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In Exercises 27–30, find the value(s) of t so that the tangent line to the given curve contains the given point. 27. r(t) = t²i + (1 + t)j + (2t – 3)k; (-8, 2, – 1) k; (0,3,–8/3) 29. r(t) = 2ti + ²j – t²k; (0, –4, 4) 30. r(t) = -ti + t°j + (ln t)k; (2,-5,–3) 28. r(t) = ti + 3j + (2, –5, – 3) %3D

In Exercises 27–30, find the value(s) of t so that the tangent line to the given curve contains the given point. 27. r(t) = t²i + (1 + t)j + (2t – 3)k; (-8, 2, – 1) k; (0,3,–8/3) 29. r(t) = 2ti + ²j – t²k; (0, –4, 4) 30. r(t) = -ti + t°j + (ln t)k; (2,-5,–3) 28. r(t) = ti + 3j + (2, –5, – 3) %3D

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In Exercises 27–30, find the value(s) of t so that the tangent line to the given curve contains the given point. 27. r(t) = t²i + (1 + t)j + (2t – 3)k; (-8, 2, – 1) k; (0,3,–8/3) 29. r(t) = 2ti + ²j – t²k; (0, –4, 4) 30. r(t) = -ti + t°j + (ln t)k; (2,-5,–3) 28. r(t) = ti + 3j + (2, –5, – 3) %3D
Transcribed Image Text:In Exercises 27–30, find the value(s) of t so that the tangent line to the given curve contains the given point. 27. r(t) = t²i + (1 + t)j + (2t – 3)k; (-8, 2, – 1) k; (0,3,–8/3) 29. r(t) = 2ti + ²j – t²k; (0, –4, 4) 30. r(t) = -ti + t°j + (ln t)k; (2,-5,–3) 28. r(t) = ti + 3j + (2, –5, – 3) %3D
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