# Wording of growth rates questions

• Jan 9th 2013, 11:30 AM
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.
• Jan 9th 2013, 11:48 AM
MarkFL
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:

$P(t)=P_0e^{kt}\,\therefore\,k=\ln(1.02)$
• Jan 9th 2013, 11:54 AM
ebaines
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)
• Jan 9th 2013, 03:11 PM
Re: Wording of growth rates questions
So if you invest your money at 2% per year continuously compounded, isn't that e^{.02t}?
• Jan 9th 2013, 03:26 PM
Re: Wording of growth rates questions
Wait a minute! Are you guys (and/or gals) telling me that power usage compounds not continuously?
• Jan 9th 2013, 03:35 PM
MarkFL
Re: Wording of growth rates questions
Quote:

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:

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).
• Jan 9th 2013, 06:09 PM
Re: Wording of growth rates questions
Quote:

Originally Posted by MarkFL2
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
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).

• Jan 9th 2013, 06:54 PM
MarkFL
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?
• Jan 10th 2013, 04:43 AM
ebaines
Re: Wording of growth rates questions
Quote: