Figure 3.2: Explicit quadratic surfaces z=2(ax+ By").(a)Left: Hyperbolic paraboloid (a=-3, B=1).(b) Right: Elliptic paraboloid(a=l, B=3) In vector notation r=r(u,U where r=(a, g, a),r(u,v)=((u,u),y(u, u), 2(u, u) x=+ 1 y=u-}→ eliminate,→2=5(x2+y2) paraboloid 3.2 Curves on a surface Let r=r(u, v) be the equation of a surface, defined on a domain D(i.e, uI <usu U1 <U<U2). Let B(t=(u(t), u(t)be a curve in the parameter plane. Then r=ru(t), v(t)) is a curve lying on the surface, see Figure 3. 3. a tangent vector of curve B(t)is given by B(t=(i(t),i(t)) a tangent vector of a curve on a surface is given by By using the chain rule dr(u(t), v(t)) ar du ar du rui(t)+ri(tFigure 3.2: Explicit quadratic surfaces z = 1 2 (αx2 + βy 2 ). (a) Left: Hyperbolic paraboloid (α = −3, β = 1). (b) Right: Elliptic paraboloid (α = 1, β = 3). In vector notation: r = r(u, v) where r = (x, y, z), r(u, v) = (x(u, v), y(u, v), z(u, v)) Example: r = (u + v, u − v, u 2 + v 2 ) x = u + v y = u − v z = u 2 + v 2    ⇒ eliminate u, v ⇒ z = 1 2 (x 2 + y 2 ) paraboloid 3.2 Curves on a surface Let r = r(u, v) be the equation of a surface, defined on a domain D (i.e., u1 ≤ u ≤ u2, v1 ≤ v ≤ v2). Let β(t) = (u(t), v(t)) be a curve in the parameter plane. Then r = r(u(t), v(t)) is a curve lying on the surface, see Figure 3.3. A tangent vector of curve β(t) is given by β˙(t) = (u˙(t), v˙(t)) A tangent vector of a curve on a surface is given by: dr(u(t), v(t)) dt (3.1) By using the chain rule: dr(u(t), v(t)) dt = ∂r ∂u du dt + ∂r ∂v dv dt = ruu˙(t) + rvv˙(t) (3.2) 3