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: Solutions for PSet 3 1. First we note that T ( P ) is again a parallelepiped. To see this notice that if x ∈ T ( P ) then x = T ( i n =1 c i v i ) where ≤ c i ≤ 1. By linearity x = i c i T ( v i ) for ≤ c i ≤ 1. That is, T ( P ) = { x ∈ R n  x = c i T ( v i ) , ≤ c i ≤ 1 } and this i is precisely the definition of a parallelepiped. Now, we determine the volume. If V denotes the matrix formed such that the i th row is v i , then: V ol ( P ) =  det ( V )  Let W denote matrix with w i = m ( T ) v i as the i th row, then by the definition of T ( P ) above: V ol ( T ( P )) =  det ( W )  =  det ( m ( T ) V )  =  det ( m ( T )) det ( V )  =  det ( m ( T ))  vol ( P ) . 2. (14.4:23) Since F is continuous, we can apply the first fundamental theorem of calculus to determine a derivative for the expression on the right hand side of the equation. Thus F ( x ) = e x A + xe x A + x 1 F ( x ) − x 1 2 1 x F ( t ) dt = 2 e x A + xe x A ∀ x > . Now we apply the second fundamental theorem of calculus and the above result to get: x x F ( x ) − F (1) = F ( t ) dt = A (2 e t + te t ) dt 1 1 Integrating by parts we have: x x F ( x ) − F (1) = A (2 e t + te t − e t ) = A ( e t + te t ) = A ( e x + xe x ) − 2 Ae 1 1 Substituting F (1) = eA we observe F ( x ) = Ae x +...
View
Full
Document
This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.
 Spring '11
 ChristineBreiner
 Multivariable Calculus

Click to edit the document details