Posted by **alice** on Sunday, December 2, 2012 at 11:29am.

Could someone explain how to find which months during the year the maximum/minimum points occur for this problem occurs? I think it's around May and Oct, but I don't get how to determine that on the graph

The National Oceanic and Atmospheric Administration (NOAA) has been measuring atmospheric

carbon dioxide concentations (in parts per million) at Mauna Loa, Hawaii since 1958. The data

closely follow the pattern H(t) = 0.013t2 + 0.81t + 316 + 3.5 sin 2πt, where t = 0 represents the

year 1960. (Complete dataset available at ftp://ftp.cmdl.noaa.gov/ccg/co2/trends/c…

1. Explore the CO2 concentration model for the period 1960 – 1962.

a) Graph H for 0 ≤ t ≤ 3 and 300 ≤ H ≤ 340

d) When during the year do the maximum and minimum occur?

- Math -
**alice**, Sunday, December 2, 2012 at 11:33am
The formula should actually be: H(t) = 0.013t2 + 0.81t + 316 + 3.5 sin 2πt

- Math -
**alice**, Sunday, December 2, 2012 at 11:34am
H(t) = 0.013t^2 + 0.81t + 316 + 3.5 sin 2πt

Ah, the formula should have an exponent!

- Math -
**Reiny**, Sunday, December 2, 2012 at 12:03pm
I hope you are taking Calculus

H ' (t) = .026t + .81 + 7πcos (2πt)

= 0 for a max or min

7πcos (2πt) + .026t + .81 = 0

Wow, that's a tough one to solve

Let's try Wolfram, one of the best pages for math.

http://www.wolframalpha.com/input/?i=solve+7πcos+%282πt%29+%2B+.026t++%2B+.81+%3D+0

using the first two positive values of

t = .2559 and t = .744

we get

H(.2559) = appr 319.7

H(.744) = appr 312.7

so I would say : .2559(12) = 3 or March of 1960 has a max of 319.7

.744(12) = 8.9 or then end of August of 1960 has a min of 312.7

repeat by taking the next 4 positive solutions from Wolfram's webpage

Without the help of pages such as the link above, solving equations like this are very onerous. Methods such as Newton's Method work, but require lots of tedious steps and lots of patience.

I want you to image doing this 40 years ago, when we had not calculators or the aid of webpages, doing this with only pencil and paper. We did it.

- Math -
**alice**, Sunday, December 2, 2012 at 1:21pm
ah, thanks... i'm only taking "liberal arts math", so have only done a little bit of calculus.

how did you get this part?

H(.2559) = appr 319.7

H(.744) = appr 312.7

- Math -
**Reiny**, Sunday, December 2, 2012 at 4:04pm
I substituted .2559 and .744 into the original equation.

Make sure the calculator is set to radians, not degrees.

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