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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-6

Vector Functions — Question 4

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Question 4

Show that r→1(t)=⟨cost,sint,t⟩\vec r_1(t)=\left\langle\cos t,\sin t,t\right\rangle and r→2(u)=⟨cos(2u),sin(2u),2u⟩\vec r_2(u)=\left\langle\cos(2u),\sin(2u),2u\right\rangle have the same image. Compare their orientations and speeds of traversal at matching points. Then give a reparameterization with the opposite orientation.

Original worksheet page 1: question and worked solution for 6-6-004
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Question 4 – Solution

Strategy Relate parameters by matching all three components.

See the diagram in the original worksheet below.

Same image Setting t=2ut=2u makes r→1(t)=r→2(u)\vec r_1(t)=\vec r_2(u), so both trace the same helix. As each parameter increases, zz increases, so orientations agree.

Traversal rate A change Δu\Delta u produces twice the angular and vertical change of the same Δt\Delta t; equivalently, r→2(u)=r→1(2u)\vec r_2(u)=\vec r_1(2u) traverses the image twice as fast with respect to its parameter.

Reverse orientation For example, q→(s)=⟨coss,−sins,−s⟩=r→1(−s)\boxed{\vec q(s)=\left\langle\cos s,-\sin s,-s\right\rangle=\vec r_1(-s)}.

Original worksheet page 2: question and worked solution for 6-6-004

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

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