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: MA 242 Test 2 Solutions 1. [10 points] Given the vectorvalued function, r ( t ) = < cos( t ) , ln( t ) , t 3 > where t > 0. Find the velocity and acceleration vectors. Solution: v ( t ) = r ( t ) = < sin( t ) , 1 t , 3 t 2 > a ( t ) = r 00 ( t ) = < cos( t ) , 1 t 2 , 6 t > 2. [10 points] Find and sketch the domain of the function f ( x, y ) = p x 2 + y 2 4 + ln(25 x 2 y 2 ). Solution: We need to have x 2 + y 2 4 0 so x 2 + y 2 4 which is the region in R 2 outside of the circle with radius 2 centered at the origin (the boundary of the circle is included). We also need to have 25 x 2 y 2 > 0 so x 2 + y 2 < 25 which is the region in R 2 inside of the circle with radius 5 centered at the origin (the boundary of the circle is not included). Therefore the domain is D = { ( x, y ) R 2  4 x 2 + y 2 < 25 } which is the annular region shown below. 3. [15 points] Find the following limit, if it exists, or show that the limit does not exist. (If you think theFind the following limit, if it exists, or show that the limit does not exist....
View
Full
Document
This note was uploaded on 03/31/2009 for the course MA 242 taught by Professor Bliss during the Spring '08 term at N.C. State.
 Spring '08
 Bliss
 Vectors

Click to edit the document details