- #1

- 71

- 0

## Homework Statement

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 2cos(t)

y = sin(t)

z = t

At the point (0,1,pi/2)

## Homework Equations

## The Attempt at a Solution

r(t) = <2cos(t),sin(t),t>

r'(t) = <-2sin(t),cos(t),1>

**The parameter value corresponding to the point (0,1,pi/2) is t=pi/2, so the tangent vector there is r'(pi/2) = <-2,0,1>.**The tangent line through (0,1,pi/2) parallel to the vector <-2,0,1>, so its parametric equations are:

x=-2t

y=1

z = (pi/2)+t

I know this is correct, but the part in bold is what I do not understand. Why does the value of t=pi/2 correspond to the point (0,1,pi/2)?