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  Ñ ( 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 = ± 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 ) Î R ² ½  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 ^{– 3} ,  f _{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

Ñ f ( a , b , c ) = l Ñ 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,  Ñ f ( x , y , z ) = ( 4 y + z , 4 x + 2 z , x + 2 y )  and  Ñ 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 ¹ 0 ,  y ¹ 0  and  z ¹ 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 Î [ 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 Î [ 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,

= .