The number of bacteria, n, in a dish, after t minutes is given by n=800e^0.13t.

Find the value of n when t=0. Find the rate at which n is increasing when t=15

Question

The number of bacteria, n, in a dish, after t minutes is given by n=800e^0.13t.

Find the value of n when t=0. Find the rate at which n is increasing when t=15

anonymous · Answer

n when t=0 is 8800*e^0=800

rate at which it is increasing
dn/dt=800*0.13t*e^0.13t=104*e^0.13t
rate of increase at t=15
dn/dt=104*e^(0.13*15)=

anonymous · Answer

what is dn/dt?

darkprince14 · Answer

dn/dt

derivative of n with respect to t...

I think?

anonymous · Answer

dn/dt is derivate of n w.r.t t which gives the rate of change of n

darkprince14 · Answer

Calculus!!!

anonymous · Answer

i didnt learn derivitives yet.. lol

anonymous · Answer

Well the rate at which it is increasing will just be $800(0.13)e^{0.13t}$. That's why exponential curves are interesting. Plug in $t=15$.