Explanation Given: The parametrized surface and the point is given as G ( u , v ) = ( 2 u + v , u − 4 v , 3 u ) ; u = 1 , v = 4 Calculation: The tangent vectors are the following vectors T u = ∂ ϕ ∂ u = ∂ ∂ u ( 2 u + v , u − 4 v , 3 u ) T u = ( 2 , 1 , 3 ) T v = ∂ ϕ ∂ v = ∂ ∂ v ( 2 u + v , u − 4 v , 3 u ) T v = ( 1 , − 4 , 0 ) The normal vector n ( u , v ) is the following cross product: n ( u , v ) = | i j k 2 1 3 0 − 4 0 | = | 1 3 − 4 0 | i − | 2 3 1 0 | j +

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, Problem 7E

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

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Chapter 17.4, Problem 6E

Chapter 17.4, Problem 8E

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Chapter 17 Solutions

CALCULUS (CLOTH)

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T u , T v , n ( u , v ) and tangent plane equation at the point given.

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

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

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Chapter 17 Solutions

T u , T v , n ( u , v ) and tangent plane equation at the point given.

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

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Tu,Tv,n(u,v) and tangent plane equation at the point given.

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Chapter 17 Solutions

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