Answer:
[tex]x = t[/tex]
[tex]y = 1 - t[/tex]
[tex]z = 2t[/tex]
Step-by-step explanation:
Given
[tex]x=t[/tex]
[tex]y=e^{-t}[/tex]
[tex]z=2t-t^2[/tex]
(0, 1, 0)
The vector equation is given as:
[tex]r(t) = (x,y,z)[/tex]
Substitute values for x, y and z
[tex]r(t) = (t,\ e^{-t},\ 2t - t^2)[/tex]
Differentiate:
[tex]r'(t) = (1,\ -e^{-t},\ 2 - 2t)[/tex]
The parametric value that corresponds to (0, 1, 0) is:
[tex]t = 0[/tex]
Substitute 0 for t in r'(t)
[tex]r'(t) = (1,\ -e^{-t},\ 2 - 2t)[/tex]
[tex]r'(0) = (1,\ -e^{-0},\ 2 - 2*0)[/tex]
[tex]r'(0) = (1,\ -1,\ 2 - 0)[/tex]
[tex]r'(0) = (1,\ -1,\ 2)[/tex]
The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)
It should be noted that:
The equation of a line through position vector a and parallel to vector v is given as:
[tex]r(t) = a + tv[/tex]
Such that:
[tex]a = (0,1,0)[/tex] and [tex]v = r'(0) = (1,-1,2)[/tex]
The equation becomes:
[tex]r(t) = (0,1,0) + t(1,-1,2)[/tex]
[tex]r(t) = (0,1,0) + (t,-t,2t)[/tex]
[tex]r(t) = (0+t,1-t,0+2t)[/tex]
[tex]r(t) = (t,1-t,2t)[/tex]
By comparison:
[tex]r(t) = (x,y,z)[/tex] and [tex]r(t) = (t,1-t,2t)[/tex]
The parametric equations for the tangent line are:
[tex]x = t[/tex]
[tex]y = 1 - t[/tex]
[tex]z = 2t[/tex]
