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Thursday, 15 December 2005

PATEL 4U EX 5E, Polynomials over the complex numbers, (Q7, 8, 9, 10, 12b, 13, 14, 19, 24, 27, 28, 29, 30, 32, 33, 34, 35)

QUESTIONS: Q7, 8, 9 (at back too) , 10, 12b, 13, 14, 19, 24, 27, 28, 29, 30, 32, 33, 34, 35, 9 (better)

 












EX 5E - Q9 ================== 

Friday, 9 December 2005

Concept check. Match the rational function with the aspect described.



ANSWERS

A 36
B 30
C 34
D 29
E 33
F 35
G 31
H 32  



Vectors : Arc length of a space curve Use the integration capabilities of you graphing utility to approximate the arc length of the space curve given by: vector r(t) = t i + t^3 j + 3k from the point (1,1,3) to the point (2,8,6). Round your answer to the nearest three decimal places.

QUESTION: Vectors : Arc length of a space curve
Use the integration capabilities of you graphing utility to approximate the arc length of the space curve given by: vector 
r(t) = t i + t^3 j + 3k
from the point (1,1,3) to the point (2,8,6). Round your answer to the nearest three decimal places. 

SOLUTION  =======================
Theory. 
If you have a curve r(t) = x(t) i + y(t) j + z(t) k  then the arc length between two points where t=a to t=b is given by 

s = \int^b_a \sqrt{ x'(t)^2 + y'(t)^2 + z'(t)^2 } dt
( this is the integral between t=a and t=b
========================================== 

It is clear that the two points given arise from t=1 and t=2

since when t=1, r(1) = 1 i + 1 j + 3 k which corresponds to (1,1,3) 
and when t=2, r(1) = 2 i + 8 j + 6 k which corresponds to (2,8,6) 

For us, 
x(t) = t 
y(t) = t^3 
z(t) = 3t

so 

x'(t) = 1
y'(t) = 3t^2 
z'(t) = 3

and s = \int^b_a \sqrt{ 1 + 9t^4 + 9} dt 

      s = \int^b_a \sqrt{ 10 + 9t^4} dt

when I use the graphic program to approximate this integral I obtain, 

   s = 7.763 to 3 decimal places. 

cheers 
Jan Hansen

Sunday, 20 March 2005

For these rational functions: find asymptotes, sketch as indicated.

Questions.
Give the equation of any horizontal, vertical, or oblique asymptotes.

Q38.  f(x)=\displaystyle -{6\over x+9}
Q42.  f(x)=\displaystyle {x^2+4\over x-1}

Sketch

Q52.  f(x)=\displaystyle {x-5\over x+3}
Q54.  f(x)=\displaystyle {2x+1\over x^2+6x+8}
Q62.  f(x)=\displaystyle {x^2-7x+10\over x^2+9}
Q64.  f(x)=\displaystyle {2x^2+3\over x-4}
Q68.  f(x)=\displaystyle {x^2-16\over x+4}

SOLUTIONS ======
Q38. Vertical asymptotes is x=-9, horizontal asymptotes is y=0.