Answer to Question #122826 in Differential Geometry | Topology for Ojugbele Daniel

Question #122826
Determine the unit tangent vector at the point (2,4,7) for the circle with parametric equations x=2u, y=u^2 +3 and z=2u^2 +5
1
Expert's answer
2020-06-23T14:54:02-0400

Let’s consider "r(u)=\\langle2u, \\ u^2+3, \\ 2u^2+5\\rangle"

"r(1)=\\langle2,4,7\\rangle"

The unit tangent vector is "\\frac{r^{\\prime}(u)}{|r(u)|}" .

"r^{\\prime}(u)=\\langle 2,\\ 2u,\\ 4u\\rangle, \\ \\ r^\\prime (1)=\\langle 2,\\ 2, \\ 4\\rangle, \\ \\ |r^\\prime (1)|=\\sqrt{2^2+2^2+4^2}=2\\sqrt{6}"


"\\frac{r^{\\prime}(1)}{|r(1)|}=\\frac{1}{2\\sqrt{6}}\\langle 2,\\ 2,\\ 4\\rangle =\\frac{1}{\\sqrt 6}\\langle 1,\\ 1,\\ 2\\rangle" .


Answer: the unit tangent vector at the point "(2,4,7)"  is "\\frac{1}{\\sqrt 6}\\langle 1,\\ 1,\\ 2\\rangle" .



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