Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 1 + 8??t, y = t^5 - t, z = t^5 + t; (9, 0, 2) x(t), y(t), z(t) =
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We can use any of the parametric equations to find t: x(t) = 9 = 1 + 8\sqrt{t} 8\sqrt{t} = 8 \sqrt{t} = 1 t = 1 Now, we can find the tangent vector at t = 1: dx/dt(1) = 4(1)^{-1/2} = 4 dy/dt(1) = 5(1)^4 - 1 = 4 dz/dt(1) = 1 So, the tangent vector is (4, 4, 1). Show more…
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