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 ================== 

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

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 $y=0$.