calculus

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when finding the tangent line to the curve using T(t)= r(t)prime/ /r(t)prime/ and the eq. of the curve being r(t)= (e^t, te^t, te^(t^2)), at point (1,0,0) why does t=o. How is the parameter for t found or calculated?

  • calculus -

    I think maybe you mean a surface defined by
    x(t) = e^t
    y(t) = t e^t
    z(t) = t e^(t)^2

    Now at (1,0,0)
    x(t) = 1
    y(t) = 0
    z(t) = 0
    Where is that true?
    well e^t = 1 only where t = 0
    and at t = 0, sure enough y = 0 and z = 0
    So go ahead, find tangent to the surface at t = 0

  • calculus -

    let's put this in the form
    z = f(x,y)
    I will assume that t e^t^2 means
    t e^t * e^t
    so z = x * y
    dz/dx = x dy/dx + y
    dz/dy = x + y dx/dy
    where
    dy/dx = dy/dt*dt/dx =[t+1]e^t*1/(e^t)
    dx/dy = dx/dt*dt/dy = e^t * 1/[e^t(t+1)]
    in terms of t
    dz/dx =[t+1]e^t +t e^t = [2t+1]e^t
    dz/dy = e^t + te^t/(t+1) = e^t
    The equation of the plane is then
    (x-xo)/A =(y-yo)/B =(z-zo)/C
    where
    A = dz/dx at t=0 = 1
    B = dz/dy at t = 0 = 1
    C = -1
    so
    (x-1)/1 = y/1 = z/-1
    or in terms of t
    e^t - 1 = t e^t = -t e^t * e^t

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