![Differential Geometry
For the curve π₯ = 3π‘, π¦ = 3π‘2, πππ π§ = 2π‘3find the equation
of osculating plane.
Solution:
Since π = π₯, π¦, π§ = (3π‘, 3π‘2, 2π‘3)
β πβ²
= (3,6π‘, 6π‘2
)
β πβ²β²
= (0,6,12π‘)
We know equation of osculating plane
[π
β π, πβ², πβ²β²] = 0](https://image.slidesharecdn.com/equofosculatingplane-201004124816/85/equ-of-osculating-plane-1-320.jpg)
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Equ of osculating plane
- 1. Differential Geometry For the curve π₯ = 3π‘, π¦ = 3π‘2, πππ π§ = 2π‘3find the equation of osculating plane. Solution: Since π = π₯, π¦, π§ = (3π‘, 3π‘2, 2π‘3) β πβ² = (3,6π‘, 6π‘2 ) β πβ²β² = (0,6,12π‘) We know equation of osculating plane [π β π, πβ², πβ²β²] = 0
- 2. Differential Geometry β π β π₯ π β π¦ π β π§ π₯β² π¦β² π§β² π₯β²β² π¦β²β² π§β²β² = 0 β π β 3π‘ π β 3π‘2 π β 2π‘3 3 6π‘ 6π‘2 0 6 12π‘ = 0
- 3. Differential Geometry β after solving the determinant we get the following equation of osculating plane 2π‘2 π β 2π‘π + π = 2π‘3