# A tree grows

#### noslen

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?

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#### mrjeffy321

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?

#### noslen

Yes but im trying to solve it using differentials which is what im struggling with

#### Curious3141

Homework Helper
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 :

$$\frac{dh}{dt} = \frac{h}{10}$$

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

Solve it by separation of variables :

$$\frac{dh}{h} = \frac{dt}{10}$$

$$\int_{h_0}^{h}\frac{1}{h}dh = \int_0^t\frac{1}{10}dt$$

$$\ln{\frac{h}{h_0}} = \frac{t}{10}$$

$$h = h_0e^{\frac{t}{10}}$$

Hence the answer is $10*e^{\frac{1}{2}} = 16.5$ feet (approx)

Last edited:

#### HallsofIvy

Homework Helper
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

Homework Helper
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!

#### noslen

I now understand thanks so much hall.

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