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.
1 answer:
The largest value τ can have for a given value of a = 1/2. Where the said value is described as τmax.
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 = /2a²
= 1/ 2
Thus, τmax = 1/2
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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?
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Here,
mean=0.4
variance=(Xn-mean)².P(x)n
=(0-0.4)²*(0.64)+(1-0.4)²*(0.32)+(2-0.4)²*0.04
variance=0.32
Standard deviation=√(variance)
=√0.32
=0.566
The standard deviation will be 0.566 where variance in 0.32 and mean is 0.40 and given sets of probability.
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