Mathematics Assignment

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