Explanation Given info. r → ( t ) = ( 1 + cos t , sin t , 2 sin ( t 2 ) ) Concept used: k ( t ) = ‖ r ' ( t ) × r " ( t ) ‖ ‖ r ' ( t ) ‖ 3 Calculation: r → ( t ) = ( 1 + cos t , sin t , 2 sin ( t 2 ) ) r → ' ( t ) = ( − sin t , cos t , cos t 2 ) ( d d t ( 1 + cos t ) = − sin t d d t ( sin t ) = cos t d d t ( 2 sin ( t 2 ) ) = cos ( t 2 ) ) r → " ( t ) = ( − cos t , − sin t , − 1 2 sin t 2 ) ( d d t ( − sin t ) = − cos t , d d t ( cos t 2 ) = − 1 2 sin t 2 d d t ( cos t ) = − sin t ) r → ' ( t ) × r → " ( t ) = ( − sin t , cos t , cos t 2 ) × ( − cos t , − sin t , − 1 2 sin t 2 ) ( using determinant to solve vector product ) = | i ∧ j ∧ k ∧ − sin t cost cos t 2 − cos t − sin t − 1 2 sin t 2 | Expanding with respect to first row. = i ∧ | cos t cos t 2 − sin t − 1 2 sin t 2 | − j ∧ | − sin t cos t 2 − cos t − 1 2 sin t 2 | + k ∧ | − sin t cost − cos t − sin t | Expanding second order determinant = i ∧ ( ( − 1 2 sin t 2 . cos t ) − ( − sin t . cos t 2 ) ) − j ∧ ( 1 2 sin t 2 . sin t + cos t . cos t 2 ) + k ∧ ( sin 2 t + cos 2 t ) = i ∧ ( − 1 2 sin t 2 . cos t + sin t . cos t 2 ) − j ∧ ( 1 2 sin t 2 . sin t + cos t . cos t 2 ) + k ∧ ( sin 2 t + cos 2 t = 1 ) ‖ r ' ( t ) × r " ( t ) ‖ = ( − 1 2 sin t 2 . cos t + sin t . cos t 2 ) 2 + ( 1 2 sin t 2 . sin t + cos t . cos t 2 ) 2 + 1 2 Expanding the above expression ( a + b ) 2 = a 2 + b 2 + 2 a b = ( 1 4 sin 2 t 2 ...

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Chapter 14.4, Problem 83E

To determine

Show that the curvature of Viviane’s curve, given by $\vec{r} (t) = (1 + \cos t, \sin t, 2 \sin (\frac{t}{2}))$ is.

$k (t) = \frac{\sqrt{13 + 3 \cos t}}{{(3 + \cos t)}^{\frac{3}{2}}}$

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Chapter 14.4, Problem 82E

Chapter 14.4, Problem 84E

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Consider the following vector function. r(t) = (4t, 1/2t^2,t^2) (a) Find the unit tangent and unit normal vectors T(t) and N(t). (b) Use this formula to find the curvature. formula k(t)= |T'(t)|/|r'(t)| this is the only information given Thank you

Consider the plane curve parametrzed by F(1) -ti + (In(com(0))3, /a

Chapter 14 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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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Show that the curvature of Viviane’s curve, given by r → ( t ) = ( 1 + cos t , sin t , 2 sin ( t 2 ) ) is. k ( t ) = 13 + 3 cos t ( 3 + cos t ) 3 2

Solution Summary: The author explains that the curvature of Viviane's curve, given by r ( t ) = ( 1 + cos, sin, and 2 sin) is.

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Show that the curvature of Viviane’s curve, given by $\vec{r} (t) = (1 + \cos t, \sin t, 2 \sin (\frac{t}{2}))$ is.

$k (t) = \frac{\sqrt{13 + 3 \cos t}}{{(3 + \cos t)}^{\frac{3}{2}}}$

Want to see the full answer?

Chapter 14 Solutions

Show that the curvature of Viviane’s curve, given by r → ( t ) = ( 1 + cos t , sin t , 2 sin ( t 2 ) ) is. k ( t ) = 13 + 3 cos t ( 3 + cos t ) 3 2

Solution Summary: The author explains that the curvature of Viviane's curve, given by r ( t ) = ( 1 + cos, sin, and 2 sin) is.

Videos

Show that the curvature of Viviane’s curve, given by r→(t)=(1+cost,sint,2sin(t2)) is. k(t)=13+3cost(3+cost)32

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Chapter 14 Solutions

Show that the curvature of Viviane’s curve, given by $\vec{r} (t) = (1 + \cos t, \sin t, 2 \sin (\frac{t}{2}))$ is.

$k (t) = \frac{\sqrt{13 + 3 \cos t}}{{(3 + \cos t)}^{\frac{3}{2}}}$