- #1
monea83
- 20
- 0
Given is a one-parameter family of planes, through
[tex]x \cdot n(u) + p(u) = 0[/tex]
with normal vector n and base point p, both depending on the parameter u.
Two planes with parameters [tex]u_1[/tex] and [tex]u_2[/tex], with [tex]u_1 < u_2[/tex], intersect in a line (planes are assumed to be non-parallel). This line also lies in the plane
[tex]x \cdot (n(u_1) - n(u_2)) + p(u_1) - p(u_2) = 0[/tex]
Now, the book I am reading claims that, "by Rolle's theorem, we get:"
[tex]x_1 n_1'(v_1) + x_2 n_2'(v_2) + x_3 n_3'(v_3) + p'(v_4) = 0[/tex] with [tex]u_1 \leq v_i \leq u_2[/tex].
However, I don't see how the theorem applies here... for starters, I don't see anything of the form [tex]f(a) = f(b)[/tex], as required by the theorem.
[tex]x \cdot n(u) + p(u) = 0[/tex]
with normal vector n and base point p, both depending on the parameter u.
Two planes with parameters [tex]u_1[/tex] and [tex]u_2[/tex], with [tex]u_1 < u_2[/tex], intersect in a line (planes are assumed to be non-parallel). This line also lies in the plane
[tex]x \cdot (n(u_1) - n(u_2)) + p(u_1) - p(u_2) = 0[/tex]
Now, the book I am reading claims that, "by Rolle's theorem, we get:"
[tex]x_1 n_1'(v_1) + x_2 n_2'(v_2) + x_3 n_3'(v_3) + p'(v_4) = 0[/tex] with [tex]u_1 \leq v_i \leq u_2[/tex].
However, I don't see how the theorem applies here... for starters, I don't see anything of the form [tex]f(a) = f(b)[/tex], as required by the theorem.