Section 2.5 #15
A population of bacteria can be modeled by the equation:
$P = 700[1-\frac{3}{(t^2+2)^2}]$
Find the rate of change with respect to $t$ when $t=1$
So we have to take the derivative with respect to $t$ and then plug in $1$.
$\frac{dP}{dt} = \frac{d}{dt}700[1-\frac{3}{(t^2+2)^2}]$
$\frac{dP}{dt} = 700[\frac{d}{dt}1-\frac{d}{dt}\frac{3}{(t^2+2)^2}]$
$\frac{dP}{dt} = 700[0-\frac{d}{dt}\frac{3}{(t^2+2)^2}]$
$\frac{dP}{dt} = 700[-\frac{d}{dt}\frac{3}{(t^2+2)^2}]$
Okay, so here is the hard part, trying to evaluate $-\frac{d}{dt}\frac{3}{(t^2+2)^2}$. Let's use the properties of exponents to move the denominator up into numerator and then use the chain rule.
$\frac{dP}{dt} = 700[-\frac{d}{dt}\frac{3}{(t^2+2)^2}]$
$\frac{dP}{dt} = 700[-\frac{d}{dt}3(t^2+2)^{-2}]$
The $3$ is just a constant so we don't have to worry about it while taking the derivative, and so
$\frac{dP}{dt} = 700[-3\frac{d}{dt}(t^2+2)^{-2}]$
$\frac{dP}{dt} = 700[-3\frac{d}{dt}(t^2+2)^{-2}]$
$\frac{dP}{dt} = 700[-3(-2)(t^2+2)^{-3}\frac{d}{dt}(t^2+2)]$
$\frac{dP}{dt} = 700[6(t^2+2)^{-3}\frac{d}{dt}(t^2+2)]$
$\frac{dP}{dt} = 700[6(t^2+2)^{-3}2t$
$\frac{dP}{dt} = 8400t(t^2+2)^{-3}$
$\frac{dP}{dt} = \frac{8400t}{(t^2+2)^{3}}$
Now we plug in $t=1$ and get:
$\frac{8400(1)}{(1^2+2)^{3}}=311.11111 \approx 311.1$
If we wanted to plug in $t=2$ then:
$\frac{8400(2)}{(2^2+2)^{3}}=77.7777777 \approx 77.8$