Recall the Mean Value Theorem: If f : [a, b] → R is continuous on [a, b], and differentiable on (a, b), then there exists c ∈ (a, b) such that f(b) − f(a) = f 0 (c)(b − a). Show that this is generally not true for vector-valued functions by showing that for r(t) = costi + sin tj + tk, there is no c ∈ (0, 2π) such that r(2π) − r(0) = 2πr 0 (c).
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