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: x = 2 tt 2 , y = √ t , and the yaxis. Solution: In order to ﬁnd the points of intersection, set x = 0: 2 tt 2 = 0 ⇒ t (2t ) = ⇒ t = 0 , 2 and y = 0 , √ 2. Since x = 2 tt 2 ≤ 0 for 0 ≤ t ≤ 2, AREA = Z √ 2 xdy = Z 2 ( 2 tt 2 ) ± 1 2 √ t dt ¶ = 1 2 Z 2 ( 2 tt 2 ) t1 / 2 dt = 1 2 Z 2 ‡ 2 t 1 / 2t 3 / 2 · dt = 1 2 ± 2 · 2 3 t 3 / 22 5 t 5 / 2 ¶ﬂ ﬂ ﬂ ﬂ 2 = 1 2 ± 4 3 2 3 / 22 5 2 5 / 2 ¶ = 4 √ 2 ± 1 31 5 ¶ = 8 √ 2 15 . t 0.5 1.0 1.5 2.0 x 0.2 0.4 0.6 0.8 1.0 x as a function of t t 0.5 1.0 1.5 2.0 y 0.2 0.4 0.6 0.8 1.0 1.2 1.4 y as a function of t x 0.2 0.4 0.6 0.8 1.0 y 0.2 0.4 0.6 0.8 1.0 1.2 1.4 parametric plot of x and y as functions of t...
View
Full
Document
This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.
 Spring '09
 EGCDD

Click to edit the document details