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
[tex]P=490(1+\frac{4t}{50+t^{2}})[/tex]
We have to find the rate at which population is growing. To do so differentiate above equation w.r.to 't'
[tex]\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[/tex]
Rate of growth of bacteria when t=2 is 3.09 bacteria/hour
