For The Function R(t) = Secti + Tantj, Find The Pa

LondonJoe

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For the function r(t) = secti + tantj, find the parametric equations for the tangent line at t = (Pi/4)

 

Answer No.1

Anonymous

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For the function r(t) = secti + tantj, find the parametric equations for the tangent line at t = (π/4) r'(t) = sect tant , sec^2(t) at t = pi/4 point is (√2 , 1) r'(pi/4) = √2 , 2 then, the tangent line is L = (x,y) = (√2,1) + (√2,2)k = (√2+√2k , 1+2k)

Answer No.2

Anonymous

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Y - 1 = 1(x - √2) y = x + 1-√2

Answer No.3

SmartBrainy8

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Dy/dx = (dy/dt) / (dx / dt) dy/dx = (sec t)^2 / (sec t * tan t) = cosec t at t = pi/4 dy/dx = sqrt(2) = m y = mx + c It passes through (sqrt(2) , 1) 1 = sqrt(2)*sqrt(2) + c c = -1.... parametric equation is r(t) = t i + (sqrt(2)*t -1) j....

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