**Theorem:** The necessary and sufficient condition for a curve to be a helix is that its curvature and torsion are in a constant ratio, that is, τ/k = ± cotα, where α is a constant angle.

**Proof:** Let the curve x = x(s) be a helix. We have to show that τ/k = ± cotα, where α is a constant angle.

Let t be a unit tangent to the helix and a be the unit vector in the fixed direction. Then by definition of helix

t . a = |t| |a| cos α = cos α ——————————-(i)

where α is a constant angle between t and a.

#### Differentiating (i) with respect to s, we get

t′ . a + a . 0 = 0

⟹ Ӄn . a = 0 [∵t′ = Ӄn]

⟹ n . a = 0 [Ӄ ≠ 0]

##### That is, a is perpendicular to n and hence a lies in the plane of t and b.

Let a = At + Bb ——————————-(ii)

Where A, B are scalars.

Now t .a = A = cosα [by (i)]

Thus, from (ii) we have

A = t cos α + Bb

Now, (t cos α + Bb) . (t cos α + Bb) = 1 [∵ a . a =1]

⟹ cos^{2}α + B^{2} = 1

⟹ B^{2} = sin^{2}α

⟹ B = ±sinα

### Putting the values of A and B in (ii), we get

a = t cos α ± b sin α

Again differentiating this with respect to s, we get

0 = t′ cos α ± b′ sin α

⟹ 0 = Ӄn cos α + τ n sin α [∵t′ = Ӄn, b′ = τn]

⟹ 0 = (Ӄ cos α + τ sin α )n

⟹ (Ӄ cos α + τ sin α ) = 0

⟹ (Ӄ cos α = ±τ sin α )

⟹ τ / Ӄ = ± cot α

# Sufficient condition: Let us suppose that τ / Ӄ = ± cot α, where α is constant. We have to show that the curve is a helix.

Now τ / Ӄ = ± cot α

⟹τ / Ӄ = ± cos α / sin α

⟹ ± τ sin α – Ӄ cos α = 0

⟹ (∓ τ sin α + Ӄ cos α)n = 0

⟹ Ӄn cos α ± τn sin α = 0

⟹ t′ cos α ± b′ sin α = 0 [∵t′ = Ӄn, b′ = τn]

Integrating both side we get

t cos α ± b sin α = a, where a is a constant vector.

⟹ t (t cos α ± b sin α) = t . a

⟹ t . a = cos α [∵ a . a =1]

*This shows that the tangent t to the curve makes a constant angle α with the fixed direction a. Thus the curve is a helix.*