UNIVERSITY OF TORONTO

DEPARTMENT OF MATHEMATICS

MAT 237 Y - MULTIVARIABLE CALCULUS

FALL-WINTER 2005-06

TEST #2. SOLUTIONS

1. Given that the equation  z 3  x z  y = 0  defines  z  as a function of  x  and  y , show that

.

Soln.    and   .

Then,   .

2. Determine all the points on the ellipsoid  x 2 + 2 y 2 + 3 z 2 = 18  where the tangent plane is parallel to the plane

x  y  3 z = 1 , and find equations of each of the corresponding tangent planes.

Soln. A vector normal to the given plane is  n = ( 1 ,  1 ,  3 )  and a vector normal to the given ellipsoid at a point

( x , y , z )  on the ellipsoid is  ( x 2 + 2 y 2 + 3 z 2 ) = ( 2 x , 4 y , 6 z ) .

For these two vectors to be parallel, we need  ( 2 x , 4 y , 6 z ) = l ( 1 ,  1 ,  3 )  or

x = l / 2 ,  y =  l / 4  and  z =  l / 2 .

Then,  ( l / 2 ) 2 + 2 (  l / 4 ) 2 + 3 (  l / 2 ) 2 = 18  and  l = ฑ 4 .

The points on the ellipsoid are  ( 2 ,  1 ,  2 )  and  (  2 , 1 , 2 ) .

Equations for the corresponding tangent planes are  x  y  3 z = 9  and  x  y  3 z =  9 .

3. Suppose that the functions  f : R R  and  g : R R  are both twice continuously differentiable and let

v ( x, y ) = f ( x 2 + 3 y ) + g ( x 2  3 y ) .

Show that the function  v  satisfies the equation  x 3  = k (   x  ) ,

where  k  is a constant, and find the value of the constant  k .

Soln.   ,

and

 x  =  .

Also,   ,   , and

.

Therefore,  x 3  = (   x  )  and  k =  9 / 4 .

4. Consider the function  f ( x , y ) = 2 x y + x  1 + 4 y  1  defined over the region

{ ( x , y ) R 2   x > 0  and  y > 0 }. Find all the critical points of the function  f  .

Classify each of the critical points of the function  f  as a local maximum, a local minimum or a saddle point.

Show that the function  f  does not have an absolute maximum over the given region.

Soln.  f x ( x , y ) = 2 y  x  2  and  f y ( x , y ) = 2 x  4 y  2 .

The condition  f x ( x , y ) = 0 = f y ( x , y )  implies  y = x  2 / 2  and  2 x = 16 x 4  with  x > 0 .

Then  ( x , y ) = ( 1 / 2 , 2 )  is the only critical point.

Now,  f x x ( x , y ) = 2 x  3f x y ( x , y ) = 2  and  f y y ( x , y ) = 8 y  3 .

At  ( x , y ) = ( 1 / 2 , 2 )  we have  f x x ( 1 / 2 , 2 ) = 16 > 0  and

( f x x  f y y  ( f x y ) 2 ) ( 1 / 2 , 2 ) = ( 16 ) ( 1 )  2 2 = 12 > 0 .

So, the function  f  has a local minimum at  ( 1 / 2 , 2 ) .

This function does not have an absolute maximum over the given region.

In effect, for any  M > 0 , take  x = y = M  1 . Then,  f ( M  1 , M  1 ) = 2 M  2 + 5 M > M .

5. a) Suppose that  f : R 3 R  and  g : R 3 R  are both differentiable at a point  ( a , b , c )  and that

f ( a , b , c )  is an extremum of the function  f  subject to the constraint  g ( x , y , z ) = k , where  k  is a constant.

Show that

and

.

Soln. If  f ( a , b , c )  is an extremum of the function  f  subject to the constraint  g ( x , y , z ) = k , then

f ( a , b , c ) = l g ( a , b , c ) , that is:

,    and

.

Using the first and third of the above equations to eliminate the parameter l , we obtain:

.

Similarly, using now the second and the third equations to eliminate the parameter l , we obtain:

.

b) Use part (a) above to find all extrema of  f ( x , y , z ) = 4 x y + x z + 2 y z ,

subject to the constraint  x y z = 27 .

Soln. In this case,  f ( x , y , z ) = ( 4 y + z , 4 x + 2 z , x + 2 y )  and  g ( x , y , z ) = ( y z , x z , x y ) .

Using part (a), we obtain the system of equations:

(i)  ( 4 y + z ) ( x y )  ( x + 2 y ) ( y z ) = 0  and   (ii)  ( 4 x + 2 z ) ( x y )  ( x + 2 y ) ( x z ) = 0 ,

which simplify to  (i)  2 y 2 ( 2 x  z ) = 0  and   (ii)  x 2 ( 4 y  z ) = 0 .

But  x 0 ,  y 0  and  z 0  because  x y z = 27 , then

(i)  implies that  z = 2 x ,  (ii)  implies that  z = 4 y  and  ( x , y , z ) = ( 3 , 3 / 2 , 6 ) .

The only extremum of the given function under the given constraint is  f ( 3 , 3 / 2 , 6 ) = 54 .

6. a) Compute the volume of the solid region enclosed between the

paraboloids  z = 3 x 2 + 3 y 2  and  z = 12  x 2  y 2 .

Soln. We use polar coordinates.

At the intersection of the two paraboloids,  3 r 2 = 12  r 2  and   .

Then,

=  = 18 p .

b) Evaluate  .

Hint for part (b): Remember that   .

Soln. The region of integration is   ,

which can also be described as   . Then,

.

7. a) Suppose that the function  f  is continuous over the interval  [ a , b ]  and the function  g  is continuous over the interval  [ c , d ] . Let  h ( x , y ) = f ( x ) g ( y )  .

Show, using Riemann sums, that the function  h  is integrable over the rectangle  R = [ a , b ] [ c , d ]

and that   .

Note: The text of this exercise requires the use of Riemann sums in the proof. So, you should not submit a proof in which the given property is viewed as an immediate consequence of Fubinis theorem.

Soln. Let  P  be a partition of the rectangle  R  and let  S m n  denote any of the Riemann sums of the function  h  over  R  with the partition  P , then

S m n =  ,

where   with  x 0 = a < x 1 < x 2 <  < x m = b  and   ,

with  y 0 = c < y 1 < y 2 <  < y n = d .

The functions  f  and  g  are both continuous over the corresponding intervals  [ a , b ]  and  [ c , d ] .

Therefore, each of the intervals  [ x i  1 , x i ]  contains numbers  t i  and  T i  such that

f ( t i ) ≤ f ( x ) ≤ f ( T i )  for all  x [ x i  1 , x i ]  and each of the intervals  [ y j  1 , y j ]

contains numbers  w j  and  W j  such that  g ( w j ) ≤ g ( y ) ≤ g ( W j )  for all  y [ y j  1 , y j ] .

Then,

,

where

are both Riemann sums of the function  f  over  [ a , b ]  and

are both Riemann sums of  g  over  [ c , d ] .

Recall that    is just the limit of  S m n  as

max ( x i  x i  1 ) 0  and  max ( y j  y j  1 ) 0 .

The functions  f  and  g  being continuous over  [ a , b ]  and  [ c , d ] , respectively, are integrable over these intervals. Then, as  max ( x i  x i  1 ) 0  and  max ( y j  y j  1 ) 0 :

,

also

,

and therefore

.

That is: the function  h  is integrable over  R  and

.

b) Use part (a) above to evaluate the integral  .

Soln. Notice that    therefore,

= .

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