A pollutant has been leaking steadily into a river. An environmental group undertook a clean-up of the river. The number of units of the pollutant in the river t years after the clean-up began is given by the equation
N(t) = 2t + (1/10t+1)
a) How many units of the pollutant were in the river when the clean-up began?
b) After how many years is the number of units a minimum?
c) What may have happened at this point?
(a) N(0) = ___
(b) find where dN/dt = 0
(c) what do you think?
Thanks for Oobleck's hint.
I use the derivative and set it equal to 0, but the answer is negative. The answer to the textbook is about 11 months.
a) To find the number of units of the pollutant in the river when the clean-up began, we need to substitute t = 0 into the equation N(t) = 2t + (1/10t+1):
N(0) = 2(0) + (1/10(0)+1) = 0 + (1/1) = 1 unit
Therefore, there was 1 unit of the pollutant in the river when the clean-up began.
b) To find when the number of units is a minimum, we need to find the value of t that minimizes the function N(t). We can do this by taking the derivative of N(t) and setting it equal to zero.
N(t) = 2t + (1/10t+1)
N'(t) = 2 - (1/(10t+1)^2)
Setting N'(t) equal to zero and solving for t:
2 - (1/(10t+1)^2) = 0
1/(10t+1)^2 = 2
(10t+1)^2 = 1/2
10t+1 = ±√(1/2)
10t+1 = ±(1/√2)
Solving for t in each equation:
When 10t+1 = 1/√2:
10t = 1/√2 - 1
t = (1/√2 - 1)/10
When 10t+1 = -1/√2:
10t = -1/√2 - 1
t = (-1/√2 - 1)/10
So, the values of t at which the number of units is a minimum are:
t = (1/√2 - 1)/10 and t = (-1/√2 - 1)/10
c) At this point, where the number of units is at a minimum, it suggests that the clean-up efforts have been successful in reducing the pollutant levels to their lowest point. The pollutant is being controlled effectively during this time, and further efforts might be needed to maintain this low level and prevent any potential pollution from getting worse.
To answer these questions, we need to analyze the given equation and find the values that satisfy the conditions.
a) To determine the number of units of the pollutant in the river when the clean-up began, we can substitute t = 0 into the equation N(t).
N(0) = 2(0) + (1/10(0+1))
N(0) = 0 + (1/10)
N(0) = 1/10
So, there were 1/10 units of the pollutant in the river when the clean-up began.
b) To find when the number of units is at a minimum, we need to find the point where the derivative of N(t) equals zero. So, let's find the derivative of N(t) with respect to t:
N'(t) = 2 - (1/10(t+1)^2)
Setting N'(t) = 0:
2 - (1/10(t+1)^2) = 0
To solve this equation, we can isolate (1/10(t+1)^2):
(1/10(t+1)^2) = 2
Multiplying both sides by 10:
(t+1)^2 = 20
Taking the square root of both sides:
t + 1 = √20
Subtracting 1 from both sides:
t = √20 - 1
So, the value of t that gives the number of units a minimum is approximately t = √20 - 1.
c) At this point (√20 - 1), we may conclude that the number of units of the pollutant in the river is at a minimum. This suggests that the clean-up efforts have been successful in reducing the pollution levels to the lowest point.