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Math Help - Wording of growth rates questions

  1. #1
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    Wording of growth rates questions

    Example in book: Energy consumption for some city is 7000 megawatts (MW) and increases at the rate of 2% per year.
    a. Find the function which gives power consumption

    The book gives
    P(1)=1.02
    P_0 = 1.02(7000)=7140
    P(1)=P_0e^{kt}
    7140=7000e^{k}
    k=.0198
    P(t)=7000e^{.0198t}

    Which means the growth rate is .0198 or 1.98%. But the growth rate (and I copied pretty much the exact question wording) is 2%. The question says it grows 2% every year. That means the growth rate is 2%! End of story.

    But k is the growth rate. The equation is e raised to the growth rate times t. Shouldn't the answer be
    P(t)=7000e^{.02t}

    What gives?
    Thanks.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Wording of growth rates questions

    We could write:

    P(t)=P_0(1+0.02)^t=P_0e^{t\ln(1.02)}

    The growth rate r is found from:

    P(t)=P_0(1+r)^t\,\therefore\,r=0.02=2\%

    The growth constant k is found from:

    P(t)=P_0e^{kt}\,\therefore\,k=\ln(1.02)
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Wording of growth rates questions

    What gives is that the exponent is NOT the same as the growth rate in decimal terms. If you evaliuate e^0.0198 it is close to 1.02, whereas e^0.02 is more like 1.0202. Hence e^(0.0198t) does indeed yield a 2% growth rate, and e^0.02T is close to a growth rate of about 2.02%.

    To prove this try writing out a table of 7000(1.02)^t, and see how it compares to 7000(e^0.0198t)
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    Re: Wording of growth rates questions

    So if you invest your money at 2% per year continuously compounded, isn't that e^{.02t}?
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    Re: Wording of growth rates questions

    Wait a minute! Are you guys (and/or gals) telling me that power usage compounds not continuously?
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    MHF Contributor MarkFL's Avatar
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    Re: Wording of growth rates questions

    Quote Originally Posted by mathDad View Post
    So if you invest your money at 2% per year continuously compounded, isn't that e^{.02t}?
    Yes, but your investment will grow by slightly more than 2%. 2% growth is what you would get if the interest was only compounded once (annually).

    Quote Originally Posted by mathDad View Post
    Wait a minute! Are you guys (and/or gals) telling me that power usage compounds not continuously?
    No, the growth is continuous, but the growth constant here is ln(1.02) rather than 0.02. Here we want P(t+1) = 1.02P(t).
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    Re: Wording of growth rates questions

    Quote Originally Posted by MarkFL2 View Post
    Yes, but your investment will grow by slightly more than 2%. 2% growth is what you would get if the interest was only compounded once (annually).
    But isn't 2%, that is, the interest rate, the growth rate? Am I just getting caught up in terminology? If so, how does one differentiate what the question means? (It's possible I may already know these answers and just be getting frustrated).

    Quote Originally Posted by MarkFL2 View Post
    No, the growth is continuous, but the growth constant here is ln(1.02) rather than 0.02. Here we want P(t+1) = 1.02P(t).
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Wording of growth rates questions

    In terms of an investment, if r is the annual percentage rate, then the investment value with n compounding periods per year is:

    A(t)=A_0\left(1+\frac{r}{n} \right)^{nt}

    Only with n = 1 will \frac{A(t+1)}{A(t)}=1+r

    The APR tells you exactly how much your investment will increase with n = 1. As n increases, you get more return though, limited by the amount returned by continuous compounding, which is:

    A(t)=A_0\lim_{n\to\infty}\left(1+\frac{r}{n} \right)^{nt}=A_0e^{rt}

    Here, r is the growth constant, not the growth rate, in which there is a subtle but distinct difference. The growth rate here is e^r-1.

    If we differentiate, we find:

    \frac{dA}{dt}=rA

    We see r is the constant of proportionality in the differential equation governing exponential growth. If we differentiate:

    A(t)=A_0(1+r)^t we find:

    \frac{dA}{dt}=\ln(1+r)A

    So, we say the growth rate in this case is r, but the growth constant is \ln(1+r).

    In other words, in the form:

    A(t)=A_0(1+r)^{t}

    r is the growth rate, but in the form:

    A(t)=A_0e^{rt}

    r is the growth constant. Observe, that we may write:

    A(t)=A_0(1+r)^{t}=A_0e^{\ln(1+r)t} and:

    A(t)=A_0e^{rt}=A_0(1+(e^r-1))^t

    Do you see now the difference?
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  9. #9
    MHF Contributor ebaines's Avatar
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    Re: Wording of growth rates questions

    Quote Originally Posted by mathDad View Post
    But isn't 2%, that is, the interest rate, the growth rate? Am I just getting caught up in terminology? If so, how does one differentiate what the question means? (It's possible I may already know these answers and just be getting frustrated).
    When told that something increases at 2% per year (as this question weas worded), then the amount you have after one year is 2% greater than the amount you staretd with. What's confusing you is that in finance banks will advertise interest rates as either the annual rate, quarterly rate, monthly rate, or continuous rate. As already discussed a 2% rate compounded continuously is equivalent to about 2.02% compounded annually. But this problem says nothing about compounding monthly, quarterly, continuously; they only give the growth rate per year. And that growth rate is 2%/year, not 2.02%.
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