{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 13-1 - Solutions to Homework Assignment 8 3 lim t → t ln...

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
This is the end of the preview. Sign up to access the rest of the 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 π units (instead of 2 π 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

{[ snackBarMessage ]}

### Page1 / 2

13-1 - Solutions to Homework Assignment 8 3 lim t → t ln...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online