Length Of A Helix

Calculus Level 4

Consider the cylindrical helix in R 3 \mathbb{R}^3 whose equations are

x = 3 cos ⁡ ( t ) y = 3 sin ⁡ ( t ) z = 4 t x = 3\cos(t) \hspace{.4cm} y = 3\sin(t) \hspace{.4cm} z = 4t

What is the length of this helix from t = 0 t = 0 to t = 403 t = 403 ?


The answer is 2015.

Guillermo Templado by Guillermo Templado , 46, Spain

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3 solutions

Otto Bretscher
Nov 2, 2015

A solution without calculus: If you roll out the cylinder, the helix turns into a straight line that is the hypotenuse of a right triangle with horizontal side 3 ∗ 403 3*403 and vertical side 4 ∗ 403 4*403 . Thus the length of the helix is 5 ∗ 403 = 2015 5*403=\boxed{2015}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This "rolling out of the cylinder" is the standard trick for the competition problem about the length of a vine which grows by winding around a stick. It presents a simple geometric interpretation of this "3-d" length.

Even I have to admit that sometimes the solutions without calculus are the easiest and prettiest.

Andrew Ellinor - 5 years, 7 months ago

f(t) = (x(t),y(t),z(t)) = (3 cos(t), 3 sin(t), 4t); f ' (t) = (-3 sin(t), 3 cos(t), 4)

Lenght (helix) = ∫ 0 403 ∣ ∣ f ′ ( t ) ∣ ∣ d t \int_0^{403} || f ' (t)|| dt = ∫ 0 403 9 s i n 2 ( t ) + 9 c o s 2 ( t ) + 16 d t \int_0^{403} \sqrt {9 sin^2(t) + 9 cos^2(t) + 16} dt = ∫ 0 403 25 d t \int_0^{403} \sqrt{25} dt = ∫ 0 403 5 d t \int_0^{403} 5 dt = 5 ⋅ 403 5 \cdot 403 = 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Nov 3, 2015

0 Helpful 0 Interesting 0 Brilliant 0 Confused

It is quite risky not to think in Calculus unless we have made a good mind set after a careful thought. If relation between x, y and z become more complicated, then a lack of caution could easily make a mistake. An opened-up triangle is a special case. Just be more careful, my advice to everyone.

Lu Chee Ket - 5 years, 7 months ago

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