) Determinar o comprimento do arco das curvas: a) y = 1 – ln(senx) de x = π/6 a x = π/4 b) y = 0,5(e-x + ex ) de x = 0 a x = 1

-> C = ∫ √[ 1 + (dy/dx)^2 ] dx

a)

-> y = 1 - ln(senx)

-> dy/dx = d[ 1 - ln(senx) ]/dx

-> dy/dx = - 1/senx*d(senx)/dx

-> dy/dx = - 1/senx*cosx

-> dy/dx = - cosx/senx

-> dy/dx = - cotgx

Então, o arco C é:

-> C = ∫ √[ 1 + (dy/dx)^2 ] dx

-> C = ∫ √[ 1 + (- cotgx)^2 ] dx

-> C = ∫ √[ 1 + (cotgx)^2 ] dx

-> C = ∫ √[ (cossecx)^2 ] dx

-> C = ∫ cossecx dx

-> C = - ln| cossecx + cotgx |

-> C = - ln| 1/senx + cosx/senx |

-> C = - ln| (1 + cosx)/senx |

-> C = ln| senx/(1 + cosx) |

Para π/6 ≤ x ≤ π/4:

-> C = ln| senπ/4/(1 + cosπ/4) | - ln| senπ/6/(1 + cosπ/6) |

-> C = ln| (1/√2)/( 1 + (1/√2) ) | - ln| 0,5/(1 + (√3/2) |

-> C = ln| 1/( √2 + 1) | - ln| 1/(2 + √3) |

-> C = ln| 1/( √2 + 1) | + ln| (2 + √3) |

-> C = ln| (2 + √3)/( √2 + 1) |

-> C = 0,4356

b)

-> y = 0,5( e^(-x) + e^x )

-> dy/dt = 0,5d( e^(-x) + e^x )/dt

-> dy/dt = 0,5( - e^(-x) + e^x )

Então, o arco C é:

-> C = ∫ √[ 1 + (dy/dx)^2 ] dx

-> C = ∫ √[ 1 + 0,25( - e^(-x) + e^x )^2 ] dx

-> C = ∫ √[ 1 + 0,25( e^(-2x) - 2 + e^(2x) ) ] dx

-> C = ∫ √[ 1 - 0,5 + 0,25e^(-2x) + 0,25e^(2x) ) ] dx

-> C = ∫ √[ 0,5 + 0,25e^(-2x) + 0,25e^(2x) ) ] dx

-> C = ∫ √[ ( 0,5e^(-x) + 0,5e^x )^2 ] dx

-> C = ∫ [ 0,5e^(-x) + 0,5e^x ] dx

-> C = [ - 0,5e^(-x) + 0,5e^x ]

Para 0 ≤ x ≤ 1:

-> C = 0,5[ - e^(-x) + e^x ]

-> C = 0,5[ - e^(-1) + e^1 ] - 0,5[ - e^(-0) + e^0 ]

-> C = 0,5[ - 1/e + e ] - 0,5[ - 1 + 1 ]

-> C = 0,5[ e - 1/e ] - 0

-> C = 1,175

Se gostou, dá um joinha!