is the rate of growth, y must be the height of the tree and x the time. Since the rate of grown is 1/10 the height your equation should be dy/dx= y/10, not dy/dx= x/10!

Curious3141 went ahead and used h for "h"eight and t for "t"ime. Good idea!
 
26
0
I now understand thanks so much hall.
 

Related Threads for: A tree grows

  • Posted
Replies
2
Views
1K
  • Posted
Replies
2
Views
4K
  • Posted
Replies
2
Views
7K
  • Posted
Replies
19
Views
728
  • Posted
Replies
9
Views
2K
  • Posted
Replies
1
Views
1K
  • Posted
Replies
2
Views
1K
  • Posted
Replies
1
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top