Explanation Given: r ( s ) = ( ln s ) i + s − 1 j + 9 s k s = 1 Formula used: a ( s ) = r ( s ) + s r ' ( s ) Calculation: The tangent line has the parametrization at given point: a ( s ) = r ( s ) + s r ' ( s ) ⇒ a ( s ) = r ( 2 ) + s r ' ( 2 ) r ( s ) = ( ln s ) i + s − 1 j + 9 s k Given, s = 1 ⇒ r ( 1 ) = ( ln 1 ) i + 1 − 1 j + ( 9 × 1 ) k = ( 0 ) i + 1 1 j + ( 9 × 1 ) k [ ∵ ln 1 = 0 ] = j + 9 k ⇒ r ' ( s ) = 1 s i − s − 2 j +

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Author: Jon Rogawski, Colin Adams, Robert Franzosa

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Chapter 13.2, Problem 36E

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Find a parametrization of the tangent line, r(s)=(lns)i+s−1j+9sk at the point s=1.

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Chapter 13.2, Problem 35E

Chapter 13.2, Problem 37E

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Chapter 13 Solutions

Calculus: Early Transcendentals

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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Find a parametrization of the tangent line, r ( s ) = ( ln s ) i + s − 1 j + 9 s k at the point s = 1.

Solution Summary: The author explains that the tangent line has the parametrization at the point s=1.

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Find a parametrization of the tangent line, r(s)=(lns)i+s−1j+9sk at the point s=1.

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