13_1.pdf - 1 (t l 9 t,and2‘arealldefinedwhent l>0:5...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern