Math Help Forum: Derivative of population

Now, I KNOW I'm being idiotic when it comes to this problem. I'm sure it's very simple I'm just not sure how to go about it.


The population, P, of China, in billions, can be approximated by the function P = 1.267(1.007)^t, where t is the number of years since the start of 2000. According to this model, approximately how fast is the population growing at the start of 2000 and at the start of 2002? Give your answers in millions of people per year to 1 decimal place. (Note: For this problem use h = 0.01 in your calculations.)

Beginning of 2000= _____ million people/year

Beginning of 2002= _____ million people/year

Originally Posted by melliep

Now, I KNOW I'm being idiotic when it comes to this problem. I'm sure it's very simple I'm just not sure how to go about it.


The population, P, of China, in billions, can be approximated by the function P = 1.267(1.007)^t, where t is the number of years since the start of 2000. According to this model, approximately how fast is the population growing at the start of 2000 and at the start of 2002? Give your answers in millions of people per year to 1 decimal place. (Note: For this problem use h = 0.01 in your calculations.)

Beginning of 2000= _____ million people/year

Beginning of 2002= _____ million people/year

Are you expected to approximate the instaneouous rate of change by where t is the number of years from the given year and h = 0.01 ....? If so, please show your work and say where you're stuck.

Well it didn't specify. That's just the problem.
I did:
[1.267(1.007)^(0+.01) - 1.267(1.007)^0] / 0.01

That was wrong.

Originally Posted by melliep

Well it didn't specify. That's just the problem.
I did:
[1.267(1.007)^(0+.01) - 1.267(1.007)^0] / 0.01

That was wrong.

What numerical answer did you get?

Mmm...
.0088

Originally Posted by melliep

Mmm...
.0088

And the unit ....?? (After answering this, I suggest you go back and read the original question very carefully).

Ooooh is that billions? I need to change it to millions.