The curvature of the helix r(t)=(acost)i+(asint)j+(bt)k (a, b is greater than or equal to 0) is k=a/(a^2+b^2). What is the largest value k can have for a given value of b?

Accepted Solution

Answer:   1/(2b)Step-by-step explanation:The value of k will be a maximum where its derivative with respect to 'a' is zero. The derivative of k with respect to 'a' is ...   [tex]k'=\dfrac{(a^2+b^2)(1)-a(2a)}{(a^2+b^2)^2}=\dfrac{b^2-a^2}{(a^2+b^2)^2}[/tex]We want to find 'a' when k' = 0. k' will be zero when the numerator of the rational function is zero:   b^2 -a^2 = 0   (b -a)(b +a) = 0   a = ±bThen the value of k is ...   k = ±b/(b^2 +b^2) = ±1/(2b)The largest value k can have is 1/(2b).