MATH 106 MIDTERM 3

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  

You are constructing the newest resort at the bottom of Lake Tahoe (which is 1654 feet deep).  Your resort will feature a giant vertical circular window of radius 50 feet such that the bottom of the window lies on the bottom of the lake.  How much total fluid force will the window have to withstand?  (The density of fresh water is 62.4 pounds per cubic foot).       

Solution

We first calculate the area

        Area  =  2xDy  =  2Dy

Since the depth of the lake is 1654 feet and the circle has a radius of 50 feet, the center of the circle is at a depth of 1604 feet.  Hence setting the origin at the center of the circle we get 

        Depth  =  1604 - y

Integrating the pressure times the area times the depth gives

       
(The first integral is the area of a semicircle and the second integral is 0 since it is an odd function)

        =  786,101,880 pounds

PROBLEM 2 (17 Points)

A semicircular piece of glass of constant density and radius 6 inches is to be attached to a string so that the string balances the glass perfectly.  Where should the string be attached?

Solution

We have 
        f ²(x)  =  36 - x²

        M  =  1/2 pr²  =  18p 

Dividing M_x by M gives 

        x  =  144/18p  =  8/p  

We put the pin horizontally at the center of the circle and 8/p units above the base.

PROBLEM 3  (15 Points Each)

Integrate the following:

1.        
      
   Solution
   We use partial fractions:
   
           2x + 1                 A          Bx + C
                              =             +                        
           x(x² + 1)              x            x² + 1
   
           A(x² + 1) + (Bx + C)x  =  2x + 1
   
   Let  x  =  0 then
           
           A  =  1
   
   So that 
   
           x² + 1 + Bx² + Cx  =  2x + 1
   
   Hence Cx  =  2x so that C  =  2 and     
   
           x² + Bx²  =  0 
   
           B  =  -1
   
   Now integrate
   
          
   
            =  ln|x|  -  1/2 ln(x² + 1)  +  2tan^-1 x  + C
   
   

     

    

    

    

    

    

    

    

    

    

    

   
   

2.  
   
   Solution 
   
   We use the trig identity
   
           tan²(2x)  =  sec²(2x) - 1
   
   So that 
   
           tan⁴(2x)  =  (sec²(2x) - 1) tan²(2x)  =  (sec²(2x)tan²(2x)) -  (tan²(2x))
   
           =  tan²(2x)sec²(2x) - sec²(2x) + 1
   
   Now let
   
            u  =  2x    du  =  2 dx
   
    

   

            v  =  tan u    dv  =  sec²u du

    

   

   
   
   
           
   
           
   
   

     

    

    

    

    

    

    

    

    

    

    

     

    

    

    

    

    

    

    

    

    

    

   
   

3. 
   
   Solution
   Use integration by parts:
   
           u  =  x²        dv  =  e^5xdx
   
           du  =  2x dx    v  =  1/5 e^5x 
   
   This produces 
   
          
   
   We use integration by parts again
   
           u  =  x        dv  =  e^5xdx
   
           du  =  dx    v  =  1/5 e^5x 
   
   Which gives
   
          
   
   Finally we get
   
           1/5 x²e^5x - 2/25 xe^5x + 2/125 e^5x + C
   

     

    

    

    

    

    

    

    

    

    

    

   
   

4. 
   
   Solution
   Let 
   
           u  =  4 - 9x⁴        du  =  -36x³ 
   
   We get
   
          
   
   

     

    

    

    

    

    

    

    

    

    

    

   
   

5. 
   
   Solution
   
   We use the trig substitution
   
           x  =  sinq        dx  =  cosq dq
   
   We get
   
           
   
   

PROBLEM 4  Calculate the following limits if it exists:

A. (10 Points)   

Solution

         

B. (10 Points) 

  Solution

         

PROBLEM 5 Find the following integrals if they converge. 

A. (10 Points)   

Solution

B. (10 Points)    

Solution

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