This preview shows pages 1–3. 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: Math 265 (Butler) Practice Midterm I B (Solutions) 1. Find the projection of h 2 s, 1 ,s 1 i onto the vector h 2 t, 5 t 2 , 4 t i . The formula for projection of a vector u onto a vector v is u v v v v . Applying it to our case with u = h 2 s, 1 ,s 1 i and v = h 2 t, 5 t 2 , 4 t i we have that the projection is h 2 s, 1 ,s 1 i h 2 t, 5 t 2 , 4 t i h 2 t, 5 t 2 , 4 t i h 2 t, 5 t 2 , 4 t i h 2 t, 5 t 2 , 4 t i = (2 s )( 2 t ) + (1)(5 t 2 ) + ( s 1)(4 t ) ( 2 t ) 2 + (5 t 2 ) 2 + (4 t ) 2 h 2 t, 5 t 2 , 4 t i = 4 st + 5 t 2 + 4 st 4 t 4 t 2 + 25 10 t 2 + t 4 + 16 t 2 h 2 t, 5 t 2 , 4 t i = 5 4 t t 2 t 4 + 10 t 2 + 25 h 2 t, 5 t 2 , 4 t i = 5 4 t t 2 ( t 2 + 5) 2 h 2 t, 5 t 2 , 4 t i . (On a side note we see that this projection only depends on t , even though the vector we were projecting involved s .) 2. A particle travels along the parametric curve h e t cos t,e t sin t i starts at (1 , 0) at time t = 0 and then spirals into the origin (0...
View
Full
Document
This note was uploaded on 02/06/2012 for the course MATH 265 taught by Professor Gregorac during the Spring '08 term at Iowa State.
 Spring '08
 Gregorac
 Math

Click to edit the document details