calculus_solutions_16_6 - r ( u , v )= u cos v i + u sin v...

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Unformatted text preview: r ( u , v )= u cos v i + u sin v j + u 2 k x = u cos v , y= u sin v , z = u 2 . ( x , y , z ) x 2 + y 2 = u 2 cos 2 v + u 2 sin 2 v = u 2 = z z = x 2 + y 2 z r ( u , v )=(1+2 u ) i +( u +3 v ) j +(2+4 u +5 v ) k = 1, 0, 2 + u 2, 1,4 + v 0, 3, 5 . (1, 0, 2) a = 2, 1, 4 b = 0, 3, 5 . a b = i 2 j 1 3 k 4 5 = 17 i 10 j +6 k 17( x 1) 10( y 0)+6( z 2)=0 17 x 10 y +6 z = 5 r ( x , )= x , cos , sin x = x , y =cos , z =sin . ( x , y , z ) y 2 + z 2 =cos 2 +sin 2 =1 x = k y 2 + z 2 =1 x = k x = x 1 x r ( x , )= x , x cos , x sin x = x , y = x cos , z = x sin . ( x , y , z ) y 2 + z 2 = x 2 cos 2 + x 2 sin 2 = x 2 x = x x 2 = y 2 + z 2 x r ( u , v )= u 2 +1, v 3 +1, u + v 1 u 1 1 v 1 x = u 2 +1 y = v 3 +1 z = u + v 1 u 1 1 v 1 u u x = u 2 +1 yz v y = v 3 +1 xz 1. , so the corresponding parametric equations for the surface are For any point on the surface, we have . Since no restrictions are placed on the parameters, the surface is , which we recognize as a circular paraboloid opening upward whose axis is the axis. 2. From Example 3, we recognize this as a vector equation of a plane through the point and containing vectors and If we wish to find a more conventional equation for the plane, a normal vector to the plane is and an equation of the plane is or . 3. , so the corresponding parametric equations for the surface are For any point on the surface, we have , so any vertical trace in is the circle , . Since with no restriction, the surface is a circular cylinder with radius whose axis is the axis. 4. , so the corresponding parametric equations for the surface are For any point on the surface, we have . With and no restrictions on the parameters, the surface is , which we recognize as a circular cone whose axis is the axis. 5. , , . The surface has parametric equations , , , , . If we keep constant at , , a constant, so the corresponding grid curves must be the curves parallel to the plane. If is constant, we have , a constant, so these grid curves are the curves parallel to the plane. 1 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.6 Parametric Surfaces and Their Areas r ( u , v )= u + v , u 2 , v 2 1 u 1 1 v 1 x = u + v y = u 2 z = v 2 1 u 1 1 v 1 u = u y = u 2 = xz v = v z = v 2 = xy r ( u , v )= cos 3 u cos 3 v , sin 3 u cos 3 v , sin 3 v . x =cos 3 u cos 3 v y =sin 3 u cos 3 v z =sin 3 v u v 2 v = v z =sin 3 v xy u = u x =cos 3 u cos 3 v y =sin 3 u cos 3 v z =sin 3 v z r ( u , v )= cos u sin v , sin u sin v , cos v +ln tan ( v /2) . x =cos u sin v y =sin u sin v z =cos v +ln tan ( v /2) u 2 6. , , ....
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This note was uploaded on 04/23/2011 for the course JAPAN 7b taught by Professor Wallace during the Spring '11 term at University of California, Berkeley.

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calculus_solutions_16_6 - r ( u , v )= u cos v i + u sin v...

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