Explanation Given: r → ( t ) = ( Cosh t , Sinh t , t 2 ) A point given t = 0 Formula used: r → ' ( t ) = d r ( t ) d t u ( t ) = ‖ r → ' ( t ) ‖ r " ( t ) = d 2 ( r ( t ) ) d t 2 u ' ( t ) = ( d u ( t ) ) / d t Calculation: r → ( t ) = ( Cosh t , Sinh t , t 2 ) r → ' ( t ) = d r → ( t ) d t = d d t ( Cosh t , Sinh t , t 2 ) r → ' ( t ) = ( Sinh t , cosh t , 2 t ) r → ' ( 0 ) = ( Sinh 0 , cosh 0 , 2.0 ) r → ' ( 0 ) = ( 0 , 1 , 0 ) − − − − ( i ) u ( t ) = ‖ r → ( t ) ‖ = ‖ ( Sinh t , cosh t , 2 t ) ‖ = ( sinh t ) 2 + ( cosh t ) 2 + ( 2 t ) 2 u ( t ) = Sinh 2 t + cosh 2 t + 4 t 2 u ( 0 ) = Sinh 2 0 + cosh 2 0 + 4 ( 0 ) 2 = 0 + ( 1 ) 2 + 0 u ( 0 ) = 1 − − − − − − ( 2 ) u ' ( t ) = d d t ( ( sinh t ) 2 + ( cosh t ) 2 + ( 2 t ) 2 ) = 1 2 ( sinh t ) 2 + ( cosh t ) 2 + ( 2 t ) 2 × ( 2 sinh t ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Vector Calculus

Vector calculus is an important branch of mathematics and it relates two important branches of mathematics namely vector and calculus.

Question

Chapter 14.4, Problem 52E

To determine

To calculate:

Calculate the normal unit vector $(\overset{\land}{N})$ at the point $t = 0$ of the curve

$(Cosh t, Sinh t, t^{2})$ .

$\vec{N} (t) = \frac{u (t) \vec{r} " (t) - u' (t) \vec{r}' (t)}{‖ u (t) \vec{r} " (t) - u (t) \vec{r}' (t) ‖}$

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Chapter 14.4, Problem 51E

Chapter 14.4, Problem 53E

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Chapter 14 Solutions

CALCULUS (CLOTH)

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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Calculate the normal unit vector ( N ∧ ) at the point t = 0 of the curve ( Cosh t , Sinh t , t 2 ) . N → ( t ) = u ( t ) r → " ( t ) − u ' ( t ) r → ' ( t ) ‖ u ( t ) r → " ( t ) − u ( t ) r → ' ( t ) ‖

Solution Summary: The author calculates the normal unit vector (N ) at the point t = 0 of the curve.

Concept explainers

To calculate:

Calculate the normal unit vector $(\overset{\land}{N})$ at the point $t = 0$ of the curve

$(Cosh t, Sinh t, t^{2})$ .

$\vec{N} (t) = \frac{u (t) \vec{r} " (t) - u' (t) \vec{r}' (t)}{‖ u (t) \vec{r} " (t) - u (t) \vec{r}' (t) ‖}$

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Chapter 14 Solutions

Calculate the normal unit vector ( N ∧ ) at the point t = 0 of the curve ( Cosh t , Sinh t , t 2 ) . N → ( t ) = u ( t ) r → " ( t ) − u ' ( t ) r → ' ( t ) ‖ u ( t ) r → " ( t ) − u ( t ) r → ' ( t ) ‖

Solution Summary: The author calculates the normal unit vector (N ) at the point t = 0 of the curve.

Concept explainers

To calculate: Calculate the normal unit vector (N∧) at the point t=0 of the curve (Cosht,Sinht,t2). N→(t)=u(t)r→"(t)−u'(t)r→'(t)‖u(t)r→"(t)−u(t)r→'(t)‖

Want to see the full answer?

Chapter 14 Solutions

To calculate:

Calculate the normal unit vector $(\overset{\land}{N})$ at the point $t = 0$ of the curve

$(Cosh t, Sinh t, t^{2})$ .

$\vec{N} (t) = \frac{u (t) \vec{r} " (t) - u' (t) \vec{r}' (t)}{‖ u (t) \vec{r} " (t) - u (t) \vec{r}' (t) ‖}$