Name                                    

MATH 107 PRACTICE MIDTERM II

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.     (15 Points)  If A, B, and C are points, v is the vector from A to B, w is the vector from B to C, and v x w  =  0 , then A, B and C are collinear.

Solution

True,  since 

        v x w  =  ||v|| ||w|| sin q

Hence

        sin q  =  0

So the angle is 0, thus they are collinear.

 

B.      (15 Points) If x  =  x(t),  y  =  y(t) are parametric equations of a line then dx/dt  is a constant.  

 

 

 

 

 

 

 

 

 

 

  Solution

False, consider the line

        x  =  t3,     y  =  t3

For this line

        dx/dt  =  3t2

which is not a constant.

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 2 (21 Points)

Consider the surface x2 + z2 - e2y  =  0 .  This surface is formed by revolving a generating curve about an axis.  Find an equation of this generating curve and state the axis of revolution.

  Solution

We write

        x2 + z2  =  e2y  =  (ey)2

So that the radius is 

        r(y)  =  ey 

We conclude that the axis of revolution is the y-axis.  There are two choices for the generating curve, either

        z  =  ey 

or

        x  =  ey    (y  =  lnx)

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 3  (21 Points) 

Use vectors to find the equation of the line that passes through the point (2,3,4) and is perpendicular to the plane 5x - 4y + 2z = 7.

Solution

This line is parallel to the normal vector to the plane.  We need to find the equation of the line that passes through the point (2,3,4) and is parallel to the vector <5,-4,2>.  The vector equation is 

        <x,y,z>  =  <2,3,4> + <5,-4,2>t

The parametric equations are 

        x  =  2 + 5t        y  =  3 - 4t        z  =  4 + 2t

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 4  (21 Points) 

Find all points (if any) of horizontal and vertical tangency.  Make sure to present your answer by listing the points not just the values of q.

        x  =  cos q        y  =  2sin(2q)

 Solution

We have 

              dy               dy/dt             4cos(2q)
                        =                    =                              
              dx                dx/dt              -sin q

The horizontal tangent lines occur when the derivative is zero.  We set the numerator to zero

        4cos(2q)  =  0

        2q  =  p/2 + kp

        q  =  p/4 + kp/2

        q  =  p/4 , 3p/4, 5p/4, 7p/4, 9p/4, 11p/4, ...

Plugging these values into the original equation gives

        (/2,2)    (-/2,-2)    (-/2,2)    (/2,-2)

To find the points of vertical tangency, we set the denominator equal to zero

        -sin q  =  0

        q  =  kp

        q  =  0, p, 2p, ...

Plugging these values into the original equation gives

        (1,0)    (-1,0)

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 5 (21 Points) 

Determine the area of the first quadrant loop of  r  =  3sin(2q)

Solution

The first loop occurs from the first time the radius equals 0 to the second time it equals 0.

         0  =  3sin(2q)

        q  =  0    and q  =  p/2

 Now integrate

       

 

 

 

 

 

 

 

 

 

 

PROBLEM 6  (21 Points) 

Show that the polar equation for the hyperbola

           x2          y2
                   -            =  1          
           a2          b2  

   is

                        -b2
            r2  =                      
                      1 - e2 cos2 q 

given that 

                              b2
            e2  =  1 +            
                               a2 

  Solution

We convert to polar coordinates using

        x  =  r cos q        y  =  r sin q

We have

        

           r2 cos2 q           r2 sin2 q
                               -                    =  1          
                  a2                   b2  

Now solve for r2

                cos2 q              sin2 q
       r2 (                       -                   ) =  1          
                     a2                    b2  

 

                b2cos2 q  - a2 sin2 q
       r2 (                                           ) =  1          
                            a2 b2  

 

                              a2 b2                                                  -b2
       r2  =                                                 =                                                
                      b2cos2 q  - a2 sin2 q                    sin2 q  -  (b2 / a2)cos2 q

 

                              - b2                                                  -b2
        =                                                     =                                                
               1 - cos2 q  -  (b2 / a2)cos2 q             1 - (1 + b2 / a2)cos2 q

                      -b2
          =                              
                1 - e2 cos2 q 

        

 

 

 

 

 

 

 

 

 

 

PROBLEM 7  (21 Points) 

Use vectors to determine if the triangle with vertices (1,0,1), (2,1,0), (0,0,4) is a right triangle.

Solution

If a triangle has a right angle, then the cosine of one of the angles equals 0, hence the dot product of two of the displacement vectors is zero.  We find displacement vectors by subtracting points.

        u  =  <2 - 1, 1 - 0, 0 - 1>  =  <2,1,-1>

        v  =  <0 - 1, 0 - 0, 4 - 1>  =  <-1,0,3>

        w  =  <0 - 2, 0 - 1, 4 - 0>  =  <-2,-1,4>

Now take the dot products

        u . v  =  -2 + 0 - 3  =  -5

        u . w  =  -4 - 1 - 4  =  -9

        v . w  =  2 + 0 + 12  =  14

Since none of the dot products are zero, the triangle is not a right triangle

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 8  (21 Points) 

Find parametric equations for the a particle moves along the line through (1,4,2) and (3,5,7) such that it is at the point (1,4,2) when t = 0 is at the point (3,5,7) when t = 2 and is speeding up as time progresses

Solution

If we just wanted a parametric equation of a line we would just use the formula

    v + wt

where v is the vector from the origin to (1,4,2) and w is the vector from (1,4,2) to (3,5,7).  However, we want to end when t  =  2, so we use

        v + wt/2

This almost works, but does not speed up as time progresses.  We instead use a t2 so that it is speeding up.  Since 22  =  4, we use 

        v + wt2/4

or

        <1,4,2> + <2,1,5>t2/4

        x  =  1 + t2/2    y  =  4 + t2/4    z  =  2 + 5t2/4