Difference between revisions of "Length of a curve using numerical methods"

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g(t) = sqr(1+(30e<sup>−0.3t</sup>)<sup>2</sup>
 
g(t) = sqr(1+(30e<sup>−0.3t</sup>)<sup>2</sup>
 +
 
<span>&#916;</span>t = (5-0)/10 = 1/2
 
<span>&#916;</span>t = (5-0)/10 = 1/2
 +
 
the subintervals are {0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4, 9/2, 5}
 
the subintervals are {0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4, 9/2, 5}
  
 
and the answer is thus = (1/6)(f(0) + 4f(1/2) + 2f(1) + 4f(3/2) + 2f(2) + 4f(5/2) + 2f(3) + 4f(7/2) + 2f(4) + 4f(9/2) + f(5))
 
and the answer is thus = (1/6)(f(0) + 4f(1/2) + 2f(1) + 4f(3/2) + 2f(2) + 4f(5/2) + 2f(3) + 4f(7/2) + 2f(4) + 4f(9/2) + f(5))

Revision as of 19:01, 1 April 2021

Simpson's Rule is used to calculate integrals numerically.

From the Length of curve page, the length can be calculated using the integral: sqr(1+f '(x)2)dx

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Instead of evaluating this integral with a table, a numerical method called Simpson's rule can be used:

the length of the curve after n iterations = Ln = (Δx/3)(f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + ...+ 2f(xn-2) + 4f(xn-1) + f(x0))

where Δx = (b-a)/n with a and b being the boundaries and n the number of iterations of the calculation.

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The length of curve equation used in the example was sqr(1+(30e−0.3t)2)dt

using boundary conditions of 0 and 5 minutes, and using 10 iterations, the problem becomes:

g(t) = sqr(1+(30e−0.3t)2

Δt = (5-0)/10 = 1/2

the subintervals are {0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4, 9/2, 5}

and the answer is thus = (1/6)(f(0) + 4f(1/2) + 2f(1) + 4f(3/2) + 2f(2) + 4f(5/2) + 2f(3) + 4f(7/2) + 2f(4) + 4f(9/2) + f(5))