**Unformatted text preview: **2 ) ( ) ( 2 ) ( t t t y t x t z--=--= This yields . ) sin cos 2 ( sin cos ) ( k j i r t t t t t--+ + = (b) . ) sin (cos sin cot ; ) cos (sin cos sin k j i r k j i r t t t t t t t + +--=-+ +-= & & & At the point (0,1,1). , 2 / = t and hence . ) 2 / ( , ) 2 / ( k j r k i r +-= +-= & & & Consequently, . ) 2 / ( ) 2 / ( j i k r r---= × & & & Hence the curvature at the given point is . 2 3 ) 2 ( 3 ) 2 / ( ) 2 / ( ) 2 / ( ) 2 / ( 2 / 3 2 / 1 3 3 = = × = r r r & & & & (c) Note that 2 = + + z y x defines a plane and hence the curve is planar. The curve is an ellipse since the intersection of a plane and a circular cylinder yields an ellipse (which could be a circle—not the case here)....

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- Spring '10
- BLUMAN
- General Relativity, Osculating circle