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MATH 107 PRACTICE MIDTERM III

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.

PROBLEM 1  Please answer the following true or false.  If false, explain why or provide a counter example.  If true, explain why.

A) Let f(x,y) be a function of two variables and P be a point.  If the value of the function tends towards 5 for every line segment that ends at P, then    

Solution

False, the limit has to be the same for all curves not just lines.  For example, if P is the origin (0,0) and 

                              xy² + x²y
            f(x,y)  =                              
                                x⁴ + y²

Then it is easy to check that the limit is zero along all line segments ending at the origin, but along y  =  x² the limit is 1/2.

B)  If  z  =  f(x - y) and f is a continuous function then  

  Solution

True.  Use the chain rule with u  =  x - y:

        z_x  =  f '(u)(u_x)  =  f '(u)(1)  =  f '(u)

        z_y  =  f '(u)(u_y)  =  f '(u)(-1)  =  -f '(u)

and

        f '(u) + (-f '(u))  =  0

PROBLEM 2 Find the dimensions of the rectangular box with the largest volume in the first octant such that one of the vertices is at the origin and the opposite vertex lies on the ellipsoid 

        x² + 2y² + 3z²  =  6

  Solution

The volume of such a box is given by 

        V  =  xyz

We use LaGrange Multipliers

        grad F  =  <2x,4y,6z>  =  l<yz,xz,xy>  =  l grad V

This gives us four equations 

        2x  =  lyz        4y  =  lxz        6z  =  lxy        x² + 2y² + 3z²  =  6  

Solving the first three equations for l gives

       l  =  2x/yz        l  =  4y/xz        l  =  6z/xy

Setting the equations equal to each other and multiplying by xyz gives

        2x²  =  4y²  =  6z² 

or

        x²  =  2y²  =  3z² 

Substituting into the fourth equation gives

        x² + x² + x²  =  6

        x²  =  2

        x  = 

Resubstituting gives

        2  =  2y²  

        y  =  1

and

        2  =  3z²

        z  =  (2/3)^0.5 

PROBLEM 3 Consider the function 

                  5
        z  =           
                  xy

A.     Use a calculator to sketch the level curves corresponding to
z = (-2,-1,0,0.5,1,2,5,10).   Draw them on a whole piece of paper.  

Solution

We solve for y

                  5
        y  =           
                  xz

  The contour map is pictured below

B.    Suppose that z represents the altitude function, and you are to travel on this surface above the unit circle in a counter-clockwise direction.  Discuss your travels.  

Solution

     Beginning at the point (1,0) we are infinitely high.  Then we slide steeply down, reaching a local minimum elevation at the point (/2, /2) of 5/2.  Then steeply rise heading towards an infinite elevation at the point (0,1).  Then suddenly appear at an infinitely low chasm and begin climbing to an elevation of -5/2 at the point (-/2, /2).  Then we again slide down to an infinitely low chasm at the point (-1,0)  We appear on the other side of this point at an infinitely high cliff and again slide down to the altitude of 5/2 at the point (-/2, -/2) .  Then we rise up again to the infinitely high cliff at the point (0,-1).  We instantly appear on the other side in another infinitely low chasm and rise up to a local high point at (/2, -/2) .  Then slide again to the infinitely low chasm at (1,0)

PROBLEM 4  Find the following limits if they exist. 

A.    

Solution

We consider the path to the origin from the positive x axis (y  =  0).  We get 

                       x⁴ 
        lim                    =    1
        x ^-> 0      x⁴ 

Now approach the origin via y  =  x.  We have

                       x⁴  + 2x⁴ + x⁴                 4x⁴
        lim                                        =                  =  2         
        x ^-> 0            x⁴ + x⁴                      2x⁴ 

Since the two limits approach two different values, we can conclude that the limit does not exist.

B.    

Solution

For this limit we use polar coordinates

                      r⁵cos⁵q - r⁴sin⁴q
        lim                                            = lim      r³cos⁵q - r²sin⁴q  =  0
        r ^-> 0                r²                       r ^-> 0  

       Therefore, the limit exists and is equal to 0

PROBLEM 5  Let  

        f(x,y)  =  x² + xy,     x(u,v)  = u²v,  and     y(u,v)  =  u - v

Find

A)      

Solution

      f_x  =  2x + y 

B)    Without calculating any derivatives, write the appropriate chain rule for 

Solution

        f_uu  =  (f_u)_u  =  (f_xx_u + f_yy_u)_u  =  f_xux_u + f_xx_uu + f_yuy_u + f_yy_uu 

        =  (f_xxx_u + f_xyy_u)x_u + f_xx_uu + (f_xyx_u + f_yyy_u)y_u + f_yy_uu 

C)   (7 Points)Find   

Solution

  Now calculate:

        x(1,2)  =  1²(2)  =  2                    y(1,2)  =  1 - 2  =  -1

        x_u  =  2uv  =  2(1)(2)  =  4            x_uu  =  2v  =  2(2)  =  4

        y_u  =  1            y_uu  =  0            f_xx  =  2            f_xy  =  1            f_yy  =  0

        f_x  =  2x + y  =  2(2) + (-1)  =  3            f_y  =  x  =  2

Plugging in, we get

        [(2)(4) + (1)(1)](4) + (3)(4) + [(1)(4) + (0)(1)](1) + (2)(0)

        =  (9)(4) + 12 + (4)(1)  =  36 + 12 + 4  =  52

PROBLEM 6  If R is the total resistance of three resistors, connected in parallel, with resistances R₁, R₂, and R₃, then 

           1                1              1               1
                    =               +              +                
            R              R₁             R₂             R₃

If the resistances are measured as R₁ = 25 ohms, R₂ = 40 ohms, and R₃ = 50 ohms, with possible errors of 0.5% in each case, use differentials to estimate the maximum error in the calculated value of R. 

Solution

We use the differential formula

        -R^-2 dr  =  -R₁^-2 dR₁ - R₂^-2 dR₂ - R₃^-2 dR₃ 

         R^-2 dr  =  .005(R₁^-2 R₁ + R₂^-2 R₂ + R₃^-2 R₃)

         R^-2 dr  =  .005(R₁^-1  + R₂^-1  + R₃^-1 )  =  .005R

Hence

        dr  =  .005R³ =  .005(1/25 + 1/40 + 1/50)³  =  .000425

PROBLEM 7  Find parametric equations for the tangent line of the curve of intersection of the paraboloid z  =  x² + y² and the ellipsoid  4x² + y² + z²  =  9 at the point (-1,1,2).

Solution

This line is on the tangent plane of both surfaces.  Thus is is perpendicular to both normal vectors.   We compute the two normal vectors.  The normal vector for the parabola is 

        <2x, 2y, -1>  =  <-2, 2, -1>

The normal vector for the ellipsoid is

        <8x, 2y, 2z>  =  <-8, 2, 4>

Now cross the two vectors to find a vector perpendicular to both normal vectors.

        10i + 16j + 12k

Now use the formula for parametric equations of a line given a point and a parallel vector

        x(t)  =  -1 + 10t        y(t)  =  1 + 16t        z(t)  =  2 + 12t