# 237q1s - 1[4 marks Find the shortest distance from the...

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1. [4 marks] Find the shortest distance from the point ) 4 , 1 , 1 ( P to the line tangent to the curve at the point . ) 6 , 3 , 2 ( ) ( t t e e t t + = r ) 5 , 2 , 0 ( 0 P The point corresponds to the parameter value ) 5 , 2 , 0 ( 0 P 0 = t . So the tangent vector to the curve at this point is If we construct the vector , then the shortest distance will be the height h of the parallelogram spanned by the ) 1 , 1 , 2 ( ) , , 2 ( ) 0 ( 0 0 = = = e e r v. P P 0 vectors v and which is: area of the parallelogram A divided by the length of the base. P P 0 Now 2 3 ) 3 , 3 , 0 ( ) 1 , 1 , , 1 ( ) 1 , 1 , 2 ( 0 = = × = × = P P A v and 6 = v , so the shortest distance is 3 6 2 3 = = h . 2. (a) [5 marks] Let S be the subset of 2 R defined by . } 1 : ) , 0 {( \ } 4 : ) , {( 2 2 > < + = y y y x y x S (i) Is S open, closed or neither ? Justify your choice, no marks for guessing. This is just a version of the homework question 1(f),sec.1.2. The set S is neither closed nor open. It contains the boundary point (0,1) but no other boundary points. (ii) Describe in the set notation: = int S )} 1 , 0 {( \ S = S } 4 : ) , {( 2 2 + y x y x = S } 2 1 : ) , 0 {( } 4 : ) , {( 2 2 = + y y y x y x (b) [3 marks] Prove that if x is in the closure of then there is a sequence of points n R S in S that converges to x. This is the second part of the proof of Th. 1.14 in the text (assignment). 2

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3. (a) [4 marks] Evaluate the limit or prove that it does not exist: 2 2 2 3 ) 0 , 0 ( ) , ( lim y x y x y x + + . This is a version of the homework question 1(a),sec1.3, exponents in nominator are higher by 1.The limit then exists. You may, for example use polar coordinates
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## This note was uploaded on 10/21/2010 for the course MATHEMATIC MAT237Y1 taught by Professor Romauldstanczak during the Fall '09 term at University of Toronto.

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237q1s - 1[4 marks Find the shortest distance from the...

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