The population $P$ of a bacteria with respect to time $t$ can be modeled by the equation: $P=500(4+\frac{5t}{70+t^2})$

Question

Test 2 Review question #7

The population $P$ of a bacteria with respect to time $t$ can be modeled by the equation:

$P=500(4+\frac{5t}{70+t^2})$

Caluclate the rate of change of the population with respect to time when $t=2$

Verify the solution:

$Solution$:

So, we need to calculate $\frac{dP}{dt}$ and then plug in $t=2$.

$\frac{d}{dt}P=\frac{d}{dt}(500(4+\frac{5t}{70+t^2}))$

$\frac{dP}{dt}=500(\frac{d}{dt}4+\frac{d}{dt}\frac{5t}{70+t^2})$

$\frac{dP}{dt}=500(0+\frac{d}{dt}\frac{5t}{70+t^2})$

$\frac{dP}{dt}=500(\frac{d}{dt}\frac{5t}{70+t^2})$

We are going to have to use the quotient to compute $\frac{d}{dt}\frac{5t}{70+t^2}$

$\frac{dP}{dt}=500(\frac{(70+t^2)(\frac{d}{dt}5t)-(5t)(\frac{d}{dt}(70+t^2))}{(70+t^2)^2})$

$\frac{dP}{dt}=500(\frac{(70+t^2)(5)-(5t)(2t)}{(70+t^2)^2})$

$\frac{dP}{dt}=500(\frac{(350+5t^2)-10t^2}{(70+t^2)^2}$

$\frac{dP}{dt}=500(\frac{(350-5t^2}{(70+t^2)^2})$

Cool, now we just have to plug in $t=2$

$\frac{dP}{dt}=500(\frac{(350-5(2)^2}{(70+2^2)^2})$

$\frac{dP}{dt}=500(\frac{(350-5(4)}{(70+4)^2})$

$\frac{dP}{dt}=500(\frac{(330}{(74)^2})$

$\frac{dP}{dt}=500(\frac{(330}{5476}) \cong 30.13$

Answer 1

Nothing much for me to say: your answer is quite correct by my checking!

Answer 2

Your answer is correct, and in the spirit of teaching a man to fish, here's how you can check yourself the next time: https://www.wolframalpha.com/input/?i=d%2Fdt%28500%284%2B5t%2F%2870%2Bt%5E2%29%29%29+when+t%3D2