Chapter 2

Practice Test

Instructions.  Show your work and/or explain your answers.

1. Find the domain of the function

   f( x,y) = Öy+   x2-1

   Is the domain open, closed, or neither? Bounded or unbounded? Connnected or not connected? Solution: dom( f) = { ( x,y) |  y ³ 0  and  x² ³ 1} = { ( x,y) |  x £ -1,  y ³ 0} È{ ( x,y) |  x ³ 1,  y ³ 0} . Since domain contains its boundaries y = 0, x = -1, and x = 1, it is closed. Since x and y can approach infinity within the domain, it is unbounded. Since no path exists from { ( x,y) |  x £ -1,  y ³ 0} to { ( x,y) |  x ³ 1,  y ³ 0} that stays in the domain, the domain is not connected.
2. Show the following limit does not exist by showing that different paths through the origin lead to different limits:

   lim ( x,y) ® ( 0,0)    (x+y) 2 x2-y2

   Solution: Along y = 0, the limit is 1. Along x = 0, the limit is -1.
3. Does the following limit exist?

   lim ( x,y) ® ( 0,0)    (x+y) 2 x2+y2

   Solution: No. Along x = 0 and y = 0, the limit is 1. However, along y = x the limit is 2.
4. Find the linearization of f( x,y) = x+e^xy at (1,0) Solution: f_x( x,y) = 1+ye^xy, f_y(x,y) = xe^xy. Thus, L( x,y) = 2+1( x-1)+1( y-1) .
5. Find the second order derivatives of

   f( x,y) = x2+exy

   Solution: f_x = 2x+ye^xy, f_y = xe^xy, f_xx = 2+y²e^xy, f_yy = x²e^xy, f_xy = e^xy+xye^xy
6. Find the separated solution of

   ¶u ¶t +  ¶u ¶x = u

   Solution: u( x,t) = f( x) T(t) implies that fT^¢+f^¢T = fT, so that

   T¢ T +  f¢ f = 1,     T¢( t) T( t) =  f¢( x) f( x) +1

   Thus, T^¢( t) = -kT( t) and f^¢( x) = ( -1-k) f( x) , so that the separated solution is

   f( x) T( t) = Pe-kte( -k-1) x

7. Find ¶_uz when z = x²+y³ and x = u²+uv, y = u³v Solution: The chain rule implies that

   ¶z ¶u = 2x  ¶x ¶u +3y2  ¶y ¶u = 2( u2+uv) ( 2u+v) +3( u3v)2( 3u2v) = 4u3+6u2v+2uv2+9u8v3

8. Prove that the derivative of a sum is the sum of the derivatives by applying the chain rule for 2 variables to

   w = x+y

   where x = f(t) and y = g(t) . Solution: The chain rule implies that

   dw dt =  ¶w ¶x  dx dt +  ¶w ¶y  dy dt

   However, w_x = 1 and w_y = 1, and also w = f( t) +g(t) , so that

   d dt ( f( t) +g( t) ) =  dw dt =  dx dt +  dy dt = f¢( t) +g¢(t)

   thus completing the proof.
9. Find the gradient of the function g( x,y) = x²+y², and then show that it is normal to the curve

   x2+y2 = 25

   at the point ( 3,4) . Solution: The curve x²+y² = 25 is a circle. Ñg = á 2x,2y ñ , so Ñg( 3,4) = á 6,8 ñ . However, Ñg( 3,4) = á 6,8 ñ is parallel to the radius á3,4 ñ and thus must be perpendicular to the tangent line.
10. In what direction is the function f( x,y) = x²+y³ decreasing the fastest at the point ( 1,3) ? Solution: The gradient of f is Ñf = á2x,3y² ñ , so that Ñf( 1,3) = á2,27 ñ . This is the direction in which f is increasing the fastest. The direction f is decreasing the fastest is thus

    -Ñf( 1,3) = á -2,-27 ñ

11. Find the extrema and saddle points of f( x,y) = x²+3xy+2y²-4x-5y. Solution: f_x = 2x+3y-4, f_y = 3x+4y-5. Thus,

    2x+3y = 4 3x+4y = 5

    Thus, the critical point is ( -1,2) . Moreover, f_xx = 2, f_xy = 3, and f_yy = 4, so that

    D = fxxfyy-( fxy) 2 = 4·2-32 = -1

    and thus there is a critical point at the saddle.
12. Find the extrema and saddle points of f( x,y) = 4x³-6x²y+3y² Solution: f_x = 12x²-12xy, f_y = -6x²+6y, Thus, 12x² = 12xy and 6x² = 6y. Since x² = xy, x = 0 or x = y. If x = 0, then x² = y implies that y = 0, and the critical point is (0,0) . If x = y, then x² = y implies that x² = x or x = 1,0. Thus, the critical points are ( 0,0) and ( 1,1) . However,

    D = ( 24x-12y) 6-( 12x) 2 = 144x-72y-144x2

    Thus, D( 0,0) = 0 and there is no info, and D( 1,1) = 144-72-144 = -72 < 0, so there is a saddle at ( 1,1) .
13. Find the point(s) on the curve xy = 1 that are closest to the origin. Solution: That is, minimize f( x,y) = x²+y² subject to xy = 1. If g( x,y) = xy, then Ñf = á2x,2y ñ and Ñg = á y,x ñ , so that

    2x = ly,    2y = lx

    Since neither x,y can be zero, we have l = 2x / y, so that

    2y =  2x y x      y2 = x2      y = x,y = -x

    If y = -x, then xy = -x² = 1 which has no solution. If y = x, then xy = x² = 1, so x = 1,-1 and the critical points are ( 1,1) and ( -1,-1) . In both cases f( 1,1) = f(-1,-1) = 2. Moreover, f( 2,1/2) = 4.25, so we must have minima at these points.
14. Use Lagrange Multipliers to solve the following: John wants to build a 500 ft² deck behind his house. His house is 50 feet long, and correspondingly, he wants the deck to be between 5 and 50 feet long. What dimensions of the deck will minimize the lengths of the rail around the 3 exposed sides of the deck? Solution: Let x be the length and y be the width of the deck. Then xy = 500. Let L denote the length of the rail. Then

    L = x+2y

    Thus, we must minimize L = x+2y subject to xy = 500 for x in [5,50] . If g( x,y) = xy, then ÑL = á1,2 ñ and Ñg = á y,x ñ , so that

    1 = ly,        2 = lx

    Since l = 1/y, substitution leads to

    2 =  1 y   x        and        2y = x

    Substituting x = 2y into the constraint yields 2y² = 500, or

    y2 = 250,        y = Ö250 =  5Ö10

    If x = 2y and y = 5Ö10, then x = 10Ö10, so that (10Ö10, 5Ö10 ) is a critical point. At that point

    L = 10Ö10  +2·5Ö10  = 20Ö10 = 63.25¢

    When x = 5, then y = 100 and at ( 5,100) the length is

    L = 5+2·100 = 205

    When x = 50, then y = 10 and at ( 50,10) , the length is

    L = 50+2·10 = 70¢

    Thus, the shortest rail occurs when x = 10Ö10 = 31.622^¢ and y = 5Ö10 =  15.811^¢
15. ** Heating of a 2 dimensional surface (such as in a sheet of metal) is modeled by the 2 dimensional heat equation

    ¶u ¶t = k2  ¶2u ¶x2 +k2  ¶2u ¶y2

    where u( x,y,t) is a function of 3 variables and k is a constant. What is the separable solution of the 2 dimensional heat equation (hint: involves 2 separation constants)? Solution: Let u( x,y,t) = f( x) r( y) T( t) . Then

    f( x) r( y) T' ( t) = k2f'' ( x) r( y) T(t) +k2f( x) r'' ( y)T( t)

    Dividing through by f( x) r( y) T(t) leads to

    f( x) r( y) T' ( t) f( x) r( y) T( t) =  k2f'' (x) r y) T(t) f( x) r( y) T( t) +  k2f(x) r'' (y) T(t) f( x) r( y) T( t)

    which simplifies to

    T' ( t) T( t) =  k2f'' ( x) f( x) +  k2r'' ( y) r( y)

    Both sides of the equation must be constant, so that

    T' ( t) T( t) = -w2       and     k2f'' ( x) f( x) +  k2r'' ( y) r( y) = -w2

    The second equation can now be written as

    k2f'' ( x) f( x) = -w2-  k2r'' ( y) r(y)

    thus implying both sides of this equation are constant (we let -l² denote this constant).

    k2f'' ( x) f( x) = -l2        and    -w2-  k2r'' ( y) r( y) = -l2

    The last equation becomes

    w2r( y) +k2r'' ( y) = l2r( y)         or    r'' ( y) +  w2-l2 k2 r( y) =

    If w² > l², then the equation is a harmonic oscillator and has a solution of

    r( y) = A1cos æ ç è y w2-l2 k ö ÷ ø +B1sin æ ç è y w2-l2 k ö ÷ ø

    If w² = l², then r'' ( y) = 0 and r( y) = A₁+B₁y.  If w² < l², then

    r( y) = A1cosh æ ç è y w2-l2 k ö ÷ ø +B1sinh æ ç è y w2-l2 k ö ÷ ø

    Moreover,

    k2f'' ( x) f( x) = -l2    implies  that  f'' ( x) +  l2 k2 f( x) = 0

    implies that

    f( x) = r( y) = A2cos æ è  l k x ö ø +B2sin æ è  l k x ö ø

    and finally, T' ( t) = -w²T( t) implies that T( t) = Pe^-w2t. Thus, for w² > l², the separated solution is

    u( x,y,t) = Pe-w2t æ è A2cos æ è  l k x ö ø +B2sin æ è  l k x ö ø ö ø æ ç è A1cos æ ç è y w2-l2 k ö ÷ ø +B1sin æ ç è y w2-l2 k ö ÷ ø ö ÷ ø

    and similar for the other two cases.