Explanation Given: The parametrized surface and the point is given as G ( r , θ ) = ( r cos θ , r sin θ , 1 − r 2 ) ; r = 1 2 , θ = π 4 Calculation: The tangent vectors are the following vectors T r = ∂ ϕ ∂ r = ∂ ∂ r ( r cos θ , r sin θ , 1 − r 2 ) T r = ( cos θ , sin θ , − 2 r ) T θ = ∂ ϕ ∂ θ = ∂ ∂ θ ( r cos θ , r sin θ , 1 − r 2 ) T θ = ( − r sin θ , r cos θ , 0 ) The normal vector n ( θ , ϕ ) is the following cross product: n ( r , θ ) = T r × T θ = | i j k cos θ sin θ − 2 r − r sin θ r cos θ 0 | = ( 2 r 2 cos θ ) i + ( 2 r 2 sin θ ) j + ( r cos 2 θ + r sin 2 θ ) k n ( r , θ ) = r 〈 2 r cos θ , 2 r sin θ , 1 〉 The tangency point and the normal at this point are, P : ϕ ( 1 2 , π 4 ) = ( 1 2 cos π 4 , 1 2 sin π 4

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, Problem 10E

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$T_{r}$ , $T_{θ}$ , $n (r, θ)$ and tangent plane equation at the point given.

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Chapter 17.4, Problem 9E

Chapter 17.4, Problem 11E

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Chapter 17 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

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T r , T θ , n ( r , θ ) and tangent plane equation at the point given.

Solution Summary: The author explains the parametrized surface and the tangent plane equation at the point given.

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$T_{r}$ , $T_{θ}$ , $n (r, θ)$ and tangent plane equation at the point given.

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Chapter 17 Solutions

T r , T θ , n ( r , θ ) and tangent plane equation at the point given.

Solution Summary: The author explains the parametrized surface and the tangent plane equation at the point given.

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Tr,Tθ,n(r,θ) and tangent plane equation at the point given.

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Chapter 17 Solutions

$T_{r}$ , $T_{θ}$ , $n (r, θ)$ and tangent plane equation at the point given.