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Show your work and give reasons for you answers on all of the following problems.
No calculators, books or notes are allowed.

1. Find, describe in words and sketch in \mathbbR² the domain of each of the following functions:

   1. f(x,y) = [²Ö(5x -6y)]
   2. f(x,y) = 1/[ [¹⁰Ö(5y -2x)]]
   3. f(x,y) = [⁹Ö(3x -5y)]
   4. f(x,y) = 1/[ [¹¹Ö(2x -5y)]]
   5. g(x,y) = ln(5x²+3y² -2)
   6. h(x,y) = 2/[ ln(5-6x²-5y²)]

   Answer: In these four functions we need to avoid one or two of the following:

   • Taking an even root of a negative number.
   • Dividing by zero.
   • Taking the log of zero or a negative number.
   1. In this case we have to avoid taking an even root of a negative number. So we need to have (5x-6y) ³ 0. Thus, the domain is

      D = {(x,y)|y £ (5/6)x }

      The domain is all points in the plane lying below or on the line y = (5/6)x.
   2. In this case we have to avoid dividing by zero as well as taking an even root of a negative number. So we need to have (5y-2x) > 0. Thus, the domain is

      D = {(x,y)|y > (2/5)x }

      The domain is all points in the plane lying above but not on the line y = (2/5)x.
   3. The root in this function is odd. We can take the root of a negative number. Moreover, we are not in danger of dividing by zero. Therefore, the domain is all of \mathbbR²,i.e. any (x,y) works.

   4. In this function the root is odd . We can take the root of a negative number. However, we do not want to divide by a negative number. Therefore, we need to have (2x-5y) ¹ 0. Thus, the domain is

      D = {(x,y)|y ¹ (2/5)x }

      The domain is all points in the plane except those on the line y = (2/5)x.
   5. For this function, we have to itavoid taking the ln itof zero or a negative number. So, we want (5x²+3y² -2) > 0. Therefore, the domain of g(x,y) is

      D = {(x,y)|5x2+3y2 > 2}

      Notice that the domain consists of all the points of the plane out side, but not on the ellipse (5x²+3y² = 2).
   6. Notice that the main difference between the function h(x,y) and and the function h(x,y) of the previous part is that we divide by a ln. So, we need to avoid dividing by zero. But lnt = 0 if and only if t = 1. Thus, we want (5-6x²-5y²) > 0 as well as (5-6x²-5y²) ¹ 1. Therefore, the domain of h(x,y) is

      D = {(x,y)|6x2+5y2 < 5,  and  6x2+5y2 ¹ 4 }

      The domain consists of all the points of the plane inside, but not on the ellipse 6x²+5y² = 5 nor on the ellipse 6x²+5y² = 4 . Notice that the second ellipse lie inside the first one.

2. Find, describe in words and sketch in \mathbbR² the domain of each of the following functions:

   1. f(x,y) = [[⁴Ö(3x -4y)]]/ [[²Ö(5y -4x)]]

   2. f(x,y) = [[⁶Ö(5x -3y)]]/ [ln(4x²+5y² -3)]

   Answer:

   1. We want to avoid taking an even root of a negative number and dividing by zero. Therefore, we need (3x -4y) ³ 0, and (5y -4x) > 0. Thus, the domain of f(x,y) is

      D = {(x,y)|y £ (3/4)x,   and    y > (4/5)x }

      The domain of f(x,y) consists of all points in the plane that lie below or on the line y = (3/4)x, but above and not on the line y = (4/5)x.
   2. For this function we should avoid the following:

      • Taking an even root of a negative number. So, we need (5x -3y ³ 0).
      • Evaluating the ln of zero or a negative number. So, we need (4x²+5y² -3 > 0).
      • Dividing by zero. Notice that ln1 = 0. So, in order not to divide by zero, we should have (4x²+5y² -3 ¹ 1).

      It follows that the domain of f(x,y) is given by

      {(x,y)|y £ (5/3)x,    4x2+5y2 > 3,  and   4x2+5y2 ¹ 4}

   The domain of this function consists of all points in the plane that lie
   outside and not on the ellipse 4x²+5y² = 3, and lie also above the line y = (5/3)x,
   but not on the ellipse 4x²+5y² = 4.

3. Find ¶f/¶x, ¶f/¶y and ¶f/¶z for

   f(x,y,z) = 4x4y4z5- 4x5z5+ 5y6+5z6 + 3y6z3

   Answer:

   fx(x,y,z) = 16x3y4z5- 20x4z5

   fy(x,y,z) = 16x4y3z5+ 30y5+18y5z3

   fz(x,y,z) = 20x4y4z4- 20x5z4+ 30z5 + 9y6z2

4. Find f_xz and f_zy for the following function:

   f(x,y,z) = x5y19/6z6

   Answer: To obtain f_xz(x,y,z) we start by computing f_x(x,y,z). To do that, we differentiate f(x,y,z) with respect to x treating y and z as constants. Thus

   fx(x,y,z) = 5x4y19/6z6

   To obtain f_xz(x,y,z) we differentiate f_x(x,y,z) with respect to z treating x and y as constants. We obtain

   fxz(x,y,z) = 30x4y19/6z5

   To obtain f_zy(x,y,z) we start by computing f_z(x,y,z). To do that, we differentiate f(x,y,z) with respect to z treating x and y as constants. Thus

   fz(x,y,z) = 6x5y19/6z5

   To obtain f_zy(x,y,z) we differentiate f_z(x,y,z) with respect to y treating x and z as constants. We obtain

   fzy(x,y,z) = (114/6)x5y13/6z5

5. Find f_xy and f_zx for the following function:

   f(x,y,z) = z3sin(x5y5)

   Answer: To obtain f_xy(x,y,z) we start by computing f_x(x,y,z). To do that, we differentiate f(x,y,z) with respect to x treating y and z as constants. This way we obtain

   fx(x,y,z) = 5x4y5z3cos(x5y5)

   To obtain f_xy(x,y,z) we differentiate f_x(x,y,z) with respect to y treating x and z as constants. We also use the multiplication rule. Thus

   fxy(x,y,z) = 25x4y4z3cos(x5y5) -25x4y9z3sin(x5y5)

   To compute f_zx(x,y,z) notice that for our function, f_zx(x,y,z) = f_xz(x,y,z). We have already computed f_x(x,y,z). So, we differentiate f_x(x,y,z) with respect to z treating x and y as constants. Thus

   fzx(x,y,z) = fz(x,y,z) = 15x4y5z2cos(x5y5)

6. Use differentials to find the linear approximation of f(x,y) at r_o = < x_o,y_o > . Then use the linear approximation you obtained to approximate f(x₁,y₁).
   

   f(x,y,z) = x5y6z7,        ro = < 1,1,1 > ,     r1 = < 1.02,.97. 96 > .

   Answer: The differential of a function f is given by

   df = Ñf ·dr

   Now we can see that the variational equation of f at r_o is

   Df » Ñf(ro)·Dr

   where

   Df » f(r) - f(ro) ,

   Ñf(ro) = < fx(ro), fy(ro), fz(ro) > ,

   Dr = r- ro = < x-xo,y-yo, z-zo >

   It follows that the linear approximation of f(x,y,z) near (x_o,y_o,z_o) is

   f(r) » f(ro)+ Ñf(ro)·Dr

   If f is a function of only two variables, we ignore the third component of each vector.
   In this case r = < x,y,z > , r_o = < 1,1,1 > and

   Ñf(x,y,z) = < 5x4y6z7,6x5y5z7,7x5y6z6 >

   Evaluating at r_o = < 1,1,1 > , we obtain

   Ñf(1,1,1) = < 5,6,7 >

   Also notice that f(1,1,1) = 1. Thus,

   f(x,y,z) » 1+ < 5,6,7 > · < x-1,y-1,z-1 >

   = 1 +5(x-1)+ 6(y-1)+7(z-1) = -17+5x+6y+7z

   Now, f(1.02,.97,.96) is approximately

   f(1.02,.97,.96) » 1 + < 5,6,7 > · < 1.02-1,.97-1, .96 - 1 >

   = 1 + < 5,6,7 > · < 2/100 , -3/100, -4/100 > = 1 +[(-36)/100] = (64)/100

7. Find an equation for the tangent plane to the following function at the given point:

   f(x,y) = 2x4 + 2x5y4 -3y4,        (2,2,1056)

   Answer: An equation of the tangent plane to the surface z = f(x,y) at a point (x_o, y_o, z_o) ( where z_o = f(x_o,y_o)) is given by

   Dz = Ñf(ro) ·Dr

   where Dz = z-z_o,Dr = r- r_o = < x-x_o, y-y_o > and Ñf(r_o) = < f_x(r_o), f_y(r_o) > . In our problem

   Ñf(r) = < 8x3+10x4y4, 8x5y3-12y3 >

   Substituting r = r0 = < 2,2 > , we obtain

   Ñf(ro ) = < 2624,1952 >

   Thus, an equation of the plane is

   z - (1056) = 2624(x- (2))+ 1952(y- (2))

8. Use differentials to approximate the number c = 4.03Ö{9.99-(1.04)³}.

   Answer: We consider the function f(x,y,z) = xÖ{y-z³} near r_o = < 4,10,1 > . We chose this point because we have the exact value f(4,10,1) = 12. Then, we try to approximate f(4.03,9.99,1.04). The differential of a function f(x,y,z) is

   df = fxdx + fydy + fzdz

   From which we obtain the variational equation

   Df » Ñf(ro)·Dr

   where

   Df » f(r) - f(ro) ,

   Ñf(ro) = < fx(ro), fy(ro), fz(ro) > ,

   Dr = r- ro = < x-xo,y-yo, z-zo >

   It follows that the linear approximation of f(x,y,y) near (x_o,y_o,z_o) is

   f(r) » f(ro)+ Ñf(ro)·Dr

   Now,

   Ñf(x,y,z) = (   æ Ö y-z3   )i+ (x/ (2   æ Ö y-z3   ))j -(3z/(2   æ Ö y-z3   )) k

   f(4,10,1) = 12,

   Ñf(4,10,1) = 3i+ (2/ 3)j-(1/2) k,

   and

   Dr = 0.03i-0.01j+0.04k

   It follows that

   f(4.03,9.99,1.04) » 12 + [(3i+ (2/ 3)j-(1/2) k) · (0.03i-0.01j+0.04k)]

   Thus,

   c = f(4.03,9.99,1.04) » 12 + [38/600]

9. Review problems 25-32, on page 793.