Midterm_Solutions

# Midterm_Solutions - Problem 1. It is easy to see that the...

This preview shows page 1. Sign up to view the full content.

Problem 1. It is easy to see that the tangent vector to the curve ( r ( t ) ( t )) in polar coordi- nates has components ˙ r and r ˙ θ . So, the length of a curve is l = Z b a q r ( t )) 2 + ( r ( t ) ˙ θ ( t )) 2 dt. Using this formula we compute length of the curve (2sin πt,πt ): l = Z 1 0 p 4 π 2 cos 2 πt + 4 π 2 sin 2 πtdt = 2 π. Problem 2. Let P = { x 0 ,...,x n } be a partition of [ a,b ]. Then we can construct partition Q = { y 1 ...,y n } of [ c,d ] such that x i = φ ( y i ). Also, for any partition of [ c,d ] we can construct corresponding partition of [ a,b ]. Since g = f φ , values of f on [ x i - 1 ,x i ] are the same as values of g on [ y i - 1 ,y i ]. So, we have U ( Q,g,β ) = U ( P,f,α ) , L ( Q,g,β ) = L ( P,f,α ) . The statement follows from these equalities and Theorem 6.6 from [Rudin]. Problem 3. We deﬁne d ( E,F ) = m (( F E ) \ ( F E )). Let d ( E,F ) = 0 and d ( F,G ) = 0. First, we see that
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/08/2010 for the course MA 108b taught by Professor Zinchenko,m during the Winter '08 term at Caltech.

Ask a homework question - tutors are online