{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

calculus_solutions_16_6

# calculus_solutions_16_6 - Stewart Calculus ET 5e...

This preview shows pages 1–3. Sign up to view the full content.

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 ² b = i 2 0 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 ³ u ³ 1 ± 1 ³ v ³ 1 x = u 2 +1 y = v 3 +1 z = u + v ± 1 ³ u ³ 1 ± 1 ³ v ³ 1 u u 0 x = u 2 0 +1 yz ± v y = v 3 0 +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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
r ( u , v )= u + v , u 2 , v 2 ± 1 ³ u ³ 1 ± 1 ³ v ³ 1 x = u + v y = u 2 z = v 2 ± 1 ³ u ³ 1 ± 1 ³ v ³ 1 u = u 0 y = u 2 0 = xz ± v = v 0 z = v 2 0 = 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 0 ³ u ³ ² 0 ³ v ³ 2 ² v = v 0 z =sin 3 v 0 xy ± u = u 0 x =cos 3 u 0 cos 3 v y =sin 3 u 0 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) 0 ³ u ³ 2 ² 6. , , . The surface has parametric equations , , , , . If is constant, constant, so the corresponding grid curves are the curves parallel to the plane. If is constant, constant, so the corresponding grid curves are the curves parallel to the plane. 7. The surface has parametric equations , , , , . Note that if is constant then is constant, so the corresponding grid curves must be the curves parallel to the plane. The vertically oriented grid curves, then, correspond to being held constant, giving , , . These curves lie in vertical planes that contain the axis.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 19

calculus_solutions_16_6 - Stewart Calculus ET 5e...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online