Find the curvature of the plane curve y - 72::2 +4 at x: -1. Round your answer to three decimal places. 0 0.855 O 0.006 O 0.005 O 0.997 O 0.009 Find the principle unit normal vector to the curve given below at the specified point. r(1) = ti +6t j, t= 4 O N(4) = -48 1 V577 1577 O N(4) = -48 i + V2305 V2305 O N(4) = -48 2305 2305 O N(4) : 48 V2305 V2305 O N(4) -48 577 1577Find the unit tangent vector T(t) and then use it to find a set of parametric equations for the line tangent to the space curve given below at the given point. r (1) = 4ti -5t' j+5tk, t= 5 O x= 20+4s, y=-125+50s, = =25+5s O x= 20+4s, y=-125-50s, = =25+5s O x=25+5s, y=-125-50s, = = 20+4s O x=20+4s, y=-125+50s, = =25-5s O x=20+4s, y=-125-50s, z=25-5s