Question

show that the torsion of the helix is what is the largest value can have for a given value of a? give reasons for your answer.

Answer 1

The largest value τ can have for a given value of a = 1/2$\left| a \right|$. Where the said value is described as τmax.

What is the reason for the above answer?

To arrive at the above conclusion, let:

r(t) = (a cos t) i + (a sin t)j + btk, a, b ≥ 0

⇒ τ = b/(a² + b²)

Given the value of τ above, we have to find τmax for given values of a, so that a is constant.

Hence:

dτ/db = [(a² + b²) * 1 - b (2b)]/ (a² + b²)² = 0

a² + b² - 2b² = 0

a² + b² = 0

a² - b² = 0

a² = b²

Hence, τmax = b/a² + b² = $\left| a \right|$/2a²

= 1/ 2$\left| a \right|$

Thus, τmax = 1/2$\left| a \right|$

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Full Question:

Show That The Torsion Of The Helix

r(T) = (A Cos t)Ii + (A Sin T) J + Btk, A, B ≥ 0

Is τ = B/(A2 + B2).

What Is The Largest Value τ can have for a given value of a?