S→ 0,y′ Figure 2.3: Vertical slopes for explicit curves involve non-polynomial functions It is easy to trace parametric curves It is relatively difficult to check if a point lies on the curve Closed and multi-valued curves are eas It is easy to evaluate tangent line to the curve when the curve has a vertical or near Axis independent.(Easy to transform to another coordinate system) Example: Folium of Descartes, see Figure 2.1, can be expressed as r(t) ∞<t<∞→ easy to trace (t)=ao solve for t= plug t into y(t)= yo need to solve a nonlinear equation to check if a point lies on the curve Explicit curve y= Va can be expressed as x=t2, y=t(t> 0) =(t2,t), unit tangent vector t att=0,t=(0,1) Therefore there is no problem representing a vertical tangent computationally 2.1.2 Space curves In 3D, a single equation generally represents a surface. For example x2+y2+22y x y = √ x; y 0 = 1/2 √ x; as x → 0, y 0 → ∞ Figure 2.3: Vertical slopes for explicit curves involve non-polynomial functions. – It is easy to trace parametric curves. – It is relatively difficult to check if a point lies on the curve. – Closed and multi-valued curves are easy to represent. – It is easy to evaluate tangent line to the curve when the curve has a vertical or near vertical tangent. – Axis independent. (Easy to transform to another coordinate system) Example: Folium of Descartes, see Figure 2.1, can be expressed as: r(t) =  3t 1+t 3 , 3t 2 1+t 3  − ∞ < t < ∞ ⇒ easy to trace x(t) = x0 ⇒ solve for t ⇒ plug t into y(t) = y0 ⇒ need to solve a nonlinear equation to check if a point lies on the curve. Explicit curve y = √ x can be expressed as x = t 2 , y = t (t ≥ 0). r = (t 2 ,t), r˙ = (2t, 1) unit tangent vector t = (2t, 1) √ 4t 2 + 1 at t = 0, t = (0, 1) Therefore there is no problem representing a vertical tangent computationally. 2.1.2 Space curves • Implicit curves In 3D, a single equation generally represents a surface. For example x 2 + y 2 + z 2 = a 2 is a sphere. 4