- #1
tangibleLime
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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)?