Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e−5t cos(5t), y = e−5t sin(5t), z = e−5t; (1, 0, 1)

Answer 1

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