**Unformatted text preview: **1. Thecomponentfunctionsln(t+l), 9 t,and2‘arealldeﬁnedwhent+l>0 :5 t>—1 and 9—t2>0 :5 —3 (t <3,sothedomainofris(—l,3). . The corresponding parametric equations for this curve are a: = sin t, y = t.
We can make a table ofvalues, or we can eliminate the parameter: t = y :>
a: = sin y, with y E R. By comparing diﬂ‘erent values oft, we ﬁnd the direction in
which t increases as indicated in the graph 11. 'I'heoorrespondingparametricequationsarez=3, y=t, z=2—t2. Eliminatingtheparameterinyandzgivesz = 2—y2_ Becausez =3,the
curve is a parabola in the vertical plane a: = 3 with vertex (3, 0, 2). 21. x=tcost, y=t, z=tsint, tZO- Atariypoint(33,111,250ont:hec11rve,.1r:2+22 =t2coszt+t23in2t=t2=yzsothe curve lies on the circular cone 1:2 + 22 = y2 with axis the y-axis. Also notice that y 2 0; the graph is II. 25. :c =cos8t, y=sin8t, z= eo'“, t Z 0. 2:2 +112 =c0328t+sin28t= l,sothecurveliesonacircularcylinderwiﬂ1
axis the z-axis. A point (z, y, 2) on the curve lies directly above the point (x, y, 0), which moves counterclockwise around the
writ circle inthe (icy-plane as t increases. The curve starts at (l, 0, 1), when t = 0, and z —r 00 (at an increasing rate) as t—>oo,sothegraphisIV. 41. Ift=—l,thenz=l, y=4, z:0,50thecurvepassesthroughthepoint(l,4,0).Ift=3,thona:=9, y=—8, 2:28,
sothecm'vepassesﬂiroughthepoint(9,—8,28).Forthepoint(4,7,—6)tobeonthecurve,werequirey=l—3t=7 => t: —2.Butthenz = 1 +(—2)3 = —7;£ —6,so(4,7,—6)isnotonthecurve. ...

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