1. ## Calc- Word Problem

Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by $P(t) = 2t + \frac {1}{162t + 1}$, $0 \leq t \leq 1$, where $t$ is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

$P'(t) = 2 - \frac {162}{(162t + 1)^2}$

$P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}$

$0 = 52 488t^2 - 160$

$160 = 52 488t^2$

$\pm 12.64911064 = 52488t$

$0.000240991 = t$

$P(0.000240991) = 0.963$
$P(0) = 0$
$P(1) = 2.006$

Therefore, the pollution in 0.000240991 year was at its lowest level of 0.963

-----

Did I take the first derivative wrong? I checked it and I don't see anything wrong with it. Please show me my mistake and fix it for me?

2. Hello,

Originally Posted by Macleef
Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by $P(t) = 2t + \frac {1}{162t + 1}$, $0 \leq t \leq 1$, where $t$ is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

$P'(t) = 2 - \frac {162}{(162t + 1)^2}$

$P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}$

[...]
You've got a problem here :

$(162t+1)^2=162^2t^2+{\color{red}2 \cdot 162t}+1$

Remember :

$(a+b)^2=a^2+2ab+b^2 \neq a^2+b^2$

3. Originally Posted by Macleef
Technicians working for the Ministry of Natural Resources have found that the amount of a pollutant in a certain river can be represented by $P(t) = 2t + \frac {1}{162t + 1}$, $0 \leq t \leq 1$, where $t$ is the time (in years) since a clean-up campaign started. At what time was the pollutant at its lowest level?

$P'(t) = 2 - \frac {162}{(162t + 1)^2}$

$P'(t) = \frac {52 488t^2 - 160}{26244t^2 + 1}$

$0 = 52 488t^2 - 160$

$160 = 52 488t^2$

$\pm 12.64911064 = 52488t$

$0.000240991 = t$

$P(0.000240991) = 0.963$
$P(0) = 0$
$P(1) = 2.006$

Therefore, the pollution in 0.000240991 year was at its lowest level of 0.963

-----

Did I take the first derivative wrong? I checked it and I don't see anything wrong with it. Please show me my mistake and fix it for me?
P(t) = 2t + 1/(162t + 1)

*use quotient rule on second term*

P'(t) = 2 - 162/(162t + 1)^2

*minimum occurs when P'(t) = 0*

2 - 162/(162t + 1)^2 = 0

2 = 162/(162t + 1)^2

2(162t + 1)^2 = 162

2(26244t^2 + 324t + 1) = 162

52488t^2 + 648t -160 = 0

t = 0.04938 or t = -0.06173

Cannot be negative, therefore t = 0.04938