UNIVERSITY OF TORONTO

DEPARTMENT OF MATHEMATICS

MAT 237 Y - MULTIVARIABLE CALCULUS

FALL-WINTER 2005-06

ASSIGNMENT #1. SOLUTIONS

1. Compute the area of the surface generated when the parametric curve

  is rotated about the line  x = 1 .

Solution:

Setting up the integral corresponding to this surface area we obtain:

A =  , where

1 – x ( t ) =  ,  x¢ ( t ) =  ,  y ¢ ( t ) =  , and

[ x¢ ( t ) ] ² + [ y ¢ ( t ) ] ² =  .

So, the area is  A =  =  =  .

2. Determine the maximum and minimum values of the curvature at points of the polar curve  r = 3 + sin q  .

Solution.

The given curve can be parametrized as:

x(q ) = rcosq  = 3cosq  + sinq cosq  = 3cosq  + sin(2q ) , and

y(q ) = rsinq  = 3sinq  + sin ²q  = 3sinq  + (1 – cos(2q )) , 0≤q ≤2p .

Now,  x¢(q ) = – 3sinq  + cos(2q ) ,  y ¢(q ) = 3 cosq  + sin(2q ) ,

x²(q ) = – 3cosq  – 2sin(2q ) ,  y ²(q ) = – 3sinq  + 2cos(2q ) ,

x¢(q ) y ²(q )–x²(q ) y ¢(q )=11+9(sin(2q )cosq –sinq cos(2q ))=11+9sinq , and

[x¢(q )] ² + [y ¢(q )] ² =10+6(sin(2q )cosq –sinq cos(2q ))=10+6sinq .

Therefore,  K (q ) =

and  K ¢(q ) = .

The possible extrema of  K (q )  are at  q = 0, p /2, 3p /2, arcsin(–1/3) .

Testing, we find  K(0)=11/10^3/2 » 0.3479 , K(p /2)=20/16^3/2 = 0.3125 ,

K(3p /2)=2/4^3/2 = and K(arcsin(–1/3))=8/8^3/2=» 0.3536 .

So, the minimum and maximum curvatures are  and  , respectively.

3. Let  p  denote a real number such that  0 < p < 1  and let  v ₁ , v ₂ , v ₃ , …  be vectors in  R³  such that    and    for all  k = 1 , 2 , 3 , …  Find the values of  p , if any, for which  .

Solution.

Notice that  

=

=  , and

=  .

Now,

=  .

We want to find the values of  0 < p < 1  such that  4 p ³ – 3 p ² + 4 p – 1 = 0 .

The polynomial function  f ( p ) = 4 p ³ – 3 p ² + 4 p – 1  is continuous and differentiable

everywhere. The function  f  has at least one zero over  ( 0 , 1 )  because  f ( 0 ) = – 1  and

f ( 1 ) = 4 . That is the only real zero of  f  because  f ¢ ( p ) = 12 p ² – 6 p + 4 > 0  for all  p .

So, there is a unique value of  p . An approximated value is  p » 0.2884 .

4. Prove or disprove each of the following propositions:

a) If  C ₁ , C ₂ , C ₃ , …  are all closed sets in  R ⁿ , then the set    is also a closed set in  R ⁿ .

Solution:

This proposition is false. We will give a counterexample.

Consider the sets  C ₁ = { 1 } , C ₂ = { 1 / 2 } ,  …  , C _k = { 1 / k } ,  … 

All of these sets are obviously closed but their union

  is not a closed set because  implies that  0  is a boundary

point of  C  and  0 Ï C .

b) If  A ₁ , A ₂ , A ₃ , …  are all non empty compact sets in  R ⁿ  such that  A _{k + 1} Ì A _k  for all  k = 1 , 2 , 3 , …  , then the set    is also non empty.

Solution:

This proposition is true.

We will prove that the assumption “A  is an empty set” leads to a contradiction.

Notice that all the sets  A _k  are closed because they are compact. Therefore, the sets    are all open. Assume now that the set  A  is empty. If the set  A  is empty then

 . It means that    is an open cover of  R ⁿ  and

therefore an open cover of  A ₁ Ì R ⁿ . Consider now any finite choice of some of these open sets, say   where  k ₁ < k ₂ < k ₃ < … < k _m . Notice that   because  therefore  but    because  Æ ¹ Ì A ₁ .

So    is an open cover of  A ₁  that does not contain a finite subcollection of open sets which also covers  A ₁ . This is a contradiction because  A ₁  is a compact set.