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Find parametric equation for the tangent line to the curve given by x(t)=e^-t cos(t), y(t) =e^-t sin(t), z(t)=e^-t and point p(1,0,1)

Sagot :

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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