You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
derivatives involving e: deriv. of 500 divided by 1 + e^-t
- Thread starter Mickey
- Start date
Re: derivatives of e
I don't know. I was under the impression you just wanted to differentiate said function.
If there is more, such as a population problem, please post the exact problem statement.
Your derivative should be \(\displaystyle \frac{500e^{-t}}{(1+e^{-t})^{2}}\)
There is no way for me to know where ln comes in without more info
I don't know. I was under the impression you just wanted to differentiate said function.
If there is more, such as a population problem, please post the exact problem statement.
Your derivative should be \(\displaystyle \frac{500e^{-t}}{(1+e^{-t})^{2}}\)
There is no way for me to know where ln comes in without more info
Re: derivatives of e
Sorry, here's the full problem.
After a particularly harsh winter, the regrowth of a popoulation of rabbits in an area is given by the funtion
P(t) = 500 divided by 1 + e^-t where t is the time in years. Determine the rate of growth in the population after 3 years.
In looking at the derivative and considering substituting 3 in for t, I figured ln must come in somewhere to get rid of e.
Sorry, here's the full problem.
After a particularly harsh winter, the regrowth of a popoulation of rabbits in an area is given by the funtion
P(t) = 500 divided by 1 + e^-t where t is the time in years. Determine the rate of growth in the population after 3 years.
In looking at the derivative and considering substituting 3 in for t, I figured ln must come in somewhere to get rid of e.
Re: derivatives of e
Nope. no 'ln's will materialise from the pages, if they aren't already there....There's nothing about 'e' that makes maths try to quench it whenever it appears - quite the opposite, in fact, it tends to keep cropping up in all sorts of unexpected places - just like pi.
Watch e, study it, get to know it, and make it your friend. It's just another number, after all - and certainly a much less strange number than, say, 2 or 0.
Mickey said:In looking at the derivative and considering substituting 3 in for t, I figured ln must come in somewhere to get rid of e.
Nope. no 'ln's will materialise from the pages, if they aren't already there....There's nothing about 'e' that makes maths try to quench it whenever it appears - quite the opposite, in fact, it tends to keep cropping up in all sorts of unexpected places - just like pi.
Watch e, study it, get to know it, and make it your friend. It's just another number, after all - and certainly a much less strange number than, say, 2 or 0.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle \text{The Amazing e}\).
\(\displaystyle \lim_{x\to \infty}(1+\frac{1}{x})^{x} = e, \;\ \text{e being irrational and = 2.71828182846......}\)
\(\displaystyle \text{However just looking at }\lim_{x\to \infty}(1+\frac{1}{x})^{x}, \;\ {\frac{1}{x}\rightarrow 0}, \;\ 1^{x}\rightarrow 1\)
\(\displaystyle \text{Hence, one would think the limit is 1 but not to be.}\)
\(\displaystyle \lim_{x\to \infty}(1+\frac{1}{x})^{x} = e, \;\ \text{e being irrational and = 2.71828182846......}\)
\(\displaystyle \text{However just looking at }\lim_{x\to \infty}(1+\frac{1}{x})^{x}, \;\ {\frac{1}{x}\rightarrow 0}, \;\ 1^{x}\rightarrow 1\)
\(\displaystyle \text{Hence, one would think the limit is 1 but not to be.}\)
Re: derivatives of e
Hello, Mickey!
\(\displaystyle \text{We have: }\;P(t) \:=\:500\left(1 + e^{-t}\right)^{-1}\)
\(\displaystyle \text{Chain Rule: }\;P'(t) \;=\;-500\left(1+e^{-t}\right)^{-2}\left(-e^{-t}\right) \;=\;\frac{500}{e^t\left(1 + e^{-t}\right)^2}\)
\(\displaystyle \text{Now find }P'(3).\)
Hello, Mickey!
After a particularly harsh winter, the regrowth of a popoulation of rabbits in an area
. . \(\displaystyle \text{is given by the funtion: }\;P(t) \:=\:\frac{500}{1 + e^{-t}}\;\text{ where }t\text{ is the time in years.}\)
Determine the rate of growth in the population after 3 years.
\(\displaystyle \text{We have: }\;P(t) \:=\:500\left(1 + e^{-t}\right)^{-1}\)
\(\displaystyle \text{Chain Rule: }\;P'(t) \;=\;-500\left(1+e^{-t}\right)^{-2}\left(-e^{-t}\right) \;=\;\frac{500}{e^t\left(1 + e^{-t}\right)^2}\)
\(\displaystyle \text{Now find }P'(3).\)