Calculate the normal curvature at the point (1,0,1) of the curve gamma on the hy

Question

Calculate the normal curvature at the point (1,0,1) of the curve gamma on the hyperbolic paraboliod sigma(u,v) = ((1/2)(u+v), (1/2)(v-u), uv) corresponding to the straight line u = v = t in the uv-plane (note that gamma is not unit-speed).

Explanation / Answer

A method to solve this question The curavature of parametric curve r(t) is given ?(t) = | r'(t)×r''(t) | / | r'(t) |³ r(t) = = => r'(t) = = < 1 , 6t^(1/2) , -2t > r''(t) = = < 0 , 3t^(-1/2) , -2 > => r'(t)×r''(t) = < 6·t^1/2·(-2) - 3t^(-1/2)·(-2t) , 0·(-2t) - 1·(-2) , 1·(3t^(-1/2) - 0·6t^(3/2) > = < 0 , 2 , 3t^(-1/2) > | r'(t)×r''(t) | = ( 0² + 2² + (3t^(-1/2)² )^(1/2) = ( 4 + 9/t)^(1/2) | r'(t) | = ( 1² + (6t^(1/2))² + (-2t)² )^(1/2) = (1 + 36t + 4t² )^(1/2) => ?(t) = ( 4 + 9/t)^(1/2) / (1 + 36t + 4t² )^(3/2) at t =1 r(t) = < 1 , 4 , -1> and ?(t) = ( 4 + 9/1)^(1/2) / (1 + 36·1 + 4·1² )^(3/2) = (13/41³)^(1/2) = (13/68921)^(1/2) = 0.0137333 · | r'(t) | = ( 1² + (6t^(1/2))

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