1 (8 points) find the equation of the tangent line to the path $c(t) = (\cos t, \sin t, t^2)$ at $t = \pi$.

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [], "new_concepts": [ "Vector-Valued Functions", "Tangent Line Equation" ], "current_concepts": [ "Vector-Valued Functions", "Tangent Line Equation", "Space Curves" ] } ## Evaluate the path and its derivative at the given parameter $$ \begin{aligned} c(t) &= (\cos t, \sin t, t^2) \ c'(t) &= (-\sin t, \cos t, 2t) \ c(\pi) &= (\cos \pi, \sin \pi, \pi^2) = (-1, 0, \pi^2) \ c'(\pi) &= (-\sin \pi, \cos \pi, 2\pi) = (0, -1, 2\pi) \end{aligned} $$ ## Formulate the vector equation of the tangent line $$ \begin{aligned} L(t) &= c(\pi) + t \cdot c'(\pi) \ L(t) &= (-1, 0, \pi^2) + t \cdot (0, -1, 2\pi) \ L(t) &= (-1, -t, \pi^2 + 2\pi t) \end{aligned} $$ ## Formulate the alternative parameterization $$ \begin{aligned} L(t) &= c(\pi) + (t - \pi) \cdot c'(\pi) \ L(t) &= (-1, 0, \pi^2) + (t - \pi) \cdot (0, -1, 2\pi) \ L(t) &= (-1, \pi - t, \pi^2 + 2\pi(t - \pi)) \ L(t) &= (-1, \pi - t, 2\pi t - \pi^2) \end{aligned} $$ $$L(t) = (-1, -t, \pi^2 + 2\pi t)$$ (Or alternatively: $L(t) = (-1, \pi - t, 2\pi t - \pi^2)$) { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Calculus", "Vector-Valued Functions" ] }

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Evaluate the path and its derivative at the given parameter

Formulate the vector equation of the tangent line

Formulate the alternative parameterization

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