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Running head: MATHS
MATHS
Name of the Student
Name of the University
Author Note

Q1 b)
Given, f(x) = (x-2)/sinx
Now, by using the quotient rule for differentiation
( d
dx ) ( f ( x ) )=
sinx∗( d
dx ) ( x−2 ) – ( x−2 )∗( d
dx ) ( sinx )
sin2 ( x )
= 1∗sinx – ( x−2 ) cos ( x )
sin2 ( x )
Q2
b) y=cos
(−1 ) x = arcos(x)
=> cos(y) = x
Differentiating both sides w.r.t x
 −sin ( y ) ( dy
dx ) =1 (applying product rule of differentiation and as y=f(x)) (1)
 dy
dx = −1
sin ( y ) =¿ −1
√ ( 1−x2 ) (as sin^(θ) + cos^(θ) = 1)
Again differentiating (1) w.r.t x
d2 y
d x2 sin ( y ) + (( dy
dx )
2
) cosy = 0
 d2 y
d x2 sin ( y ) = −( ( dy
dx )
2
) cosy = −1
1−x2 ∗x
 d2 y
d x2 √1−x2 = −1
1−x2 ∗x

 d2 y
d x2 =
−x
( 1−x2 )
3
2
c) f ( x ) =sinh ( −1 ) ( 5 x )
sinh(y) = (e^(y) - e(-y))/2 (By definition of hyperbolic function)
Now, y = f(x) = arcsinh(5x)
 sinh(y) = 5x
 (e^(y) - e(-y))/2 = 5x
 (e^(y) - e(-y)) = 10x
 (dy/dx)* (e^(y) + e(-y)) = 10 (derivating both sides w.r.t x)
 (dy/dx)* (e^(y) + e(-y))/2 = 5
 (dy/dx)*cosh(y) = 5
 (dy/dx)* √1+sinh ( y )2 = 5 (as cosh (θ)2 – sinh(θ)2=1)
 (dy/dx) = 5
√1+sinh ( y )2 = 5
√1+25 x2 (as sinh(y) = 5x)
Q3:
a) ∫ 1
2 ( 5 x−3 )6 dx,
let, ( 5 x−3 ) =z => 5dx = dz (taking natural derivative on both sides)
 dx = dz/5
Hence, ∫ 1
2 ( 5 x−3 )6 dx = ∫ 1
10 z6 dz = z7
70 + C = ( 5 x−3 ) 7
70 +C
Q4:

a)
i=50 sin(100 πt) t>=0
i = current in mA, t = time in seconds.
Average or mean value of a function f(x) over a finite interval [a,b] is given by
f avg = 1
b−a ∫
a
b
f ( x) dx
Mean value of current in t = 0 to t =10 ms is
iavg = 1
10−0 ∫
0
10
50 sin ( 100 πt ) dt = (50/(10*100 π ¿ ¿ [ cos ( 100 πt ) ] 10
0 = (1/20π ¿∗¿ = (1/20
π ¿∗(1−1) = 0 mAmps.
b) The rms value of a function f(x) over an interval [a,b] is given by,
f rms = √ 1
b−a ∫
a
b
f ( x )2 dx
Hence, irms = √ 1
10−0 ∫
0
10
( 50sin ( 100 πt ) )2 dt
= √ 1
10−0 ∫
0
10
2500 sin2( 100 πt) dt
Let, u = 100πt
 dt = (1/100π)du
Hence, 1
10−0 ∫
0
10
2500 sin2 (100 πt)dt = 250
100 π ∫
0
1000 π
sin2 (u) du
= ( 250
200 π )∗ ∫
0
1000 π
2sin2 ( u ) du = ( 250
200 π )∗ ∫
0
1000 π
(1−cos ( 2u))du

= ( 250
200 π )∗
[ ( u−1
2 sin ( 2u ) ) ]0
1000 π
= 1250 (by putting the limits and evaluating)
Hence, irms= √ 1250 = 35.355 mAmps.
Q5:
y=e
( −t ) sin ( 2 t )
t = time in seconds
y = displacement in meters.
a) Graph of y=f(t):
0 1 2 3 4 5 6
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
y=f(t)(in meters)
TIme t(in secs)
Displacment y(in meters)
b) The velocity of the particle is zero when the displacement is maximum or minimum. At
maximum or minimum velocity = dy
dt = 0
=> 2 e (−t ) cos ( 2t ) – e ( −t ) sin ( 2t ) =0
=> 2 cos ( 2 t ) – sin ( 2 t )=0 ( as e−t ≠0 at finite t≥0 )

=> tan(2t) = 2 => t = (½)( πn+arctan ( 2 ) ) (n=0,1,2…)
Hence, the first zero velocity at n=0 or t = 1
2 arctan ( 2 ) = 0.5536 secs
c) From the graph it can be seen that the velocity of the particle is maximum at time t =
0.5536 secs.
d) The displacement at first point when velocity is zero is
y = f(0.5536) = 0.514198 meters (this can be approximately observed from the above graph
also).
Q6:
f(x) = -5cos(2x) x = [0,π/4]
b) The volume generated for a curve y = f(x) when rotated about x axis over an interval
x=[a,b] given by,
V = ∫
a
b
π y2 dx, where y= f(x)
Hence, Volume generated for this particular case will be
V = ∫
0
π /4
π 25 cos2 2 x dx = 25 π
2 ∫
0
π / 4
2 cos2 2 x dx= 25 π
2 ∫
0
π / 4
(1+cos (4 x )) dx
= 25 π
2 [ x+ 1
2 sin ( 4 x ) ]0
π
4 = 25 π
2 [ π
4 + 1
2 sin ( π ) ]= 25 π 2
8 unit3