You may nd it valuable for intuition to plot your solutions in CalcPlotBD when you nish each problem. 1. Consider the parametric surface S traced out by 1"(11, 1)) - (u: 1):, u + v,u 1.1/2). Compute a parametric form for the tangent plane 31-313135. 2. Consider the parametric surface S traced out by a\": u U T(H.U)= (2cosc 231112;)- Compute a parametric form for the tangent plane Tim/5.1V; 3. Suppose you have a surface 3 containing the point P = (1, 1, U). All you know about the surface is that the two curves 1 mm = (:32: 1,1nt), T26) = (13+ 1, t + 1,.2' 1) lie on the surface. Using only this information, nd a parametric form for the tangent plane T(1.1.|3)S' Then nd an equation for that tangent plane. 4. Given a 2-variable function an y), its graph is the surface S= {mart} | 3= aw}- This surface can be parameterized using the function day) = (as, y, f(a:, y a) Use this parameterization to nd a parametric form for the tangent plane Ttmmmmmyniis' b) Find an equation for that tangent plane. c) Rewrite your equation for the tangent plane in the form 2 = a.(x so) + My yo) + c. Challenge 1 (optional!) Consider the parametric surface traced out by u, v) = (112, 111}, v2). Notice that there are exactly two choices of parameters (on, on) with Home\") = (1,0,1): they are (1, 1) and (1, 1). a) Compute an equation for T(1,0.1)S. thinking of (1, 0, 1) as being r(1,1). b) Compute an equation for Tth'l, thinking of (1,0,1) as being r(1,1). c) You will get two different answers. In CalcPlot3D, plot the point (1,0,1), and then plot the graph of r(u,v) in CalcPlotBD, first with parameter range 2 5 {11} 5 0 and then with parameter range 0 s n. u s 2. See that the tangent planes at (1, U, 1} appear different for both of these. Then plot the graph with parameter range 2 3 1m; S 2, and notice how the surface intersects itself. This shows that our idea of tangent spaces don't work well for self-intersecting objects; these are not regular surfaces