Arc Length with Vector Functions — Question 3

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Question 3

For r→(t)=⟨t2,23t3,0⟩\vec r(t)=\left\langle t^2,\frac 23t^3,0\right\rangle, find the arc length from the point corresponding to t=1t=1 to the point corresponding to t=2t=2. Also find the chord length and compare the two quantities.

Original worksheet page 1: question and worked solution for 6-9-003
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Question 3 – Solution

Strategy Use the same speed structure as in Question 2, but retain the nonzero lower limit.

See the diagram in the original worksheet below.

Arc length L=∫122t1+t2dt=23(55−22).L=\int_1^2 2t\sqrt{1+t^2}\,dt =\boxed{\frac 23(5\sqrt 5-2\sqrt 2)}.

Chord The endpoints are P=(1,2/3,0)P=(1,2/3,0) and Q=(4,16/3,0)Q=(4,16/3,0), so ∥Q−P∥=32+(143)2=2773.\|Q-P\|=\sqrt{3^2+\left(\frac{14}{3}\right)^2} =\boxed{\frac{\sqrt{277}}3}.

Comparison Numerically, L≈5.568L\approx 5.568 while the chord is about 5.5485.548. The curve length is slightly larger, as required; equality would occur only for a straight segment traversed without reversal.

Original worksheet page 2: question and worked solution for 6-9-003

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