Question

A pollutant has been leaking steadily into a river. An environmental group undertakes a clean-up of the river. The amount of pollutant in the river after t years the clean-up begins is given by the equation N(t)=0.5t+(1/(10t+1))

a) What quantity of pollutant was in the river when the clean-up began.

Ans: i tried substituting zero in the original equation but gives a value of one, i then tried finding the derivative but it gave me a negative value. Where am i going wrong?

Answer 1

st t=0? N= 1 . It probably means 100 precent. What bothers me about your equation is that as cleanup progresses, say at t=40years, N= 20.2, and at 100 years, N=50.1, so pollutant is increasing over time. THe simple solution is that clean up will never work, the pollution has to stop leaking in the river. Look at the annual flow:
annual rate increase=N'=.5-(10)/(10t+1)

so the cleanup is the second term, the first term is the leakinginto the river rate.