Question

Polluted water flows into a pond. The concentration of pollutant in the pond at the t minutes is modelled by the equation c(t)= 9-90,000(1/10000+3t), where c is measured in kilograms per cubic metre.

a) When will the concentration of pollutant in the pond reach 6 kg/m^3?

b) What will happen to the concreatration of pollutant over time?

Answer 1

I have a feeling that there are some brackets in there

e.g. is it
c(t)= 9-90,000(1/ (10000+3t) )

check and come back

Answer 2

yea there is, i forgot to add it in there.

Answer 3

so , assuming my guess was correct .....

6 = 9-90,000(1/ (10000+3t) )
-3 = - 90,000(1/ (10000+3t) )
3(10000 + 3t) = 90000
30000 + 9t = 90000
9t = 60000
t = appr 6666.7 minutes or appr 111.1 days

for the second part look at the term
90,000(1/ (10000+3t) )
as t gets larger, the denominator gets larger and as a result 1/ (10000+3t) gets smaller and smaller and eventually approaches zero.
So
-90,000(1/ (10000+3t) ) will get smaller and smaller and you are left with
c(t)= 9 - (very very small)

so the concentration will settle in at 9 kg/m^3

Answer 4

Is it not 4.6 days?

For: t = appr 6666.7 minutes or appr 111.1 days