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MATH 107 PRACTICE FINAL

 

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.       (11 Points)  Suppose that    is defined at x = 3 then f '(x) is also defined at x = 3.

Solution

False,  3 could be an endpoint of the interval of convergence.  The derivative may not preserve the endpoints.  An example of this is where

                            1
        an  =                
                    3n n2
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.       (12 Points)  Let x = x(t), y = y(t) be parametric equations for a differentiable curve such that x''(-1) = y''(-1) = 3 , then the curve is concave up at the point (x(-1),y(-1)) .

Solution
False, the second derivative is not the y''/x'', but rather

                d2y             (y'/x')'
                     =                      
          dx2                x'

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C.       (12 Points)  If f(x,y) is a differentiable function at the point P, then Dgradf(P)(P)  the directional derivative in the direction of gradf(P) cannot be negative.

  Solution

True, since 

        Dgradf(P)(P) =  gradf(P) . gradf(P) 

which is the square of the magnitude of the gradient, hence positive.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 2  Test the following series for convergence.   If applicable, determine if the series converges absolutely or conditionally.

 

A.      (17 Points) 

Solution

We use the limit comparison test with bn  =  1/n.  

  Sbn diverges since it is the harmonic series.  Now compute

       

since the limit is finite, by the limit comparison test the original series diverges also

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.       (18 Points) 

  Solution

Use the ratio test to get

       

By the ratio test, the series converges absolutely.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 3  (35 Points)  Determine the Maclaurin series for the function

                          3
        f(x)  =                      
                      2x - 4        

Solution

 We write

                          3                      3         1
        f(x)  =                      =    -                                            
                       2x - 4                  4     1 - x/2

 

          

     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

      

PROBLEM 4 (35 Points)  Find the interval of convergence of

       

  Solution

We use the ratio test first

       

        =  |5 - 2x| < 1

Now solve the absolute value inequality 

        5 - 2x  =  1        or        5 - 2x  =  -1

        x  =  2        or        x  =  3

Next test the endpoints

For x  =  2 the series becomes

       

Which diverges by the limit test since limit of the terms goes to infinity not zero.

Similarly at x  =  3, the series becomes

       

Again by the limit test the series diverges.  We can conclude that the interval of convergence is 

        (2,3)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 5    (35 Points)  Find the length of one of the petals of the graph of the curve

        r  =  4 sin(3q)
(You may use a calculator to perform the integration.)

  Solution

We use the formula

       

We have 

        r2 + r'2  =  16sin2(3q) + 144cos2(3q)

Now to find a and b, the petals all begin at the origin.  We find

        4sin(3q)  =  0

        3q  =  0, p,...

        q  =  0, p/3,... 

Now calculate the integral

       

        =  8.91

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 6 (35 Points)  Find a unit vector that is perpendicular to the two vectors

        3i + 4j - k        and        2i + j + 2k

Solution

We find the cross product of the two vectors

       

        =  9i - 8j - 5k

Now find the magnitude

       

Divide to get

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 7  (35 Points)   Consider the paraboloid  z  =  x2 + y2 

  1. Convert this equation to cylindrical coordinates.  

    Solution
    since 

                 r2  =  x2 + y2 

    we get

            z  =  r2

     

     

     

     

     

     

     

     

     

     

     

     

     

     

  2. Convert this equation to spherical coordinates.

  Solution

Adding z2 to both sides, we get

        z  + z2  =  x2 + y2 + z2 

         r cosf + r2 cos2f  =  r2

Now divide by r to get

        cosf + r cos2f  =  r

Solve for r to get

                     cosf
        r  =                        =  cotf cscf
                  1  - cos2f

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 8  (35 Points)  Let f(x,y)  =  x cos(y2) .  Use the chain rule to determine the angular rate of change of f given that x  =  r cos q,    y  =  r sin q .

  Solution

We want

        fq(r,q)

We have 

        fq  =  fxxq + fyyq  

        =  cos(y2)(-r sin q) - 2xy sin(y2)(r cos q)

        =  -r cos(r2 sin2q) sin q - 2r3 cos2q sin q sin(r2 sin2q)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 9  (35 Points)  Find the equation of the tangent plane to the surface

        x2
                + y2 - z2  =  1
        4
  

at the point (2,1,1).

  Solution

First find the gradient vector which is the normal vector to the plane by taking partial derivatives

        grad(F)  =  <x/2, 2y, -2z>  =  <1/2,2,-2>

Now use the formula for a plane given a normal vector and a point

        <1/2,2,-2> . <x - 2, y - 1, z - 1>  =  0

        1/2x - 1 + 2y - 2 - 2z + 2  =  0

       x - 2 + 4y - 4 - 4z + 2  =  0

        x + 4y - 4z  =  4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 10  (35 Points)  The temperature of a room in your factory can be modeled by the equation f(x,y)  =  exy .  There is a round table of radius  centered at the origin.  Use the method of Lagrange multipliers to determine the hottest points on the table.  

Solution

  We want to maximize 

        f(x,y)  =  exy 

under the constraint 

        g(x,y)  =  x2 + y2  -  8  =  0

Lagrange multipliers tells us that 

        gradf  =  lgradg

        <yexy, xexy>  =  l<2x, 2y>

This gives us the three equations

        yexy  =  2lx        xexy  =  2ly        x2 + y2  =  8

Solving the first two equation for l and setting them equal to each other gives

        yexy              xexy 
                   =                  
        2x                  2y

or

        y/x  =  x/y

so that 

        x2  =  y2 

Plugging this back into the third equation gives

        y2 + y2  =  8

or 

        y  =  2

and 

        x  =  2

From

        f(x,y)  =  exy 

we immediately see that the maximum occurs when the signs are the same, that is the hottest points are at

        (2,2) and (-2,-2)