Respuesta :
The parametric equation for the tangent line to the curve is x = 1 - t, y = t, z = 1 - t.
For this question,
The curve is given as
x(t)=e^-t cos(t),
y(t) =e^-t sin(t),
z(t)=e^-t
The point is at (1,0,1)
The vector equation for the curve is
r(t) = { x(t), y(t), z(t) }
Differentiate r(t) with respect to t,
x'(t) = -e^-t cos(t) + e^-t (-sin(t))
⇒ x'(t) = -e^-t cos(t) - e^-t sin(t)
⇒ x'(t) = -e^-t (cos(t) + sin(t))
y'(t) = - e^-t sin(t) + e^-t cos(t)
⇒ y'(t) = e^-t ((cos(t) - sin(t))
z'(t) = -e^-t
Then, r'(t) = { -e^-t (cos(t) + sin(t)), e^-t ((cos(t) - sin(t)), -e^-t }
The parameter value corresponding to (1,0,1) is t = 0. Putting in t=0 into r'(t) to solve for r'(t), we get
⇒ r'(t) = { -e^-0 (cos(0) + sin(0)), e^-0 ((cos(0) - sin(0)), -e^-0 }
⇒ r'(t) = { -1(1+0), 1(1-0), -1 }
⇒ r'(t) = { -1, 1, -1 }
The parametric equation for line through the point (x₀, y₀, z₀) and parallel to the direction vector <a, b, c > are
x = x₀+at
y = y₀+bt
z = z₀+ct
Now substituting the (x₀, y₀, z₀) as (1,0,1) and <a, b, c > into x, y and z, respectively to solve for the parametric equation of the tangent line to the curve, we get
x = 1 + (-1)t
⇒ x = 1 - t
y = 0 + (1)t
⇒ y = t
z = 1 + (-1)t
⇒ z = 1 - t
Hence we can conclude that the parametric equation for the tangent line to the curve is x = 1 - t, y = t, z = 1 - t.
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