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?
5 answers
The formula should actually be: H(t) = 0.013t2 + 0.81t + 316 + 3.5 sin 2πt
H(t) = 0.013t^2 + 0.81t + 316 + 3.5 sin 2πt
Ah, the formula should have an exponent!
Ah, the formula should have an exponent!
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.
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.
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
how did you get this part?
H(.2559) = appr 319.7
H(.744) = appr 312.7
I substituted .2559 and .744 into the original equation.
Make sure the calculator is set to radians, not degrees.
Make sure the calculator is set to radians, not degrees.
1 answer