Calculus with Vector Functions — Question 6

Section 13-7 · Calculus pages 2511–2512

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Question 6

For r→(t)=⟨t,t2,t3−3t⟩\vec r(t)=\left\langle t,t^2,t^3-3t\right\rangle, find all points where the tangent line is parallel to the plane 2x−y+z=42x-y+z=4. Determine whether any tangent line is perpendicular to that plane.

Question 6 – Solution

Strategy A tangent is parallel to a plane when its direction is orthogonal to the plane normal; it is perpendicular when its direction is parallel to the normal.

Diagram shown in the original SVG worksheet.

Parallel condition r→′(t)=⟨1,2t,3t2−3⟩\vec r'(t)=\left\langle 1,2t,3t^2-3\right\rangle and n→=⟨2,−1,1⟩\vec n=\left\langle 2,-1,1\right\rangle. Thus r→′(t)⋅n→=3t2−2t−1=(3t+1)(t−1).\vec r'(t)\cdot\vec n=3t^2-2t-1=(3t+1)(t-1). So t=1t=1 or t=−1/3t=-1/3, giving points (1,1,−2)\boxed{(1,1,-2)} and (−1/3,1/9,26/27)\boxed{(-1/3,1/9,26/27)}.

Perpendicular condition Parallelism of r→′(t)\vec r'(t) with n→\vec n would force scale 1/21/2 from the first component, then 2t=−1/22t=-1/2, but the third component at t=−1/4t=-1/4 is not 1/21/2. Hence none exist.