# Oak tree problem

• Feb 17th 2013, 11:53 PM
mockingjay95
Oak tree problem
Ann plants some acorns. She figures that as the tree grows, the cells will divide in such a way that the rate of change of the tree's mass will be directly proportional to the mass. She digs up a seedling after 3 months and finds that the mass of the roots/trunks/leaves is 200 grams. The original acorn were 70 grams.
1. Write an equation.
2. How long after the acorn was planted will the oak tree reach 50 kg?
3. predict mass of oak 1, 2, 3 years after the acorn was planted.
4 use the results of 3. to prove the oak grows by a constant factor each year
5. Explain to ann why her model would not give reasonable answers for the tree's mass after many years.

This is my work for part 1. dM/dt=kM, (skipping a few steps) I get M=Ce^kt
As M=70 when t=0, C=70
As M=200 when t=3, I solved M=70e^3k to get k=0.3499
M=70e^0.3499t (M is the mass)
For part 2, I converted 50kg to g, then I solved this equation 70e^0.349t=50000
I get t=304.22 (months)
Part 3 I plug in 12, 24, 36 for t
But I don't know how to do 4 and 5, help?
• Feb 18th 2013, 12:16 AM
earboth
Re: Oak tree problem
Quote:

Originally Posted by mockingjay95
Ann plants some acorns. She figures that as the tree grows, the cells will divide in such a way that the rate of change of the tree's mass will be directly proportional to the mass. She digs up a seedling after 3 months and finds that the mass of the roots/trunks/leaves is 200 grams. The original acorn were 70 grams.
1. Write an equation.
2. How long after the acorn was planted will the oak tree reach 50 kg?
3. predict mass of oak 1, 2, 3 years after the acorn was planted.
4 use the results of 3. to prove the oak grows by a constant factor each year
5. Explain to ann why her model would not give reasonable answers for the tree's mass after many years.

This is my work for part 1. dM/dt=kM, (skipping a few steps) I get M=Ce^kt
As M=70 when t=0, C=70
As M=200 when t=3, I solved M=70e^3k to get k=0.3499
M=70e^0.3499t (M is the mass)
For part 2, I converted 50kg to g, then I solved this equation 70e^0.349t=50000
I get t=304.22 (months)
Part 3 I plug in 12, 24, 36 for t
But I don't know how to do 4 and 5, help?

to #4.: You are asked to show that

$M(24)=k \cdot M(12)$ and that $M(36)=k \cdot M(24)$ for some constant k.

to #5.: Ann's model of the growth of the oak tree is only valid for a limited number of years (months). Otherwise the oak tree will have a nearly unlimited mass which has never been observed in nature so far.
• Feb 18th 2013, 12:37 AM
mockingjay95
Re: Oak tree problem
Is my equation correct?
• Feb 18th 2013, 05:26 AM
earboth
Re: Oak tree problem
Quote:

Originally Posted by mockingjay95
Is my equation correct?

I don't know which of your equation do you mean?

1. These equations are OK:

M(t)=Ce^kt

70e^0.349t=50000

2. The value of k is OK but I don't understand how you got the solution of t = 304 months. That means after more than 25 years the oak tree has a mass of 50 kg (Thinking)
• Feb 18th 2013, 06:01 AM
mockingjay95
Re: Oak tree problem
2. I converted 50kg into 50000g and then solved for t in the equation M(t)= 70e^0.349t. I know it's not logical, but what did I do wrong?
• Feb 18th 2013, 06:06 AM
mockingjay95
Re: Oak tree problem
OK, so I plugged in multiples of 10 until I found the one that would give me M value closest to 50000. It's 18.781, which gives me M=50009.443 is that good enough?
• Feb 18th 2013, 12:56 PM
earboth
Re: Oak tree problem
Quote:

Originally Posted by mockingjay95
2. I converted 50kg into 50000g and then solved for t in the equation M(t)= 70e^0.349t. I know it's not logical, but what did I do wrong?

I'm very much interested to know how you solved this correct(!) equation.

Quote:

Originally Posted by mockingjay95
OK, so I plugged in multiples of 10 until I found the one that would give me M value closest to 50000. It's 18.781, which gives me M=50009.443 is that good enough?

I'll show you a simpler and more accurate way to solve this equation:

$50000= 70 e^{0.349 t}~\implies~e^{0.349 t} = \frac{50000}{70}$

Now take natural logarithms on both sides of the equation:

$0.349 t= \ln\left( \frac{50000}{70} \right) ~\implies~t \approx \frac{\ln\left( \frac{50000}{70} \right)}{0.349} \approx 18.82889$