The tangent plane at a point Po(f(uo, vo), g(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 r,(uo, vo) × r,(uo, vo), the cross product of the tangent vectors r,(uo, v) and r„(u0, vo) at Po. In Exercises 27–30, 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. 27. Cone The cone r(r, 0) = (r cos 0)i + (r sin 0)j + rk, r > 0, 0 < 0 < 27 at the point Po(V2, V2, 2) corresponding to (r, 0) = (2, 1/4) 28. Hemisphere The hemisphere surface r(, 0) = (4 sin øcos 0)i + (4 sin o sin 0)j + (4 cos 4)k, 0 < ¢ < T/2, 0

Advanced Engineering Mathematics
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Chapter2: Second-order Linear Odes
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The tangent plane at a point Po(f(uo, vo), g(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 r,(uo, vo) × r,(uo, vo), the
cross product of the tangent vectors r,(uo, v) and r„(u0, vo) at Po. In
Exercises 27–30, 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.
27. Cone The cone r(r, 0) = (r cos 0)i + (r sin 0)j + rk, r > 0,
0 < 0 < 27 at the point Po(V2, V2, 2) corresponding to
(r, 0) = (2, 1/4)
28. Hemisphere The hemisphere surface r(, 0) = (4 sin øcos 0)i
+ (4 sin o sin 0)j + (4 cos 4)k, 0 < ¢ < T/2, 0 <o < 27,
at the point Po(V2, V2, 2V3) corresponding to (4, 0) =
(п/6, п/4)
29. Circular cylinder The circular cylinder r(0, z) = (3 sin 20)i +
(6 sin²0)j + zk, 0 < 0 < , at the point Po(3V3/2, 9/2, 0)
corresponding to (0, z) = (T/3, 0) (See Example 3.)
30. Parabolic cylinder The parabolic cylinder surface r(x, y):
xi + yj – x*k, -0 < x < o, –∞ < y < ∞, at the point
Po(1, 2, – 1) corresponding to (x, y) = (1, 2)
Transcribed Image Text:The tangent plane at a point Po(f(uo, vo), g(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 r,(uo, vo) × r,(uo, vo), the cross product of the tangent vectors r,(uo, v) and r„(u0, vo) at Po. In Exercises 27–30, 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. 27. Cone The cone r(r, 0) = (r cos 0)i + (r sin 0)j + rk, r > 0, 0 < 0 < 27 at the point Po(V2, V2, 2) corresponding to (r, 0) = (2, 1/4) 28. Hemisphere The hemisphere surface r(, 0) = (4 sin øcos 0)i + (4 sin o sin 0)j + (4 cos 4)k, 0 < ¢ < T/2, 0 <o < 27, at the point Po(V2, V2, 2V3) corresponding to (4, 0) = (п/6, п/4) 29. Circular cylinder The circular cylinder r(0, z) = (3 sin 20)i + (6 sin²0)j + zk, 0 < 0 < , at the point Po(3V3/2, 9/2, 0) corresponding to (0, z) = (T/3, 0) (See Example 3.) 30. Parabolic cylinder The parabolic cylinder surface r(x, y): xi + yj – x*k, -0 < x < o, –∞ < y < ∞, at the point Po(1, 2, – 1) corresponding to (x, y) = (1, 2)
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