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Math Help - Oak tree problem

  1. #1
    Newbie mockingjay95's Avatar
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    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?
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  2. #2
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    Re: Oak tree problem

    Quote Originally Posted by mockingjay95 View Post
    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.
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  3. #3
    Newbie mockingjay95's Avatar
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    Re: Oak tree problem

    Is my equation correct?
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  4. #4
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    Re: Oak tree problem

    Quote Originally Posted by mockingjay95 View Post
    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
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  5. #5
    Newbie mockingjay95's Avatar
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    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?
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  6. #6
    Newbie mockingjay95's Avatar
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    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?
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  7. #7
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    Re: Oak tree problem

    Quote Originally Posted by mockingjay95 View Post
    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 View Post
    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
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