This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Solutions to Homework Assignment 8 3. lim t + t ln t = lim t + ln t 1 /t H = lim t + 1 /t- 1 /t 2 = lim t + (- t ) = 0 . The overall limit is < 1 , , > . 9. Notice that y 2 + z 2 = 1 , so the curve lies on that cylinder. This will be a helix revolving around the x-axis every units (instead of 2 because of the 2 t inside the sine and cosine).-4-4-2-4-2-2 z x 2 2 2 y 4 4 4 14. We have x 2 + y 2 + z 2 = 2 , so the curve lies on that sphere. The x- and y-coordinates are equal and oscillate between 1 , so the graph is on the plane y = x. Also, the z-coordinate oscillates between 2 . We will just get the circle that is the intersection of y = x and the sphere.-2-2-1-1 y 2 1-1-2 1 z 1 x 2 2 20. Since y = x 2 , the curve must lie on that parabolic cylinder. At the same time, z is a decaying exponential. This matches II. 22. Since x 2 + y 2 = z 2 , the curve lies on that cone. It climbs the cone by wrapping around it and falling exponentially; this matches I....
View Full Document
This homework help was uploaded on 04/07/2008 for the course MTH 249 taught by Professor Starr during the Spring '08 term at Willamette.
- Spring '08