Calculus Assignment: Differentiation, Integration, and Applications

This calculus assignment solution covers a range of problems from differentiation and integration to applications such as finding average and RMS values of functions. It includes detailed solutions using techniques like the quotient rule, hyperbolic functions, and definite integrals. The assignment further explores curve sketching and volume calculation using integration, providing a comprehensive understanding of calculus concepts. The solutions are presented step-by-step, making it a useful resource for students studying calculus. The assignment includes problems involving trigonometric functions, exponential functions, and curve analysis, offering a complete review of calculus fundamentals.

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