# Calc- Word Problem

• May 31st 2008, 04:30 AM
Macleef
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

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Textbook answer: 0.049 years

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?
• May 31st 2008, 04:51 AM
Moo
Hello,

Quote:

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 :

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

(Wink)
• May 31st 2008, 04:59 AM
sean.1986
Quote:

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

-----

Textbook answer: 0.049 years

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