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do not evaluate the equation

2007-04-17 16:20:01 · 3 answers · asked by rockabilly.betty 2 in Science & Mathematics Mathematics

3 answers

Ok, let's find the perimeter of every function :

L = integrate(raiz( 1 + (dy / dx)^2))dx

x : goes from 0 to 1

For : y = x^2 : dy / dx = 2x

L = integrate(raiz( 1 + 4x^2))dx

x = 1/2*tant >>>>> dx = 1/2*sec^2t

L = integrate(sqrt(1 + tan^2t)*1/2*sec^2t

L = 1/2*integrate(sec^3tdt)

L = 1/2*(sect*tant ) / 2 + 1/2*ln(sect + tant)

x = 1/2tant

tant = 2x ; sect = sqrt(1 + 4x^2)

L = x*sqrt(1 + 4x^2) + 1/2*ln(sqrt(1 + 4x^2) + 2x)

Now, replace 0 to 1

L = sqrt(5) + 1/2*ln(sqrt(4) + 2) - 1/2*ln(3)

Hope that helps

2007-04-17 16:26:22 · answer #1 · answered by anakin_louix 6 · 0 0

Remember the arclength formula:
L = ∫√(1+(dy/dx)^2) dx
So here we have
P = ∫(0 to 1) √(1 + (2x)^2) dx + ∫(0 to 1) √(1 + [(1/2)x^(-1/2)]^2) dx
= ∫(0 to 1) √(1 + 4x^2) dx + ∫(0 to 1) √(1 + 1/(4x)) dx

2007-04-17 16:32:34 · answer #2 · answered by Scarlet Manuka 7 · 0 0

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2016-12-26 12:27:47 · answer #3 · answered by Anonymous · 0 0

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