To determine Find the value of unit tangent vector ( T ) , the principal unit normal vector ( N ) and the curvature ( κ ) . Explanation Given data: The expression of plane curve is, r ( t ) = ( cosh t ) i − ( sinh t ) j + t k (1) Formula used: Write a general expression to calculate constant vector v . v = d r d t (2) Here, d r d t is the value of change in plane curve. Write a general expression to calculate the unit tangent vector. T = v | v | (3) Write a general expression to calculate the principal unit normal vector. N = ( d T d t ) | d T d t | (4) Here, d T d t is the value of change in unit tangent vector. Write a general expression to calculate the curvature. κ = 1 | v | | d T d t | (5) Calculation: Differentiate equation (1) with respect to t . d r ( t ) d t = ( sinh t ) i − ( cosh t ) j + ( 1 ) k = ( sinh t ) i − ( cosh t ) j + k Substitute ( sinh t ) i − ( cosh t ) j + k for d r ( t ) d t in equation (2) to find v . v = ( sinh t ) i − ( cosh t ) j + k Take the magnitude of above equation to find | v | . | v | = | ( sinh t ) i − ( cosh t ) j + k | = ( sinh t ) 2 + ( − cosh t ) 2 + 1 2 = sinh 2 t + cosh 2 t + 1 = cosh 2 t + cosh 2 t { ∵ cosh 2 θ = 1 + sinh 2 θ } Simplify the above equation to find | v | . | v | = 2 cosh 2 t = 2 cosh t Substitute ( sinh t ) i − ( cosh t ) j + k for v , and 2 cosh t for | v | in equation (3) to find T . T = ( sinh t ) i − ( cosh t ) j + k 2 cosh t = 1 2 sinh t cosh t i − 1 2 cosh t cosh t j + 1 2 cosh t k = ( 1 2 tanh t ) i − 1 2 j + ( 1 2 sech t ) k { ∵ sech θ = 1 cosh θ tanh θ = sinh θ cosh θ } Differentiate above equation with respect to t

Thomas' Calculus: Early Transcendentals in SI Units

14th Edition

ISBN: 9781292253114

Author: Hass, Joel R., Heil, Christopher E., WEIR, Maurice D.

Publisher: PEARSON

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Chapter 13.4, Problem 16E

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Find the value of unit tangent vector (T), the principal unit normal vector (N) and the curvature (κ).

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Chapter 13.4, Problem 15E

Chapter 13.4, Problem 17E

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Thomas' Calculus: Early Transcendentals in SI Units

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