The tangent plane at a point Po (f(uovo) 9 (uo.vo),h(uo.vo)) on a parametrized surface r(u,v) = f(u, v) i + g(u,v) j+h(u,v) k is the plane through Po normal to the vector ru (uovo) xrv (uovo), the cross product of the tangent vectors ru (40,vo) and rv (uo.vo) at Po. Find an equation for the plane tangent to the surface at Po. Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. The circular cylinder r(0,z) = (3 sin (20))i + (6 sin²0) j+z k at the point Po 3√3 9 2 ¹2 1 -₁)₁ -1 corresponding to (0,z) = - (5-1) An equation for the plane tangent to the surface at Po is (Type an equation using x, y, and z as the variables.)

Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Vectors In Two And Three Dimensions
Section9.6: Equations Of Lines And Planes
Problem 2E
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2. A cartesian equation for the surface is?

3. Draw the graph and the tangent plane

The tangent plane at a point Po (f(uovo) 9 (uo.vo),h(uo.vo)) on a
parametrized surface r(u,v) = f(u, v) i + g(u,v) j+h(u,v) k is the plane through
Po normal to the vector ru (uovo) xrv (uovo), the cross product of the
tangent vectors ru (40,vo) and rv (uo.vo) at Po. Find an equation for the
plane tangent to the surface at Po. Then find a Cartesian equation for the
surface and sketch the surface and tangent plane together.
The circular cylinder r(0,z) = (3 sin (20))i + (6 sin²0) j+z k at the point
Po
3√3 9
2 ¹2
1
-₁)₁
-1 corresponding to (0,z) =
- (5-1)
An equation for the plane tangent to the surface at Po is
(Type an equation using x, y, and z as the variables.)
Transcribed Image Text:The tangent plane at a point Po (f(uovo) 9 (uo.vo),h(uo.vo)) on a parametrized surface r(u,v) = f(u, v) i + g(u,v) j+h(u,v) k is the plane through Po normal to the vector ru (uovo) xrv (uovo), the cross product of the tangent vectors ru (40,vo) and rv (uo.vo) at Po. Find an equation for the plane tangent to the surface at Po. Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. The circular cylinder r(0,z) = (3 sin (20))i + (6 sin²0) j+z k at the point Po 3√3 9 2 ¹2 1 -₁)₁ -1 corresponding to (0,z) = - (5-1) An equation for the plane tangent to the surface at Po is (Type an equation using x, y, and z as the variables.)
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