Question

A population of 490 bacteria is introduced into a culture and grows in number according to the equation below, where t is measured in hours. Find the rate at which the population is growing when t = 2. (Round your answer to two decimal places.)

Answer 1

Answer:

Rate of growth of bacteria when t=2 is 3.09 bacteria/hour

Step-by-step explanation:

As equation is not given so considering the Equation of growth of bacteria as

$P=490(1+\frac{4t}{50+t^{2}})$

We have to find the rate at which population is growing. To do so differentiate above equation w.r.to 't'

$\frac{dP}{dt}=\frac{d}{dt}490(1+\frac{4t}{50+t^{2}})\\\\\frac{dP}{dt}=490(\frac{d}{dt}(1)+\frac{d}{dt}(\frac{4t}{50+t^{2}}))\\\\\frac{dP}{dt}=490(0+\frac{4(50+t^{2})-(4t)(2t)}{(50+t^{2})^{2}})\\\\\frac{dP}{dt}=490(\frac{200+4t^{2}-8t^{2}}{(50+t^{2})^{2}})\\\\\frac{dP}{dt}=490(\frac{4(50-t^{2})}{(50+t^{2})^{2}})\\\\at\,\,t=2hours\\\\\frac{dP}{dt}=490(\frac{4(50-(2)^{2})}{(50+(2)^{2})^{2}})\\\\\\\frac{dP}{dt}=490(\frac{4(50-4)}{(50+4)^{2}})\\\\=3.09$

Rate of growth of bacteria when t=2 is 3.09 bacteria/hour