- #1

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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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- Thread starter noslen
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- #1

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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?

- #2

mrjeffy321

Science Advisor

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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 (

Agree?

- #3

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

- #4

Curious3141

Homework Helper

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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)

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:

- #5

HallsofIvy

Science Advisor

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mrjeffy321 said:

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

- #6

HallsofIvy

Science Advisor

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noslen said:

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

Since dy/dx is the rate of growth, y must be the height of the tree and x the

Curious3141 went ahead and used h for "h"eight and t for "t"ime. Good idea!

- #7

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I now understand thanks so much hall.

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