Arc Length with Vector Functions — Question 9

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Question 9

The catenary r→(t)=⟨t,cosht⟩\vec r(t)=\left\langle t,\cosh t\right\rangle is restricted to −a≤t≤a-a\le t\le a, where a>0a>0. Find its exact length in terms of aa. Then determine the value of aa that makes the length equal to 66.

Original worksheet page 1: question and worked solution for 6-9-009
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Question 9 – Solution

Strategy Use the identity 1+sinh⁡2t=cosh⁡2t1+\sinh^2t=\cosh^2t and the positivity of cosh⁡t\cosh t.

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Length ∥r→′(t)∥=1+sinh⁡2t=cosh⁡t,L=∫−aacosh⁡tdt=2sinh⁡a.\|\vec r'(t)\|=\sqrt{1+\sinh^2t}=\cosh t, \qquad L=\int_{-a}^{a}\cosh t\,dt=2\sinh a. Therefore L=2sinh⁡a\boxed{L=2\sinh a}.

Prescribed length Setting 2sinh⁡a=62\sinh a=6 gives sinh⁡a=3\sinh a=3, hence a=arsinh⁡(3)=ln⁡(3+10).\boxed{a=\operatorname{arsinh}(3)=\ln(3+\sqrt{10})}.

Symmetry The even speed makes the two halves have equal length; equivalently, L=2∫0acosh⁡tdtL=2\int_0^a\cosh t\,dt.

Original worksheet page 2: question and worked solution for 6-9-009

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