UNIVERSITY OF TORONTO

DEPARTMENT OF MATHEMATICS

MAT 237 Y - MULTIVARIABLE CALCULUS

FALL-WINTER 2005-06

TEST #1. SOLUTIONS

1. a) Compute the area of the region bounded by the curve   x = 2 + t 3y = 1 + t + 5 t 2  and the lines  x = 2 ,  x = 3  and  y = 0 .

Soln. At  x = 2 ,  t = 0  and at  x = 3 ,  t = 1 .

Then,  A =  =

=  =  .

b) Determine whether the curve  x = 2 t –  t 2  ,  y = 2 + 3 tt 2  is concave upward or concave downward at the point  ( 0 , 4 ) .

Soln.   , then   .

At  ( 0 , 4 ) ,  t = 2  and   .  The curve is concave downward at  ( 0 , 4 ) .

2. Compute the length of the polar curve  r = sin 3 (q / 3 ) ,  0 £ q  £ 3 p / 4 .

Soln.

=  =

=  =

=  =  .

3. Suppose that  a  and  b  are two non-zero vectors such that    and   . Compute  .

Soln. Let q  denote the angle between the vectors  a  and  b , then

,

and from the first given condition we obtain    (eq. 1) .

From the second given condition we obtain    or    (eq. 2) .

Adding  (eq. 1)  to the square of  (eq. 2) , we obtain   . Therefore,   .

4. Find the distance from the point  ( 2 , 0 , – 1 )  to the line of intersection of the planes  2 x + y = 1  and  y + z = 2 .

Soln. Let  L  denote the line of intersection of the given planes and let  P ( 2 , 0 , – 1 ) .

The point  A ( 1 , – 1 , 3 )  belongs to the line  L  and  v = ( 2 , 1 , 0 ) ´ ( 0 , 1 , 1 ) = ( 1 , – 2 , 2 )  is a vector director of the line  L .

Now,   = ( 1 , 1 , – 4 ) ,    and   .

Then, the distance  d  from the point  P  to the line  L  is   .

5. Let  L  be line that is tangent to the curve  x = 1 + t 2y = 1 – tz = 7 + tt 2 ,  at the point  ( 5 , 3 , 1 ) . Find the coordinates of the point of intersection of the line  L  and the plane  xy = 8 .

Soln.  ( x ¢ y ¢z ¢ ) = ( 2 t , – 1 , 1 – 2 t ) . At  ( 5 , 3 , 1 ) ,  t = – 2  and the vector  v = ( – 4 , – 1 , 5 )  is tangent to the curve at  ( 5 , 3 , 1 ) . Then,  x = 5 – 4 ly = 3 – l  and  z = 1 + 5 l  are parametric equations of the corresponding tangent line. At the intersection with the plane  xy = 8 ,  ( 5 – 4 l ) – ( 3 – l ) = 8  and  l = – 2 .

The coordinates of the intersection point are  ( 13 , 5 , – 9 ) .

6.  Suppose that  r ( t )  is a vector function such that  r ¢ ( t )  exists for all real numbers  t .

a) Let  f ( t ) =½r ( t )½5 . Determine the values of the constants  m  and  n , if any, for which

f ¢ ( t ) = m½r ( t )½n .

Soln.   , then

= 5½r ( t )½ 3 . Therefore,  m = 5  and  n = 3 .

b) Let  v ( t ) = t 3 r ( t ) . Determine the polynomial function  p ( t ) , if any, for which

( r ( t ) + r ¢ ( t ) ) ´ v ¢ ( t ) = p ( t ) ( r ( t ) ´ r ¢ ( t ) ) .

Soln.  v ¢ ( t ) = 3 t 2 r ( t ) + t 3 r ¢ ( t ) , then  r ( t ) ´ v ¢ ( t ) = t 3 ( r ( t ) ´ r ¢ ( t ) )  and

r ¢ ( t ) ´ v ¢ ( t ) = 3 t 2 ( r ¢ ( t ) ´  r ( t ) ) .

Therefore,  ( r ( t ) + r ¢ ( t ) ) ´ v ¢ ( t ) = ( t 3 – 3 t 2 ) ( r ( t ) ´ r ¢ ( t ) ) .

That is,  p ( t ) = ( t 3 – 3 t 2 ) .

7. Let  .

a) Show that the function  f  is not continuous at  ( 0 , 0 ) .

Soln. Approaching  ( 0 , 0 ) along the line  ( t , t )  we obtain, for  t ¹ 0 ,

,

and  f  is not continuous at  ( 0 , 0 ) .

b) Compute  f x ( 0 , 0 )  and  f y ( 0 , 0 ) , if they exist .

Soln. Notice that  g ( x ) = f ( x , 0 ) = 1  for all values of  x .

Then,  g ¢ ( x ) = f x ( x , 0 ) = 0  for all values of  x , and  f x ( 0 , 0 ) = 0 .

Similarly,  h ( y ) = f ( 0 , y ) = ( 1 – y ) 2h ¢ ( y ) = f y ( 0 , y ) = – 2 ( 1 – y ) ,  and  f y ( 0 , 0 ) = – 2 .

8. a) Prove that for any two sets  A  and  B  in  R n ,  int ( A Ç B ) = int ( A ) Ç int ( B ) .

Is it also true that for any two sets  A  and  B  in  R n ,  int ( A È B ) = int ( A ) È int ( B ) ?

Soln. We will use the fact that the interior of any set  S  is just the largest open set contained in  S .

That is, for any open set  U , if  U Í S  then  U Í int ( S ) .

Now,  int ( A Ç B ) Í ( A Ç B ) , therefore  int ( A Ç B ) Í A  and  int ( A Ç B ) Í B .

But  int ( A Ç B )  is an open set, then  int ( A Ç B ) Í int ( A )  and  int ( A Ç B ) Í int ( B ) .

Therefore  int ( A Ç B ) Í ( int ( A ) Ç int ( B ) )  (cond. 1) .

Also,  int ( A ) Í A  and  int ( B ) Í B , then  ( int ( A ) Ç int ( B ) ) Í ( A Ç B ) .

But  int ( A ) Ç int ( B )  is an open set because  both  int ( A )  and  int ( B )  are open.

Therefore  ( int ( A ) Ç int ( B ) ) Í int ( A Ç B )  (cond. 2) .

From  (cond. 1)  and  (cond. 2) , we obtain  int ( A Ç B ) = int ( A ) Ç int ( B ) .

The property does not hold if the intersections are replaced by unions. Here is a counterexample.

Take  A = [ 0 , 1 ]  and  B = [ 1 , 2 ] . Then  A È B = [ 0 , 2 ] ,  int ( A ) = ( 0 , 1 ) ,  int ( B ) = ( 1 , 2 )

and  int ( A È B ) = ( 0 , 2 ) ¹ ( 0 , 1 ) È ( 1 , 2 ) = int ( A ) È int ( B ) .

b) Prove or disprove the following proposition:

If   f : C ® R n  is continuous, where  C Í R m  and  C  is closed, then  f ( C ) Í R n  is also closed.

Soln. The given proposition is false. The following is a simple counterexample.

Take  C = [ 1 , ¥ ) Í R  and  f : C ® R  defined as  f ( x ) = 1 / x .

Notice that  C  is closed and  f  is continuous but  f ( C ) = ( 0 , 1 ] , which is not a closed set in  R .

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