Rate of Change problem?

Question

The rate of change (in thousands of people per year) of the population of a town between 2000 and 2012 can be modeled by

$$R(t) = 1.5e^{0.03t},$$

where $t$ is the number of years after 2000. Assume the population continues to grow in this manner. How many years from now (2012) will it take for the population to increase by 25,000 people?

• I'm a bit confused for this one because I know that to find an amount you use the derivative of the equation.
• I took the derivative of $R(t)$ and set it equal to 25,000 so my new equation looks like this: $25,000=(0.03)(1.5)e^{0.03t}$
• This gave me a huge answer around 440 years so I'm just not sure what I'm doing wrong.

Answer 1

If you have a value defined by a function $f$, then the rate at which this value evolves is given by its derivative $f'$ (if $f$ is differentiable of course :))

Now let's say your population is given by the function $p$. Then the rate at which the population evolves is given by $p'$.

So as @exploringnet says, $R$ is already the derivative giving the rate (i.e: $p'$), what you want is the population $p$.

Since the rate is $p'=R=1.5e^{0.03t}$, what can you say about population $p$ ?

Once you have the function giving the population, to get the growth of your population between year $2000+t_1$ and year $2000+t_2$, just do:

$$\Delta p=p(t_2)-p(t_1)$$

In your case, $t_1=12$ so $\Delta p=p(t_2)-p(12)$.

Note: this is actually what exploringnet suggested :

$$\large\Delta p=\int_{12}^{t_2}R(t)dt=\left[p(t)\right]_{12}^{t_2}=p(t_2)-p(12)$$

If you want to know in how many years $\Delta p$ will be equal to $n$ (in your case $25$ since the unit of measure is in "thousands of people"), just find the value $t_2$ that verifies the euqation $\Delta p=n$:

$$p(t_2)=n+p(12)$$