Solution: Calculus, Laplace Transform, Complex Numbers, Ellipse

This document presents a comprehensive solution to a calculus assignment, addressing various concepts within the field. It includes detailed solutions for problems involving Laplace transforms, such as finding the Laplace transforms of given expressions and the inverse Laplace transform of a given function. The solution also covers complex numbers, including calculations of magnitude and argument, and determining the value of a complex expression. Furthermore, it includes an analysis of an ellipse, deriving its equation from given conditions. The assignment solution provides step-by-step explanations and calculations, making it a valuable resource for students studying calculus and related mathematical concepts. This resource is available on Desklib, a platform offering a wide range of study materials.

Solutions
Solution-B1
(a) (i) By shifting property,
L (eat∗f ( t ) )=F (s−a)
(ii) L (∫
0
t
[ f (u )−e−2 u∗cos ( 3 u ) ] du )
¿ L ¿
As L❑ [ cos ( at ) ] =s /(s2+ a2) then by using shifting property and Laplace transform of
integrals, the Laplace transform is
¿ F ( s )
s − ( s− (−2 ) )
s ( s− (−2 ) )2 + ( 3 )2 ¿ ¿= F ( s )
s − ( s+ 2 )
s ( ( s +2 )2+ 9)
(b) For 0≤ t<2 , f ( t ) =4−t2∧for t ≥ 2 , f ( t ) =0
So, f(t) can be written as
f ( t )= ( 4−t2 ) ( u ( t )−u ( t−2 ) )
As u ( t )=1 for t ≥ 0∧u ( t−2 )=1 for t ≥2
∴
For 0≤ t<2 , u ( t ) −u ( t−2 )=1−0=1
For t ≥ 2 ,u ( t )−u ( t−2 ) =1−1=0
L ( ( 4−t2 ) ( u ( t ) −u ( t−2 ) ) ) =L ( 4 u ( t ) )−L ( 4 u ( t−2 ) ) −L ( t2 u ( t ) ) + L ( t2 u ( t−2 ) )
Let t 1=t−2 ,
so , t2 u ( t−2 ) can be written as (t 1+2 )2 u ( t 1 )
L ( 4 u ( t ) ) −L ( 4 u ( t−2 ) ) −L ( t 2∗u ( t ) ) + L ( (t 1+2)2 u ( t 1 ) )
¿ L ( 4 u ( t ) )−L ( 4 u ( t−2 ) )−L ( t2∗u ( t ) )+ L ((t 12 +2 t 1+4 )u ( t 1 ) )
¿ 4
s − 4
s e
−2 s
− 2
s3 −e−2 s
( 2
s3 + 4
s2 + 4
s )
¿ 4
s (1−2 e−2 s ) + 4
s2 e−2 s − 2
s3 (1+e−2 s )
(c ) L−1 ¿
Decomposing the given rational expression into partial fractions,

2 s +3
( s¿¿ 2+ 4)(s¿¿ 2+9)= As+ B
s2 +4 + Cs+ D
s2 +9 ¿ ¿
Equating coeff of s3, s2 , s∧constant terms
A+C = 0
B+D = 0
9A+4C = 2
9B+4D = 3
Solving for A, B, C and D, we get A = 2/5, B = 3/5, C=-2/5, D=-3/5
= L−1
( 1
5 (2 s+3
s2+4 − 2 s +3
s2 +9 ) )
= L−1
( 2
5 ( s
s2+4 )+ 3
5 ( 1
s2 +4 )− 2
5 ( s
s2+ 9 )−3
5 ( 1
s2 +9 ) )
Since L−1
( s
s2 +a2 )=cos (at) and L−1
( a
s2 +a2 ) =sin ( at), so, the L−1 of the expressionis
¿ 1
5 [2 cos (2 t)+ 3
2 sin (2 t)−2cos (3 t)−sin ( 3 t ) ] Ans.
(d )¿ z1∨¿ 2 , arg(z1 ¿=−π
3 =¿ z1=2¿
z2 = eiπ /3 => ¿ z2∨¿ 1
z3=−2+2 √3i
z1+ p∗¿ z2∨ ¿
z3
=
2 ( 1
2 +i ( − √ 3
2 ) ) + p
−2(1+ √ 3 i) ¿
¿ [ ( 1− √3 i )+ p ]
−2(1+ √3 i) *(1− √3 i)
(1− √3 i)
¿ (1−3+ p−2 √ 3i− p √ 3 i)
−8
For the expression to be purely imaginary, real part should be zero i.e.
1-3+p = 0 => p = 2
Ans. p = 2
(e) xyez + y z3−ln ( x +z )=7

Differentiating the eqn. wrt y gives,
x(y ez ∂ z
∂ y +ez ¿+(3y z2 ∂ z
∂ y + z3 ¿– 1
( x+ z )
∂ z
∂ y =0
 ∂ z
∂ y = x ez + z3
xy ez +3 y z2 − 1
( x + z )
Solution- B2
(a) z=f ( u )
∂ z
∂ x =f ' ( u )∗∂u
∂ x =f ' ( u )∗2 x y3
∂2 z
∂ x2 = ∂
∂ x ( f ' ( u )∗2 x y3 )=2 y3
[ x∗f ' ' (u ) ∂ u
∂ x + f ' ( u ) ]
¿ 2 y3 [ x∗f ' ' ( u )∗2 x y3 +f ' ( u ) ]
∂ z
∂ y =f ' ( u )∗∂ u
∂ y =f ' ( u )∗3 x2 y2
∂2 z
∂ y2 = ∂
∂ y ( f ' ( u )∗3 x2 y2 )=3 x2
[ y2∗f ' ' ( u ) ∂ u
∂ y +f ' ( u )∗2 y ]
¿ 3 x2 [ y2∗f ' ' ( u )∗3 x2 y2+ f ' ( u ) .2 y ]
¿ 9 x4 y4∗f ' ' ( u ) + 6 x2 y . f ' (u)
∂2 z
∂ x ∂ y = ∂
∂ x ( ∂ z
∂ y )= ∂
∂ x [ f ' ( u )∗3 x2 y2 ]
¿ 3 y2 [f ' ( u )∗2 x+x2∗f ' ' ( u )∗∂u
∂ x ]
¿ 3 y2 [ f ' (u )∗2 x+ 2 x3 y3∗f '' (u ) ]
¿ 6 x y2∗f ' ( u ) +6 x3 y5∗f ' ' ( u )
Therefore,
6 y2 ∂2 z
∂ x2 − y2 ∂2 z
∂ y2 −¿ xy ∂2 z
∂ x ∂ y
¿ 6 y2 [2 y3 [ x∗f ' ' ( u )∗2 x y3 + f ' ( u ) ] ]− y2 [9 x4 y4∗f ' ' ( u ) +6 x2 y . f ' (u)]−¿
xy [6 x y2∗f ' ( u ) +6 x3 y5∗f ' ' ( u ) ]
¿ 9 x4 y6∗f ' ' ( u ) Ans

(b) F(x,y,z) = x2 y z3+ln (xz )
Max rate of change of a function at point (a, b, c) is the magnitude of the gradient at
the point i.e.
|∇ f ∨¿ √(f ¿ ¿ x ( a , b , c))2 +¿❑ (f ¿¿ y ( a , b , c))2+(f ¿¿ z(a , b , c))2 ¿ ¿ ¿ ¿
¿ √ (2 xy z3 +1/ x)2 +¿❑ x4 z6+(3 x2 y❑ z2 +1/ z )2 ¿
At ( 1, 1 , 2 )
|∇ f ∨¿ √ ( 17 )2+ ¿❑ 1∗64+ ( 12.5 )2 = √509.25=22.5666 ¿ Ans
(c ) f ( x , y , z ) =4 xz +7 yz+ xy
xyz=756 -----------(1)
F ( x , y , z , λ )=f ( x , y , z )− λ ( g ( x . y . z )−K )
¿ 4 xz+7 yz+ xy −λ(xyz −756)
Fx=4 z+ y−λyz=0
F y=7 z + x−λxz=0
Fz =4 x +7 y− λxy=0
Fλ=−xyz +756=0
λ= 4
y + 1
z
λ= 7
x + 1
z
λ= 4
y + 7
x
Solving for x, y, z , we get
x=7 z , y =4 z
Putting these values in (1)
( 7 z ) ( 4 z ) ( z ) =756
 z=3
So, x=21 , y=12 , z =3
Therefore, f min=4∗21∗3+7∗12∗3+21∗12=756 Ans
Solution-A4
(a) |z|+|z −i|=2
z=x +iy
∴ √ x2+ y2 + √ x2+( y−1)2 = 2

¿ , √ x2 + y2=2− √ x2 +( y−1)2
Squaring both sides, we get
x2+ y2=4 + x2+( y−1)2−2∗2∗√ x2+( y−1)2
2 y−5=−4 √ x2 +( y −1)2
 4 y2 +25−20 y =16 x2 +16 y2−32 y +16
Or, 16 x2+12( y ¿¿ 2− y )−9=0 ¿
Comparing with standard eqn.
a x2+2hxy + b y2+2 gx+ 2 fy +c=0
we have, a = 16, 2h = 0, b = 12, 2g = 0, 2f = -12 => f= -6, c= -9
∆ = abc +2 fgh−a f 2−b g2−c h2= ( 16∗12∗−9 )−16∗144 ≠ 0
2h*2h – 4ab = 0 - 4(16*12) < 0 and a ≠ c
Therefore, locus of z is the equation of an ellipse.
Solving further, the equation of the ellipse comes out as
x2
¿ ¿ ¿
(b) w= ( S+1 ) eiθ=u+iv
z=x +iy
z +i w=0
x−iy+i(u−iv)=0
x +v =0=¿ x=−v
− y=−u=¿ y =u
z=x +iy=−v +iu=i ( u+iv ) =iw=( S+1)eiθ eiπ/ 2 as i=¿ eiπ /2
arg ( zw )=θ+θ+ π
2 =2 π
3
As −π ≤θ< π,

 θ=
2 π
3 − π
2
2 = π
12 , (−π + π
12 )=¿ θ= π
12 ,−11 π
12
Ans