Explanation Given data: The vector function is given as follows. r ( u , v ) = u 2 i + 2 u sin v j + u cos v k ; u = 1 , v = 0 Formula used: Write the expression to find tangent plane to the parametric surface with the normal vector n = 〈 a , b , c 〉 at the specified point ( x 0 , y 0 , z 0 ) . a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 (1) Write the expression to find normal vector from the tangent vectors of the parametric surface. n = | i j k a 1 b 1 c 1 a 2 b 2 c 2 | (2) Here, The vector 〈 a 1 , b 1 , c 1 〉 is a tangent vector r u of the parametric surface and The vector 〈 a 2 , b 2 , c 2 〉 is a tangent vector r v of the parametric surface. Write the expression to find the tangent vector r u of the parametric surface. r u = ∂ x ∂ u i + ∂ y ∂ u j + ∂ z ∂ u k (3) Write the expression to find the tangent vector r v of the parametric surface. r v = ∂ x ∂ v i + ∂ y ∂ v j + ∂ z ∂ v k (4) Write the vector function as follows. r ( u , v ) = u 2 i + 2 u sin v j + u cos v k ; u = 1 , v = 0 Write the parametric equations from the vector function as follows. x = u 2 , y = 2 u sin v , z = u cos v Calculation point in the surface ( x 0 , y 0 , z 0 ) : Write the expression to find point ( x 0 , y 0 , z 0 ) in the surface. ( x 0 , y 0 , z 0 ) = ( u 2 , 2 u sin v , u cos v ) Substitute 1 for u and 0 for v , ( x 0 , y 0 , z 0 ) = ( 1 2 , 2 ( 1 ) sin ( 0 ) , ( 1 ) cos ( 0 ) ) = ( 1 , 0 , 1 ) Calculation of tangent vector r u : Substitute u 2 for x , 2 u sin v for y , and u cos v for z in equation (3), r u = ∂ ( u 2 ) ∂ u i + ∂ ( 2 u sin v ) ∂ u j + ∂ ( u cos v ) ∂ u k = [ ∂ ∂ u ( u 2 ) ] i + [ ∂ ∂ u ( 2 u sin v ) ] j + [ ∂ ∂ u ( u cos v ) ] k = 2 u i + 2 sin v j + cos v k = 〈 2 u , 2 sin v , cos v 〉 Substitute 1 for u and 0 for v , r u = 〈 2 ( 1 ) , 2 sin ( 0 ) , cos ( 0 ) 〉 = 〈 2 , 0 , 1 〉 Calculation of tangent vector r v : Substitute u 2

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Chapter 16.6, Problem 37E

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To find: An equation of the tangent plane to the vector function r(u, v)=u2 i+2usinv j+ucosv k; u=1, v=0 .

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Chapter 16.6, Problem 36E

Chapter 16.6, Problem 38E

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Chapter 16 Solutions

CALCULUS,EARLY TRANS.-WEBASSIGN ACCESS

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An equation of the tangent plane to the vector function r ( u , v ) = u 2 i + 2 u sin v j + u cos v k ; u = 1 , v = 0 .

Solution Summary: The author explains how the equation of the tangent plane to the vector function is underset_-x+2z=1.

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To find: An equation of the tangent plane to the vector function r(u, v)=u2 i+2usinv j+ucosv k; u=1, v=0 .

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Chapter 16 Solutions