Answer:
x = 1 - 5t
y = t
z = 1 - 5t
Step-by-step explanation:
For the equation of a line, we need a point and a direction vector. We are given a point (1, 0, 1).
Since the line is suppose to be a tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector for which the curve passes through that point.
We have x = e^(-5t)cos5t
at t = 1, x = e^(-5)cos5
at t = 0, x = 1
y = e^(-5t)sin5t
at t = 1, y = e^(-5)sin5
at t = 0, y = 0
z = e^(-5t)
at t = 1, z = e^(-5)
at t = 0, z = 1
Clearly, the only parameter value for which the curve passes through the point (1, 0, 1) is t = 0.
In vector notation, the curve
r(t) = xi + yj + zk
= e^(-5t)cos5t i + e^(-5t)sin5t j + e^(-5t) k
r'(t) = [-5e^(-5t)cos5t - e^(-5t)sin5t] i +[e^(-5t)cos5t - 5e^(-5t)sin5t] j - 5e^(-5t) k
r'(0) = -5i + j - 5k
is a vector tangent at the point.
We get the parametric equation from this.
x = x(0) + tx'(0)
= 1 - 5t
y = y(0) + ty'(0)
= t
z = z(0) + tz'(0)
= 1 - 5t
