Approximating Error in (16.04)^4

This article explains how to approximate the error in (16.04)^4 by (16)^4 using the mean value theorem. It provides step-by-step calculations and explanations.

QUESTION 1
π1 :
( x1
x2
x3
) =
( 1
−3
2 ) + λ1
( 2
−2
1 )+ λ2
( 3
0
3 )
π2 :
(x1
x2
x3
)=
( 1
0
−1 )+μ1
( 2
1
−2 )+ μ2
(3
3
0 )
a.
n1=¿ 3 ,0,3> x <2 ,−2,1> ¿
¿
|i j k
3 0 3
2 −2 1|=| 0 3
−2 1|i−|3 3
2 1| j+|3 0
2 −2|k=6 i+3 j−6 k
Factorizing to get unit normal vector,
n1=¿ 2,1,−2>¿
n2 =¿ 3,3,0> x<2,1,−2>¿
¿
|i j k
3 3 0
2 1 −2|=|3 0
1 −2|i−|3 0
2 −2| j+|3 3
2 1|k =−6 i+6 j−3 k
Factorizing to get unit normal vector,
n2 =¿ 2,−2 ,1>¿
The elements of the normal vectors of the planes are neither the same nor multiples of each other
hence their cross product is not a zero. Therefore, the two planes are not parallel and must
intersect at a given point or line.
b. Cartesian forms of the equations
For plane π1 , the parametric equations are:
x1=1+2 λ1 +3 λ2 …(i)
x2=−3−2 λ1 …(ii)
x3=2+ λ1+3 λ2 … (iii)
¿ ( i ) ,

2 λ1=x1−1−3 λ2
Replacing this∈ ( ii ) ,
x2=−3− ( x1−1−3 λ2 ) =−2−x1 +3 λ2
So that ,
3 λ2 =x2 +x1 +2∧λ1=−3
2 − x2
2
Replacing theabove two equations∈ ( iii ) ,
x3=2+ λ1+3 λ2=2− 3
2 − x2
2 +x2+x1 +2
x3= 5
2 + x2
2 + x1
ℜ−arranging∧simplyfying the above equation ,
x1+ x2
2 −x3 + 5
2 =0
The cartesian equation of π1 is;
2 x1 + x2−2 x3+5=0
For plane π2 , the parametric equations are:
x1=1+2 μ1+3 μ2 …(i)
x2=μ1 +3 μ2 …(ii)
x3=−1−2 μ1 …(iii )
¿ ( i ) ,
3 μ2=x1−1−2 μ1
Replacing this∈ ( ii ) ,
x2=μ1 + x1−1−2 μ1=x1−1−μ1
So that ,
μ1=x1−x2−1
Replacing theabove equation∈ ( iii ) ,
x3=−1−2(x1−x2 −1)
x3=−1−2 x1 +2 x2 +2 ¿
ℜ−arranging∧simplyfying the above equation , cartesianequation of π2 is ;
2 x1 −2 x2 + x3−1=0
c. Let x1=ω,
The cartesian equation s thus reduces ¿ ;
2 ω−2 x2 +x3−1=0
2 ω+ x2−2 x3+ 5=0
Expressing x2 ∈terms of ω ,
x3=1+2 x2 −2 ω

2 ω+ x2−2 x3+ 5=0
2 ω+ x2−2 ( 1+2 x2 −2 ω ) +5=0
3 x2=6 ω+3
∴ x2=1+2 ω
Expressing x3 ∈terms of ω ,
2 ω+ x2−2 x3+ 5=0
x2=−2 x3−2 ω−5
2 ω−2 x2 +x3−1=0
2 ω−2 ( −2 x3−2 ω−5 )+ x3−1=0
3 x3=9+6 ω
x3=3+2 ω
Hence the line of intersection takes the parametric vector form
L=¿ 0,1,3>+ ω<1,2,2>¿
d.
x1=1+2 μ1+3 μ2
x2=μ1 +3 μ2
x3=−1−2 μ1
Replacing the above equations∈the cartesian equation of plane1 ,
2 x1 + x2−2 x3+5=0
2 ( 1+2 μ1+3 μ2 ) + μ1 +3 μ2−2 ( −1−2 μ1 ) +5=0
9 μ1 +9 μ2 +9=0
μ1 + μ2 +1=0
L=¿ 1,1,0>+¿1 , 2 ,2> μ
e. The parametric equations, though slightly different, represents the same line due to the
same vector direction obtained in both cases.
f.
m=n1 x n2
¿<2,1 ,−2>x<2,−2,1>¿

¿
|i j k
2 1 −2
2 −2 1 |=| 1 −2
−2 1 |i−|2 −2
2 1 | j+|2 1
2 −2|k
¿−3 i−6 j−6 k
= <1, 2, 2>
¿ show that mis∥¿ the line , cross product of thetwo vector directions isevaluated
m x lc=¿−3 ,−6 ,−6> x <1,2,2>¿
¿
| i j k
−3 −6 −6
1 2 2 |=|−6 −6
2 2 |i−|−3 −6
1 2 | j+|−3 −6
1 2 |k
= <0, 0, 0>
Hence the vector mis∥¿ the line of ∩¿
g. The line intersects both the planes with a direction vector m=n1 x n2 because m is
perpendicular to both the normal vectors n1∧n2 and is therefore parallel to the two planes.

QUESTION 2
To show that the equation, e−7 x+6 cos ( 8 x ) =0 , has a unique solution over the interval [0, π
8 ],
Let e−7 x+6 cos ( 8 x ) be a given function say f ( x )
i .e . f ( x )=e−7 x +6 cos ( 8 x )
This function , f ( x ) ,is continous
the given interval
Step 1: ¿ the given interval ,the function f ( x )=0 is first shown ¿ haveat least
one solution .
Replacing thelower ∧theupper values of the interval given ,
f ( 0 ) =e−7 (0)+6 cos ( 0 ) =7
f ( π
8 )=e−7 ( π
8 )+ 6 cos ( π
8 )=5.607
Both f ( 0 ) ∧f ( π
8 ) are greater than zero∧thus there exist no f ( c ) =0 for some
c ∈theinterval [ 0 , π
8 ] … inter−mediate valuetheorem
Step 2: f ( x ) =0is then shown ¿ have at most one solution ∈theinterval [ 0 , π
8 ]
By first assuming that more than one solution exist for
f ( x )=0 ,two arbitrary values are picked ; say x=a , x=b∧a <b
Because of the continuity of f on [ a , b ] , its differentiability on ( a ,b ) ∧the that!
f ( a )=f ( b )=0; by Roll e' s theorem, for some d ∈theinterval ( a , b ) ,follows that
f ' ( d )=0
But ,

f ' ( x )=−7 e−7 x−48 sin ( 8 x ) <0 for all values of x ∈the interval [0 , π
8 ]
i .e .sin ( 8 x ) ∧e−7 x are both positive∈the intervalthus the∑ of their negationmust yield
negative values .
Thisis an implication that f ' ( d ) can never be 0;
f ' ( d ) ≠ 0
Therefore , a contradiction o ccur depicting that f ( x ) =0 has at most one solution∈¿
the interval [ 0 , π
8 ] . Coupling this reasoning withthe that!f ( x ) =0 has got at
at least one solution ,is reasonable ¿ conclude that theequation has aunique solution∈the
interval provided .
QUESTION 3
To approximate the error in ( 16.04 )
1
4 by ( 16 )
1
4 ,
Let a function f ( x ) =x4 such that f ( 16 ) =16
1
4 =2
∴ f ' ( x )= 1
4 x
−3
4
so that ,
f ' ( 16 ) = 1
4 ( 16 )
−3
4 = 1
32
Using mean value theorem ,
Error , F ( x )=f ( 16 ) + f ' ( 16 ) ( x−16 )
¿ 2+ 1
32 ( x−16 )
F ( 16.04 )=2+ 1
32 ( 16.04−16 )
¿ 2.00125