Mathematics Assignment

This Mathematics Assignment includes solutions to problems related to unique solutions, parametric equations, Cartesian equations, and absolute error.

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
17th July 2019

1. Show that e−8 x+5 cos ( 20 x ) =0 has a unique solution from x=0 ¿ x= π
20
An equation is said to have a unique solution if it has only one solution at a particular point.
At x=0 ,
e−8 x+5 cos (20 x )=e−8(0)+5 cos ¿ ¿
¿ e0 +5 cos 0=1+5=6
At x= π
20 ,
e−8 ( π
20 )
+5 cos ( 20 ) π
20 =e−0.4 π +5 cos ( π ) =0.2846-5= -0.4.7154
Therefore the above equation has a unique s0lution within the given closed interval.
2 a ¿ π 1 : (x1
x2
x3
)=
( 3
1
−1 )+ t
( 1
2
−2 )+ p ( 3
1
−3 )
Parametric equations are
x1=3+t +3 p------------------------------i
x2=1+2 t+ p………………………………..ii
x3=−1−2t−3 p -------------------------iii
From i and ii we have;
p=0.4 x1−0.2 x2−1 And t = x2 −x1
3
Substituting p and t in iii
x3=−1−2t−3 p
x3=−1−2( x2−x1
3 )−3 ¿ )
x3=−1−2
3 x2− 2
3 x1−1.2 x1−0.6 x2−3
x3=−28
15 x1− 19
15 x2−4
−28
15 x1− 19
15 x2−x3−4=0

Normal vector to π1 is (−28
15 ,−19
15 ,−1 ¿
n1=¿ ¿(−28
15 ,−19
15 ,−1 ¿
π2 : (x1
x2
x3
)=
( 3
−3
3 )+ μ1
( 1
−2
2 )+ μ2
( 3
−3
1 )
Parametric equations are
x1=3+ μ1 +3 μ2------------------------------i
x2=−3−2 μ1−3 μ2………………………………..ii
x3=3+2 μ1+ μ2----------------------iii
From i and ii we have;
μ1=−x1−x2 and μ2= 2 x1 + x2+3
3
Substituting μ1and μ2in iii
x3=3+2 μ1+ μ2
x3=3+2 ( −x1−x2 ) +( 2 x1 + x2 +3
3 )
−4
3 x1− 5
3 x2−x3+ 4=0
−4
3 x1− 5
3 x2−x3+ 4=0
Normal vector to π2 is (−4
3 ,−5
3 ,−1 ¿
n2=¿ ¿ (−4
3 ,−5
3 ,−1 ¿
b) The Cartesian equation for π1can be found by eliminating the parameters s and t from the
parametric equations.
π1 : ( x1
x2
x3
)=
( 3
1
−1 )+ t
( 1
2
−2 )+ p ( 3
1
−3 )
Parametric equations are

x1=3+t +3 p------------------------------i
x2=1+2 t+ p………………………………..ii
x3=−1−2t−3 p -------------------------iii
From i and ii we have;
p=0.4 x1−0.2 x2−1 And t = x2 −x1
3
Substituting p and t in iii
x3=−1−2t−3 p
x3=−1−2( x2−x1
3 )−3 ¿ )
x3=−1−2
3 x2− 2
3 x1−1.2 x1−0.6 x2−3
x3=−28
15 x1− 19
15 x2−4
−28
15 x1− 19
15 x2−x3−4=0
The Cartesian equation for π1 is− 28
15 x1− 19
15 x2−x3−4=0
The Cartesian equation for π2can be found by eliminating the parametersμ1 and μ2from the
parametric equations.
π2 : ( x1
x2
x3
)=
( 3
−3
3 )+ μ1
( 1
−2
2 )+ μ2
( 3
−3
1 )
Parametric equations are
x1=3+ μ1 +3 μ2------------------------------i
x2=−3−2 μ1−3 μ2………………………………..ii
x3=3+2 μ1+ μ2----------------------iii
From i and ii we have;
μ1=−x1−x2 And μ2= 2 x1 + x2+3
3
Substituting μ1and μ2in iii

x3=3+2 μ1+ μ2
x3=3+2 ( −x1−x2 ) +( 2 x1 + x2 +3
3 )
−4
3 x1− 5
3 x2−x3+ 4=0
−4
3 x1− 5
3 x2−x3+ 4=0
The Cartesian equation for π2 is −4
3 x1− 5
3 x2−x3+4=0
c) Let x1=w
π1 : ( w
x2
x3 )=
( 3
1
−1 )+ t
( 1
2
−2 )+ p ( 3
1
−3 )
Let x1=w
Parametric equations are
w=3+t +3 p------------------------------i
x2=1+2 t+ p………………………………..ii
x3=−1−2t−3 p -------------------------iii
From i and ii we have;
p=0.4 w−0.2 x2 −1 And t = x2 −w
3
Substituting p and t in iii
x3=−1−2t−3 p
x3=−1−2( x2−w
3 )−3 ¿ )
x3=−1−2
3 x2− 2
3 w−1.2 w−0.6 x2−3
x3=−28
15 w− 19
15 x2−4
−28
15 w− 19
15 x2−x3−4=0

The Cartesian equation for π1 is− 28
15 w− 19
15 x2−x3−4=0
Let x1=w
π2 : ( w
x2
x3 )=
( 3
−3
3 )+ μ1
( 1
−2
2 )+ μ2
( 3
−3
1 )
Parametric equations are
w=3+ μ1 +3 μ2------------------------------i
x2=−3−2 μ1−3 μ2………………………………..ii
x3=3+2 μ1+ μ2----------------------iii
From i and ii we have;
μ1=−w−x2 And μ2= 2 w+ x2+ 3
3
Substituting μ1and μ2in iii
x3=3+2 μ1+ μ2
x3=3+2 (−w−x2 ) +( 2 w+ x2 +3
3 )
−4
3 w− 5
3 x2−x3 +4=0
−4
3 w− 5
3 x2−x3 +4=0
The Cartesian equation for π1 is− 28
15 w− 19
15 x2−x3−4=0…………………………………..k
The Cartesian equation for π2 is −4
3 w− 5
3 x2−x3 +4=0………………………………..m
From k the value of w=−15
28 (19
15 x2 +x3 +4 )=−19
28 x2− 15
28 x3− 15
7
From m the value of w= 3
4 ( −5
3 x2 −x3 −4 ) =−5
3 x2− 3
4 x3 −3
Hence −19
28 x2− 15
28 x3− 15
7 =−5
3 x2− 3
4 x3−3

= −19
28 x2− 15
28 x3− 15
7 + 5
3 x2 + 3
4 x3+3=0
−83
84 x2 + 3
14 x3 + 6
7 =0
Parametric vector equation of L is −83
84 j+ 3
14 k + 6
7 =0
d)
N1=¿¿(−28
15 ,−19
15 ,−1 ¿
N2=¿¿ (−4
3 ,−5
3 ,−1 ¿
Let s=N1 × N2
s= N1 × N2=
| i j k
−28
15 − 19
15 −1
−4
3 −5
3 −1 |=−2
5 i+ 8
15 j−64
45 k
−4
3 w− 5
3 x2−x3 +4=0 ……………………………………….iv
−28
15 w− 19
15 x2−x3−4=0………………………………………….v
Suppose the point (w, x2, 0) is the intersection line and the xy co-ordinate plane. By substituting
x3=0 in iv and v above we have
−4
3 w− 5
3 x2 +4=0 ……………………………………….iv
−28
15 w− 19
15 x2−4=0………………………………………….v
Solving the two equations simultaneously we get
w= -8.25 , x2=¿9
(w,x2,0)= (-8.25,9, 0)
The equation of the line of intersection L is
w+8.25
−2
5
= x2−9
8
15
= x3
−64
45

π2 : (x1
x2
x3
)=
( 3
−3
3 )+ μ1
( 1
−2
2 )+ μ2
( 3
−3
1 )
Parametric equations are
x1=3+ μ1 +3 μ2------------------------------i
x2=−3−2 μ1−3 μ2………………………………..ii
x3=3+2 μ1+ μ2----------------------iii
The Cartesian equation for π1 is− 28
15 x1− 19
15 x2−x3−4=0
= −28
15 ( 3+μ1 +3 μ2 )− 19
15 ( −3−2 μ1−3 μ2 ) −(3+2 μ1+μ2 )−4=0
=−28
5 − 28
15 μ1− 28
5 μ2+ 19
5 + 38
15 μ1+ 19
5 μ2−3−2 μ1−μ2−4=0
= - 4
3 μ1− 4
5 μ2 − 44
5 =0
Parametric vector equation is - 4
3 μ1 i− 4
5 μ2 j− 44
5 =0
e) The two equation actually represent the same line.
f) n1=¿ ¿(−28
15 ,−19
15 ,−1 ¿
n2=¿ ¿ (−4
3 ,−5
3 ,−1 ¿
m=n1 × n2
m=n1× n2=
| i j k
−28
15 − 19
15 −1
−4
3 − 5
3 −1 |=−2
5 i+ 8
15 j−64
45 k
m=−2
5 i+ 8
15 j−64
45 k…………………………………..q
−83
84 j+ 3
14 k + 6
7 =0 …………………………………r (from part c)

- 4
3 μ1 i− 4
5 μ2 j− 44
5 =0 ……………………………..s (from part d)
Comparing q and r i.e.
−2
5 i+ 8
15 j−64
45 k=g( −83
84 j+ 3
14 k + 6
7 ¿where g is a scalar.
It is clearly evident that q isn’t a scalar multiple of r hence the two vectors aren’t parallel.
Comparing q and s i.e.
−2
5 i+ 8
15 j−64
45 k=h( 4
3 μ1 i− 4
5 μ2 j− 44
5 )where h is a scalar.
It is clearly evident that q isn’t a scalar multiple of s hence the two vectors aren’t parallel.
g) For vectors to be parallel then one must be a scalar multiple of the other.
3. Let f= ( 16.022 ¿ ¿
3
4 , f 1=¿
16.022=16+X
X=0.022 Where X is the error
Absolute error=(0.022)
3
4 =0.057124