MATHS 3001: Advanced Level Modeling and Change Assignment 2 Solution

This document presents a detailed solution to MATHS 3001 Assignment 2, focusing on the analysis of two functions: f(x) = sin^2(x) and g(x, y) = x^2 - xy + y^2. The solution meticulously applies the chain rule and trigonometric identities to determine the gradient function and Taylor series approximations for f(x). It includes the evaluation of derivatives, integrals, and the application of the quotient rule for the function g(x, y). The document also explores the properties of g(x, y) in three dimensions and derives the equation of the tangent. Furthermore, the solution provides step-by-step calculations, including partial derivatives, and utilizes the Taylor polynomial for two-variable functions to find series expansions, culminating in a simplified series representation and integration with respect to y. The solution emphasizes the symmetry of the equations and provides detailed explanations for each step, making it a valuable resource for students studying calculus and mathematical modeling.

Part 2.
a.
f ( x )=sin2 x
The gradient function
dy
dx = d
dx ( sin2 x )
Based on the nature of f(x), we apply Chain rule method of differentiation.
Letting u=sinx, du
dx =cosx
From chain rule we get
f (u)=u2
f ' ( u ) =2u
u' ( x ) =cosx
f ' ( x )= d
du ( u2 ) d
dx ( sin ( x ) ) =2ucos ( x )
Substituting for the actual value of u in, we get
¿ 2 ¿
When x= π
2 , f '
( π
2 )=0
y−1=0 ¿
y=1
b.
∫
0
π
sin2 xdx
Using trigonometric identities to rewrite sin2x as
sin2 xdx= 1−cos 2 x
2
∫
0
π
sin2 xdx=∫
0
π
1−cos 2 x
2 dx=¿ π
2 =1.57079 ¿
Area under tangent line
1 × π
2 =1.57079
c.
Taylor series approximate a function to a polynomial of the form
f ( x ) P ( x )=∑
0
n f k (a)
k ! xk
The value of f(0)= ( sin2 x )
The nth order derivatives are then evaluated as shown below
f 1 ( x ) = ( f 0 ( x ) )
'
= ( sin2 x ) '
=sin ( 2 ×0 ) =0

f 2 ( x ) = ( f 1 ( x ) ) '
= ( sin 2 x ) '=2 cos ( 2 ×0 ) =2
f 3 ( x ) = ( f 2 ( x ) )
'
= ( 2 cos 2 x ) ' =−4 sin ( 2 ×0 ) =0
f 4 ( x ) = ( f 3 ( x ) )
'
= ( −4 sin 2 x ) ' =−8 cos ( 2 ×0 ) =−8
Hence f(x) is approximated as shown below
f ( x ) 0
0 ! x0+ 0
1 ! x1 + 2
2! x2 + 0
3 ! x3 + −8
4 ! x4 …
f ( x ) x2 −1
3 x4
d. The integral of f(x) between
∫
0
π
x2= π3
3 =10.33542
∫
0
π
¿ 0
∑ ¿10.33542
e. The function g(x,y) is defined as shown below
g ( x , y ) = xy
x2+ y2
Volume=∬ xy
x2+ y2 dR
Where R is defined as [0,1]×[0,1]
∬ xy
x2 + y2 dxdy
∫
0
1
xy
x2+ y2 dx= 1
2 yln( y2+ 1)− yln( y )
∫
0
1
¿ ¿
From the fundamental laws of integration.
∫
0
1
¿ ¿
∫
0
1
¿ ¿
∫
0
1
yln( y ) dy=−0.25
∬ xy
x2 + y2 dR= 1
4 ( 2 ln ( 2 )−1 ) −(−1
4 )= 1
2 ln2=0.3466 cubic units

f.
At point (1,1),
The value of g(x,y)
g ( x , y )= xy
x2+ y2 = 1
1 =1
Hence in a 3 dimensions, we have the point of interest as (1,1,1)
∂
∂ x ( xy
x2 + y2 )
Using product rule on the numerator, we have the derivative of the numerator wrt to x as
∂
∂ x ( x ) =1∧∂
∂ x ( y )=0
Hence numerator’s derivative with respect to =y
Derivative of the denominator is obtained as
∂
∂ x ( x2 + y2 ) =2 x
Using quotient rule
∂
∂ x ( xy
x2 + y2 )= y (−x2 + y2 )
( x2+ y2 ) 2 substituting x=1∧ y=1 ,
∂
∂ x ( xy
x2 + y2 ) =0
Because of the symmetry of the equation g(x,y), we can substitute x for y and have
∂
∂ y ( xy
x2+ y2 )= x (− y2 + x2 )
( x2 + y2 ) 2 at x=1 , y =1=0
Hence the equation of the tangent is obtained from
The equation of the tangent
g( x , y)−1=0 ( x−1 ) +o ( y−1)
f ( x , y )=1
g.
∬1 dR

∬
0
1
1 dxdy=∫
0
1
1 dy=1 cubic units
h. The Taylor polynomial of a two variable function is given as
G ( x , y )=g ( 1,1 ) + ∂ g
∂ x ( 1,1 ) ( x−1 )+ ∂ g
∂ y ( 1,1 ) ( y−1 ) +( 1
2 ! ) [ δ2 g
δ x2 ( 1,1 ) ( x−1 )2 +2 δ 2 g
δxδy ( 1,1 ) ( x−1 ) ( y−1 ) + δ2 g
δ y2 ( 1,1 ) (
g ( 1,1 )= 1
12+12 = 1
2
∂ g
∂ x = y ( −x2 + y2 )
( x2+ y2 )
2
when x=1 , y=1 ,
∂ g
∂ x =1 (−12 +12 )
( 12+ 12 ) 2 =0
δ2 g
δ x2 = ∂
∂ x ( y (−x2 + y2 )
( x2 + y2 ) 2 )
From quotient rule,
u' =−2 xy
z=x2 + y2
dz=2 xdx
dg=2 zdu
dg
dx = dg
dz × dz
dx =4 zx
Substituting for the value of z, we get
v'=4 x ( x2 + y2 )
Using quotient rule,
Where u=numerator and v=denominator
δ2 g
δ x2 =−2 xy ( x2+ y2 ) 2
− ( 4 x ( x2 + y2 ) y ( y2−x2 ) )
( x2 + y2 ) 4
When x=1, y=1,
δ2 g
δ x2 =−2 xy ( x2+ y2 ) 2
− ( 4 x ( x2 + y2 ) y ( y2−x2 ) )
( x2 + y2 ) 4 =−0.5

∂ g
∂ y = x ( x2 − y2 )
( x2 + y2 )
2
when x=1 , y=1 ,
∂ g
∂ y = 1 ( 12−12 )
( 12+ 12 ) 2 =0
Because of the symmetrical nature of g(x,y), we can substitute x=y
δ 2 g
δ y2 =−2 xy ( x2+ y2 ) 2
− ( 4 y ( x2 + y2 ) x ( x2− y2 ) )
( x2 + y2 ) 4
When x=1, y=1, we get
δ2 g
δ y2 =−2 xy ( x2+ y2 ) 2
− ( 4 y ( x2 + y2 ) x ( x2− y2 ) )
( x2 + y2 ) 4 =−0.5
δ2 g(x , y )
δxδy
We first differentiate with respect to y. From the previous procedure, we have
∂ g
∂ y = x ( x2 − y2 )
( x2 + y2 )
2
∂
∂ x ( x ( x2− y2 )
( x2+ y2 )2 )
Again using product rule on the numerator
u'= ( x2− y2 ) +2 x2
From chain rule, we evaluate the derivative of the denominator
Let
z=x2 + y2
dz=2 xdx
dg=2 udz
dg
dx = dg
dz × dz
dx =4 zx
Substituting for the value of z, we get
4 x ( x2 + y2 )
From quotient rule, we now have
δ2 g( x , y )
δxδy = ( 3 x2− y2 ) ( x2 + y2 )2
− ( 4 x2 ( x2 + y2 )( x2 − y2 ) )
( x2 + y2 )4
For x=1, y=1
δ2 g(x , y )
δxδy = ( 3 x2− y2 ) ( x2 + y2 )2
− ( 4 x2 ( x2 + y2 ) ( x2 − y2 ) )
( x2 + y2 )4 =0.5

Substituting the values of δ2 g( x , y )
δxδy , ∂ g
∂ y , ∂ g
∂ x , δ2 g
δ x2 , δ2 g
δ y2 ∧g(1,1) in the Taylor series formula,
we get the series as
G ( x , y )=1+ ( 1
4 ) [− ( x−1 )2+ 2 ( x−1 ) ( y−1 ) − ( y−1 )2 ]
G ( x , y )=0.5+ ( 0.25 ) [−x2 +2 xy− y2 ]
Which can be simplified to
G ( x , y ) = ( 0.5+0.25 [ −x2+ 2 xy− y2 ] )
i.
∬
0
1
( ( 0.5+0.25 [ −x2+ 2 xy− y2 ] ) ) dxdy
For x=1
∫
0
1
( 1
2 + ( 1
4 ) [ −x2 +2 xy− y2 ] ) dx= 1
2 x +( 1
4 ) [ −x3
3 + x2 y− y2 x ] =1
2 +( 1
4 ) [ −1
3 + y− y2
]
Integrating with respect to y
∫
0
1
( 1
2 + ( 1
4 ) [−1
3 + y− y2
] )dy= 1
2 y +( 1
4 ) [−1
3 y + y2
2 − y3
3 ]
For y=1
1
2 y +( 1
4 ) [−1
3 y + y2
2 − y3
3 ]=0.458 ˙3