UNIVERSITY OF TORONTO

DEPARTMENT OF MATHEMATICS

MAT 237 Y - MULTIVARIABLE CALCULUS

FALL-WINTER 2005-06

ASSIGNMENT #3. SOLUTIONS

 

1.  a)  Let  u ( x 1 , x 2 , … , x n ) = , where  a  is a constant, and  n  is given.

Find all the values of the constant  a , if any, for which the function  u  satisfies the partial differential equation   .

Solution:

Let  w = , then   ,   , and

 .

Therefore,   .

 

And this is 0  for any choice of  ( x 1 , x 2 , … , x n )  only when  a = 0  or  a = 1 .

(Except in the “trivial” case  n = 1 , where  a = 1 / 2  is also a solution.)

In effect, if  a ¹ 0  and  a ¹ 1, taking  ( x 1 , x 2 , … , x n ) = ( 1 , 0 , … , 0 ) , we have

  and  a = 1 / 2  is required.


But taking now  ( x 1 , x 2 , … , x n ) = ( 1 , 1 , 0 , … , 0 ) , we obtain,

 , which is  ¹ 0  when  a = 1 / 2 .

 

b)    Let  F : R 2 ® R  be a function whose partial derivatives are continuous and suppose that the equation  F ( r cos q , z + r sin q  ) = 0  implicitly defines  z  as a function of the variables  r  and  q  . Find all the values of the constant  b , if any, for which  z  satisfies the partial differential equation   .

Solution:

  and   .

 

Then,   .  Therefore  b = – 1 .

 

2. a)  Suppose that  E  is the ellipsoid   , where  a > 0 ,  b > 0  and  a ¹ b .

Let  d ( x , y , z )  denote the distance from the point  ( 0 , 0 , 0 )  to the line that is normal to the ellipsoid  E  at the point  ( x , y , z ) . Find the maximum value of the function  d  and determine all points  ( x , y , z )  where this maximum value is reached.

Solution:

The given ellipsoid is a surface of revolution about the  z-axis  and is also symmetric with respect to the  xy-plane . It means that the desired maximum is reached at a point  ( x , y , z )  of the given ellipsoid  if and only if it is also reached at any point  ( x cos t , y sin t  , ± z ) , where

0 £ t £ 2 p . So, it will be enough to find the unique solution to the problem for the case

y = 0 , x > 0 , z > 0  and then generalize to the whole family of solutions.

The resulting two-dimensional curve  a 2 x 2 + b 2 z 2 = 1 ,  x > 0 ,  z > 0  can be parametrized as

 . Then, the tangent line to the curve at any point  ( x , z )  follows the direction of the vector   . The distance  d  from  ( 0 , 0 )  to

 

the line  L  that passes through  ( x , z )  and is perpendicular to the curve is then given by

d = .

Now,  d ¢ =   and  d ¢ = 0  only when  a cos 2 q  – b sin 2 q  = 0 .

That is,  tan q =  , therefore  x =   and  z =  .

The maximum value for  d  is   . This maximum value is reached at any of the points

( x , y , z ) =  , where  0 £ t £ 2 p .

 

b)  A small insect is seating at the point  ( 4 , 3 , 1 )  inside a cage in the shape of the hollow rectangular region  0 £ x £ 5 ,  0 £ y £ 5 ,  1 £ z £ 4 . Suppose that the temperature at any point  ( x , y , z )  in the cage is given by the function  T ( x , y , z ) = 30 + x 2 + 2 y 2 – z 2 .

In order to cool off as soon as possible, the insect flies off from its position at the bottom of the cage in such a way that it experiences the maximum possible rate of cooling at each of the points of its trajectory. What are the coordinates of the point at which the insect reaches the top of the cage?

Solution:

Let  r ( t ) = ( x ( t ) , y ( t ) , z ( t ) )  denote the position of the insect at time  t , where  t ³ 0  and  r ( 0 ) = ( 4 , 3 , 1 ) . At any point  ( x , y , z ) , the vector  – Ñ T ( x , y , z )  points in the direction of maximum decrease of the temperature. So, the condition required for the insect’s trajectory is  r ¢ = – l Ñ T  ,  l > 0 .

Taking  l = 1 , we obtain  x ¢ ( t ) = – 2 x ( t ) ,  y ¢ ( t ) = – 4 y ( t )  and  z ¢ ( t ) = 2 z ( t ) .

Then,  x ( t ) = 4 e – 2 t  ,  y ( t ) = 3 e – 4 t  and  z ( t ) = e 2 t  . For  z ( t ) = 4 ,  t = ln 2 .

The insect reaches the top of the cage at the point  ( 1 , 3 / 16 , 4 ) .

 

3.  a)  Question 2-5 from Spivak:

Let  f : R 2 ® R  be defined by  .

 

 

Show that  f  is not differentiable at  ( 0 , 0 ) .

Solution:

Notice that    and   . So, the function  f  is differentiable at  ( 0 , 0 )  if and only if   . But taking  h = k >0 , we obtain    and the limit as  ( h , k ) ® ( 0 , 0 )  can not be 0 .

 

b)  Define  f , g : R 3 ® R  by   and

 , where  u , v , w : R 3 ® R  are continuous functions, with continuous partial derivatives, and such that for all  ( x , y , z )

u x ( x , y , z ) + v y ( x , y , z ) + w z ( x , y , z ) = 0 .

Show that  fg  and  w  satisfy the partial differential equation  f x + g y = w .

Solution:

Notice that  f x ( x , y , z ) =   and

g y ( x , y , z ) =  . Then,

f x ( x , y , z ) + g y ( x , y , z ) =

=  = = w ( x , y , z )    

 

4. a)  Question 2-24 from Spivak:

Define  f : R 2 ® R  by   .

 

 

a1) Show that  f y ( x , 0 ) = x  for all  x  and  f x ( 0 , y ) = – y  for all  y .

Solution:

f y ( x , 0 ) =   .

 

 

So,  f y ( x , 0 ) = x  for all  x . Similarly,

f x ( 0 , y ) =   .

 

 

So,  f x ( 0 , y ) = – y  for all  y .

 

a2) Show that  f x y ( 0 , 0 ) ¹ f y x ( 0 , 0 ) .

Solution:

f x y ( 0 , 0 ) = and similarly  f y x ( 0 , 0 ) = 1 .

 

b)    Question 2-32 from Spivak:

b1) Let  f : R ® R  be defined by   .

 

 

Show that  f  is differentiable at  0  but  f ¢  is not continuous at  0 .

Solution:

f ¢ ( 0 ) = , because  – 1 £  £ 1 .

Now, if  x ¹ 0 ,  f ¢ ( x ) =  . But   does not exist.

So,  f ¢  is not continuous at  0 .

 

b2) Let  f : R 2 ® R  be defined by  .

 

 

Show that  f  is differentiable at  ( 0 , 0 )  but  f x  and  f y  are both discontinuous at  ( 0 , 0 ) .

Solution:

As in (b1), it is easy to conclude that  f x ( 0 , 0 ) = 0  and  f x ( x , 0 ) =  if  x > 0 .

Similarly,  f y ( 0 , 0 ) = 0  and  f y ( 0 , y ) =  if  y > 0 . Therefore,  f x  and  f y  are both discontinuous at  ( 0 , 0 ) .

Now, with  f x ( 0 , 0 ) = 0 = f y ( 0 , 0 ) , in order to show that  f  is differentiable at  ( 0 , 0 )  we only have to prove that   .

But   .    

 

 

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