SIT291 Mathematical Methods 2013AssignmentsAssignments are a very important part of your studies. They serve twopurposes: Feedback and Diagnostic. Assignments allow students to find outhow they are...

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SIT291 Mathematical Methods 2013AssignmentsAssignments are a very important part of your studies. They serve twopurposes: Feedback and Diagnostic. Assignments allow students to find outhow they are going in developing the skills and understanding the new ideasas the unit goes along. They are also intended to help you determine whetheryou are acquiring the specific knowledge and skills which are needed, so thatyou can remedy this if necessary.Feedback and diagnostic assessment intentionally do not carry a greatdeal of weight in calculating your final mark. This means that you are notoverly penalised when you are still working towards developing certain skillsand knowledge. But don’t think a low weighting means that submittingthese forms of assessment is not important – they carry value well abovetheir weight, because they give you so much information about what youmay need to work on, and what standards are expected. Similarly, it is notintended that the time it may take you to complete an assignment is reflectedin its weighting; the work required to get to the stage where you can answerthe questions is also a major contribution to your preparation for the finalassessment.It is clearly important that feedback is provided while the task is freshin your mind and in time for you to act on it before your next assessment. Iundertake to return assignments to you in a timely manner, generally withintwo weeks of the submission deadline. I will put master solutions online oneweek after the submission due date. The number and spacing of assignmentsensures that you get feedback throughout the trimester.For each assignment, I will mark in detail some of the arguments youput forward in your solutions. You will be provided with complete modelanswers, and you will get most benefit from the assignment if youself-assessthe other parts of your work against these. You need to develop confidencein your ability to check your own work both for the final assessment, and inyour future career. You will also be assessed on how well you have plannedyour work to complete the whole assignment. Your use of writing and thegrammar of mathematics in presenting your reasoning makes up the thirdcomponent of your mark.Finally I suggest you make a copy of your submission, especially of As-signment 4, as the feedback may come after the exam date, which is unknownat this stage.
Assignment 1Due 5 p.m. Monday 8 April, 20131. Show that the equation represents a sphere and find its centre andradius3x2+ 3y2+ 3z2-12x+ 6z-10 = 02. Given ¯a=<1,-2,5>and ¯b=<-1,0,3>, find:(a) ¯a+ ¯b,(b) 3 ¯a+ 2 ¯b,(c)| ¯a|,(d)| ¯a- ¯b|,(e) ¯a· ¯b,(f) ¯c= ¯a× ¯b,(g)proj ¯a ¯b,(h) determine whether ¯aand ¯bvectors are orthogonal, parallel or nei-ther.(i) verify that ¯cis orthogonal to both ¯aand ¯b.3. Find an equation of the plane that includes the pointP(3,0,4) and isparallel to the plane 3x-y+ 4z= 1.4. Write down the equation in spherical and cylindrical coordinatesx2+y2+z2= 55. If ¯r(t) =, find the curvature at the point(0,0,1).6. Find the length of the curve ¯r(t) =<3,43t3,t2>, 0=t=2.
Answered Same DayDec 23, 2021

Answer To: SIT291 Mathematical Methods 2013AssignmentsAssignments are a very important part of your studies....

David answered on Dec 23 2021
115 Votes
Sol: (1)
2 2
We are asked for the extreme values of subject to the constraint
( , ) 2. Using Lagrange multipliers, we solve the equations
, ( , ) 2, which can be written as
f
g x y x y
f g g x y
  
  

 , , , 2,x x y yf g f g g x y   
Or as
 
2 2
8 2 .................. (1)
2 1 2 ................. (2)
2 .................. (3)
x x
y y
x y



 
 

From equation (1),
 
 
8 2 0
2 4 0
0, 4
x x
x
x



 
 
 

1
If 0, then equation (3) gives 2. If =4, then from equation (2).
3
17
So then equation (3) gives . Therefore, has possib le extreme values
3
17 1 1
at the points (0, 2), (0, 2), , ,
3 3
x y y
x f
    
 
 
   
 
7 1
, . Evaluating at
3 3
these four points, we find that
f
 
 
 
 
 
2
2
2
2
(0, 2)= 2 1 3 2 2
(0, 2)= 2 1 3 2 2
17 1 4 17 1 84
, 1
3 3 9 3 9
17 1 4 17 1 84
, 1
3 3 9 3 9
f
f
f
f
  
    
    
        
  
    
         
  

2 2 17 1 84Therefore, the naximum value of on the circle 2, is , =
3 3 9
and the minimum value is (0, 2)=3 2 2.
f x y f
f
 
    
 

Sol: (2)
4 3 4 3
1 1 1 1
4 3
11
4
1
4
1
4
1
1
ln
ln 3 ln 1
ln 3
ln 3
ln 3
x
x
x
x
x
x
e
dydx e dx dy
y y
e dx y
e dx
e dx
e

   
   
   
      

   



 
4
3 1 ln 3
e e
e e
     
   
Sol: (3)
   
 
   
   
   
   
 
   
2 3
0 0
2 3
00
2
0
2
0
2
0
2
0
2
2
0
0
2 2
0 0
cos cos
sin
sin 3 sin
sin 2 sin
sin 2 sin
sin 2 sin
cos 2
cos
2
1
cos 2 cos
2
1
cos cos
2
x
R
x
x y dA x y dydx
dx x y
dx x x x
dx x x
x x dx
x x dx
x
x
x x








 

  
    
     
     
     
   
 
    
 
  
 ...
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