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?
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.
To find the quantity of pollutant in the river when the clean-up began, you need to substitute t = 0 into the equation N(t) = 0.5t + (1/(10t+1)):
N(0) = 0.5(0) + (1/(10(0)+1))
N(0) = 0 + (1/(0+1))
N(0) = 1/(1)
N(0) = 1
So, the quantity of pollutant in the river when the clean-up begins is 1.
Regarding your attempts with substitution and differentiation, let's discuss them:
1. Substituting zero into the original equation would give you:
N(0) = 0.5(0) + (1/(10(0)+1))
N(0) = 0 + (1/(0+1))
N(0) = 0 + (1/1)
N(0) = 1
So, substituting zero into the equation correctly gives you the correct value of 1.
2. Differentiation is used to find the rate of change of a function with respect to its independent variable. In this case, you are trying to find the initial amount of pollutant, not its rate of change. Differentiation wouldn't help in this specific case.
In conclusion, the correct answer is indeed 1 for the quantity of pollutant in the river when the clean-up began.
To find the quantity of pollutant in the river when the clean-up began, we need to evaluate the function N(t) at t = 0. However, substituting t = 0 directly into the equation doesn't give the correct result.
Let's take a closer look at the equation N(t) = 0.5t + (1 / (10t + 1)). To find the initial quantity of pollutant, we want to determine the value of N(t) when t is very close to, but not equal to, zero.
One approach to finding the quantity of pollutant when the clean-up began is by taking the limit as t approaches zero. We can calculate this as follows:
lim (t->0) N(t) = lim (t->0) (0.5t + (1 / (10t + 1)))
To simplify the expression before applying the limit, we can multiply the entire equation by (10t + 1) to eliminate the denominator:
lim (t->0) (0.5t*(10t + 1) + 1)
= lim (t->0) (5t^2 + 0.5t + 1)
= 0*(0) + 0*(0) + 1
= 1
The limit of N(t) as t approaches zero is 1, which means that the initial quantity of pollutant in the river when the clean-up began is 1.
Therefore, the correct answer is that there was 1 unit of the pollutant in the river when the clean-up started.