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Century Growth Problem: Rate of Change

  • Thread starter simplemuse
  • Start date
  • #1
[SOLVED] Century Growth Problem: Rate of Change

1. Problem: Because Earth's rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 27 centuries, what is the total (in hours) of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day, etc.)?



2. Homework Equations :
Sum of An Arithmetic Sequence:
Sn={n[1+(n-1)d]}/2



3. Attempt:
So my train of thought is to use the formula for sum of an arithmetic sequence, by having n=985500, which is the amount of days in 27 centuries. Since the difference every century is by 1 millisecond, I did this to find the change in time increase per day:
1ms/(100 years*365 days). This value is 2.74e-5. which I used as d. Plugging in the rest of the values in the formula, I determined there was a net change of 1.38e7ms over the course of 27 centuries.
S(985500)=[985500(1+(985499*.0000274)]/2
Converting this value to hours yields a change of 3.83 hours over the course of 27 centuries. However, my answer is incorrect. Any help is appreciated!
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
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2. Homework Equations :
Sum of An Arithmetic Sequence:
Sn={n[1+(n-1)d]}/2
I believe the sum should be

Sn = (n/2) · [2·a1 + (n-1)·d] .

So I get a slightly larger value. (Note: this is the sum of the time gains, not the amount by which the day would be lengthened. So this result will not have a physical interpretation.]
 
  • #3
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What does the "a" represent in the arithmetic sequence you have above?
 

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