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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-6

Vector Functions — Question 10

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Question 10

Find a parametrization of the curve where the sphere x2+y2+z2=25x^2+y^2+z^2=25 meets the plane z=3z=3. Give both orientations and explain why the parametrization covers the entire intersection exactly once on your chosen interval.

Original worksheet page 1: question and worked solution for 6-6-010
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Question 10 – Solution

Strategy A horizontal plane cuts the sphere in a circle; determine its radius.

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Circle Substituting z=3z=3 gives x2+y2=16x^2+y^2=16, a circle of radius 4 centered at (0,0,3)(0,0,3). A counterclockwise parametrization viewed from positive zz is r→(t)=⟨4cost,4sint,3⟩,0≤t<2π.\boxed{\vec r(t)=\left\langle 4\cos t,4\sin t,3\right\rangle},\quad 0\le t<2\pi. The opposite orientation is q→(t)=⟨4cost,−4sint,3⟩\boxed{\vec q(t)=\left\langle 4\cos t,-4\sin t,3\right\rangle}.

Coverage Sine and cosine generate every radius-4 circle point; the half-open interval prevents duplication of the initial point at 2π2\pi.

Original worksheet page 2: question and worked solution for 6-6-010

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