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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....
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 Spring '08
 Bliss
 Derivative, Vectors

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