Calculus

if x=e^t and y=(t-3)^2 , find an equation y=mx+b of the tangent to the curve at (1,9).

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  1. dy/dx=2(t-3)dt/dx

    but
    dx/dt=e^t, so

    dy/dx=2(t-3)e^-t
    so m= that. Now when x=1, (x=e^t), t must be zero, so

    m=2(-3)=-6

    y=-6x+b

    now, at point 1,9
    9=-6+b,or b=15, so equation for line must be

    y=-6x+15

    check that.

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    bobpursley
  2. I just figured it out! thank you!

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