1
$\begingroup$

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.
|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ It's already a derivative of number of people(rate of change), annd you need number of people . So, you need to ......... it. $\endgroup$ – ABC Apr 24 '13 at 16:34
  • $\begingroup$ @exploringnet thanks for your comment. that helped me realize that I need to integrate it. but when I integrate it and set it equal to 25,000 I still get a large # around 200. is this still wrong? $\endgroup$ – Surat Apr 24 '13 at 16:53
  • $\begingroup$ By integrating you'll get the number of people increase after #2000 but you need to calculate the number of people after #2012.So, to avoid this you can directly put limits from #2012 to #x year in integral and you put that change equal to what you want. $\endgroup$ – ABC Apr 24 '13 at 17:03
  • $\begingroup$ @Surat I can't comment since I don't have enough reputation yet (so I'm posting this as an "answer"). It seems to me that your integral is correct, but now your problem is one of units of measure. Remember that your rate of change function is given in "thousands of people" for the unit.... The units of measure issue should probably be included in the answer provided by Dolma because otherwise that answer is very good. $\endgroup$ – agktmte Apr 24 '13 at 19:21
  • $\begingroup$ I don't know if you'll see this but thank you so much! that was exactly the issue I was having & I didn't even see it. thanks again $\endgroup$ – user74202 Apr 24 '13 at 19:33
2
$\begingroup$

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 $\Delta p=n$:

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Why u posted the full soln with answer?Homework questions should not to be given full answer. $\endgroup$ – ABC Apr 24 '13 at 16:50
  • $\begingroup$ @exploringnet Better ? ;) $\endgroup$ – Dolma Apr 24 '13 at 16:53
  • $\begingroup$ Hmm... deleted most of the useful part too. Now seems like a comment. I wanted to say like you don't need to calculate. But can post each step's detail. $\endgroup$ – ABC Apr 24 '13 at 17:05
  • $\begingroup$ @exploringnet Oh ok, Well I'm actually pretty new here so I might not know all the subtle details. Thanks for the comment I'll edit it again $\endgroup$ – Dolma Apr 24 '13 at 17:08
  • $\begingroup$ +!, Great answer. $\endgroup$ – ABC Apr 24 '13 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.