Equ of osculating plane
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The document provides the solution to finding the equation of the osculating plane for the curve x = 3t, y = 3t^2, z = 2t^3. It first finds the first and second derivatives of the curve, r' and r''. It then uses these derivatives and the formula for the equation of an osculating plane to determine the final equation is 2t^2X - 2tY + Z = 2t^3.
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![Differential Geometry
For the curve 𝑥 = 3𝑡, 𝑦 = 3𝑡2, 𝑎𝑛𝑑 𝑧 = 2𝑡3find the equation
of osculating plane.
Solution:
Since 𝑟 = 𝑥, 𝑦, 𝑧 = (3𝑡, 3𝑡2, 2𝑡3)
⇒ 𝑟′
= (3,6𝑡, 6𝑡2
)
⇒ 𝑟′′
= (0,6,12𝑡)
We know equation of osculating plane
[𝑅 − 𝑟, 𝑟′, 𝑟′′] = 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 𝑥 = 3𝑡, 𝑦 = 3𝑡2, 𝑎𝑛𝑑 𝑧 = 2𝑡3find the equation of osculating plane. Solution: Since 𝑟 = 𝑥, 𝑦, 𝑧 = (3𝑡, 3𝑡2, 2𝑡3) ⇒ 𝑟′ = (3,6𝑡, 6𝑡2 ) ⇒ 𝑟′′ = (0,6,12𝑡) We know equation of osculating plane [𝑅 − 𝑟, 𝑟′, 𝑟′′] = 0
- 2. Differential Geometry ⇒ 𝑋 − 𝑥 𝑌 − 𝑦 𝑍 − 𝑧 𝑥′ 𝑦′ 𝑧′ 𝑥′′ 𝑦′′ 𝑧′′ = 0 ⇒ 𝑋 − 3𝑡 𝑌 − 3𝑡2 𝑍 − 2𝑡3 3 6𝑡 6𝑡2 0 6 12𝑡 = 0
- 3. Differential Geometry ⇒ after solving the determinant we get the following equation of osculating plane 2𝑡2 𝑋 − 2𝑡𝑌 + 𝑍 = 2𝑡3