Explanation Given data: The expression of the curve is, r ( t ) = ( 3 cosh 2 t ) i + ( 3 sinh 2 t ) j + 6 t k , t = ln 2 (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 the general expression to calculate the binormal vector B . B = T × N (5) Write a general expression to calculate the curvature. κ = 1 | v | | d T d t | (6) Calculation: Substitute the equation (1) in equation (2) to find v . v = d d t ( ( 3 cosh 2 t ) i + ( 3 sinh 2 t ) j + 6 t k ) = ( 3 ( sinh 2 t ) ( 2 ) ) i + ( 3 ( cosh 2 t ) ( 2 ) ) j + 6 ( 1 ) k = 6 sinh 2 t i + 6 cosh 2 t j + 6 k Take the magnitude of above equation to find | v | . | v | = | 6 sinh 2 t i + 6 cosh 2 t j + 6 k | = ( 6 sinh 2 t ) 2 + ( 6 cosh 2 t ) 2 + 6 2 = 36 sinh 2 2 t + 36 cosh 2 2 t + 36 = 6 2 cosh 2 t Substitute 6 sinh 2 t i + 6 cosh 2 t j + 6 k for v , and 6 2 cosh 2 t for | v | in equation (3) to find T . T = 6 sinh 2 t i + 6 cosh 2 t j + 6 k 6 2 cosh 2 t = 6 sinh 2 t 6 2 cosh 2 t i + 6 cosh 2 t 6 2 cosh 2 t j + 6 6 2 cosh 2 t k = 1 2 sinh 2 t cosh 2 t i + 1 2 cosh 2 t cosh 2 t j + 1 2 1 cosh 2 t k T = 1 2 tanh 2 t i + 1 2 j + 1 2 sech 2 t k (7) Substitute ln 2 for t in equation (7) to find T ( ln 2 ) . T ( ln 2 ) = 1 2 tanh 2 ( ln 2 ) i + 1 2 j + 1 2 sech 2 ( ln 2 ) k = 5 17 2 i + 1 2 j + 8 17 2 k Differentiate the equation (7) with respect to t . d T d t = d d t ( 1 2 tanh 2 t i + 1 2 j + 1 2 sech 2 t k ) = 1 2 ( sech 2 2 t ) ( 2 ) i + ( 0 ) j + 1 2 ( − sech 2 t tanh 2 t ) ( 2 ) k = ( 2 2 sech 2 2 t ) i − ( 2 2 sech 2 t tanh 2 t ) k Substitute ln 2 for t in above equation to find d T d t ( ln 2 ) . d T d t ( ln 2 ) = ( 2 2 sech 2 2 ( ln 2 ) ) i − ( 2 2 sech 2 ( ln 2 ) tanh 2 ( ln 2 ) ) k = ( ( 2 2 ) ( 8 17 ) 2 ) i − ( ( 2 2 ) ( 8 17 ) ( 15 17 ) ) k = ( 128 289 2 ) i − ( 240 289 2 ) k Take magnitude for above equation to find | d T d t ( ln 2 ) |

Pearson eText for Thomas' Calculus: Early Transcendentals -- Instant Access (Pearson+)

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ISBN: 9780137399185

Author: Joel Hass, Christopher Heil

Publisher: PEARSON+

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Chapter 13, Problem 20PE

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Find the value of unit tangent vector (T), the principal unit normal vector (N), the binormal vector (B) and the curvature (κ) at the given value of t.

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Chapter 13, Problem 19PE

Chapter 13, Problem 21PE

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Pearson eText for Thomas' Calculus: Early Transcendentals -- Instant Access (Pearson+)

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Find the value of unit tangent vector ( T ) , the principal unit normal vector ( N ) , the binormal vector ( B ) and the curvature ( κ ) at the given value of t .

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Find the value of unit tangent vector (T), the principal unit normal vector (N), the binormal vector (B) and the curvature (κ) at the given value of t.

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Chapter 13 Solutions