A manufacturer of corrugated metal roofing has some scrap sheets of metal laying around the shop. The width of material is w . The standard manufacturing process produces corrugated sheets having a profile given by y = sin ( π x/ 7 ) .
1) How thick (T ) is a finished panel?
2) If the given material has a width w = 50, how wide (L ) will a finished panel be?
Calculus 2 Arc Length Formula HELP?
2018-02-20 3:44 am
回答 (2)
2018-02-21 5:52 am
The difficulty here is that the integral of
sqrt[1 + cos^2(x)] dx
leads to an elliptic integral of the 2nd kind,
a function that is TABULATED in various places,
but which cannot be represented in terms of "elementary" functions
such as algebraic functions, trig or hyperbolic functions, or logarithms.
It is obvious that the length of a sine wave (0,0) to (pi,0) is more than
2*[sqrt(pi/2)^2 + 1^2] = 3.724 but less than 1 + pi + 1 = 5.142.
It might turn out to be easier to program a calculator to make a good
approximation of the arc length of a sine wave, than to figure out the
many different and confusing tables of elliptic integrals that have been
published.
Next approximation of the arc length of one arch of sin(x) dx:
2*sqrt[(1/2)^2 + (pi/6)^2] + 2*sqrt[(sqrt(3)/2 - 1/2)^2 + (pi/6)^2] +
+ 2*sqrt[(1 - sqrt(3)/2)^2 + (pi/6)^2] = 3.807.
The "answer" to my simplified question is probably less than 3.9,
considering how little was added by subdividing the arch into 6
pieces instead of 2.
To find the arc length of sin(pi*x/7), one can
fiddle around with a change of variables...or note that the horizontal
length of one arch is 7, and then find sin(pi*x/7) for values of x
such as 0,1,2,3,4,5,6,7 and use them to construct the sum of
seven straight-line segments, analogous to what I did above.
sqrt[1 + cos^2(x)] dx
leads to an elliptic integral of the 2nd kind,
a function that is TABULATED in various places,
but which cannot be represented in terms of "elementary" functions
such as algebraic functions, trig or hyperbolic functions, or logarithms.
It is obvious that the length of a sine wave (0,0) to (pi,0) is more than
2*[sqrt(pi/2)^2 + 1^2] = 3.724 but less than 1 + pi + 1 = 5.142.
It might turn out to be easier to program a calculator to make a good
approximation of the arc length of a sine wave, than to figure out the
many different and confusing tables of elliptic integrals that have been
published.
Next approximation of the arc length of one arch of sin(x) dx:
2*sqrt[(1/2)^2 + (pi/6)^2] + 2*sqrt[(sqrt(3)/2 - 1/2)^2 + (pi/6)^2] +
+ 2*sqrt[(1 - sqrt(3)/2)^2 + (pi/6)^2] = 3.807.
The "answer" to my simplified question is probably less than 3.9,
considering how little was added by subdividing the arch into 6
pieces instead of 2.
To find the arc length of sin(pi*x/7), one can
fiddle around with a change of variables...or note that the horizontal
length of one arch is 7, and then find sin(pi*x/7) for values of x
such as 0,1,2,3,4,5,6,7 and use them to construct the sum of
seven straight-line segments, analogous to what I did above.
2018-02-20 12:15 pm
S=int(1+f'^2)dx
收錄日期: 2021-04-24 00:59:46
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