Chapter 1 Topics in Analytic Geometry

MA112 Section 750001: Prepared by Dr.Archara Pacheenburawana 18Solution .........Example 2.6 The curve represented by the parametric equationsx = t 2 y = t 3 (−∞ < t < +∞)is called a semicubical parabola. The parameter t can be eliminated by cubing x andsquaring y, from which it follows that y 2 = x 3 .yxx = t 2 , y = t 3 (−∞ < t < +∞)The graph of this equation consists of two branches: an upper branch obtained by graphingy = x 3/2 and a lower branch obtained by graphing y = −x 3/2 . The two branches meet atthe origin, which corresponds to t = 0 in the parametric equations. This is a singular pointbecause the derivatives dx/dt = 2t and dy/dt = 3t 2 are both zero there. ✠Example 2.7 Without eliminating the parameter, find dy/dx and d 2 y/dx 2 at (1,1) and(1,−1) on the semicubical parabola given by the parametric equations in Example2.6Solution .........Arc Length of Parametric CurveThe following result provides a formula for finding the arc length of a curve from parametricequations for the curve.Arc Length Formula for Parametric CurveIf no segment of the curve represented by the parametric equationsx = x(t), y = y(t) (a ≤ t ≤ b)is traced more than once as t increase from a to b, and if dx/dt and dy/dt are continuousfunctions for a ≤ t ≤ b, then the arclength L of the curve is given by√∫ b (dx ) 2 ( ) 2 dyL = + dt (2.3)dt dtaExample 2.8 Find the circumference of a circle of radius a form the parametric equationsSolution .........x = acost, y = asint (0 ≤ t ≤ 2π)