Calculus Techniques for Solved Assignments, Essays, Dissertations

This document contains solved assignments on Calculus Techniques including differentiation, integration, and more. It also includes a graph and calculations for temperature and voltage. The subject and course code are not mentioned.

Running head: CALCULUS TECHNIQUES 1
Calculus Techniques
Name
Institution

CALCULUS TECHNIQUES 2
Task 1-Part a
Question a
y=4 x3−9 x2 +3 x−4
dy
dx =4 ( 3 ) x3−1−9 ( 2 ) x2−1 +3 ( 1 ) x1−1−0=12 x2 −18 x +3
Question b
s=4 e3 t −2 e−3 t
ds
dt =4 ( 3 ) e3 t−2 ( −3 ) e−3 t =12 e3 t −6 e−3 t
Question c
y=4 cos ( 5 x ) −3 sin (2 x )
dy
dx =4 d
dx cos ( 5 x )−3 d
dx sin (2 x )
In evaluating d
dx ( 4 cos ( 5 x ) ) ,let 5 x=u
cos ( 5 x )=cosu
du
dx =5 , d
du ( cosu )=−sinu
d
dx (4 cos ( 5 x ) )=4 du
dx × d
du =−4(5)sinu =−20 sin (5 x )
In evaluating d
dx ¿let 2 x=m

CALCULUS TECHNIQUES 3
sin ( 2 x ) =sin ( m )
dm
dx =2, d
dm ( sin ( m) )=cos ( m)
d
dx ¿
dy
dx =−20 sin ( 5 x ) −6 cos (2 x )
Task 1-Part b
Question d
∫(8¿ x3 +5 x2 −4 x )dx=8∫ x3 dx +5∫ x2 dx−4∫ x dx ¿
¿ 8 x3 +1
3+1 +5 x2 +1
2+1 −4 x1+1
1+1
¿ 8 x4
4 +5 x3
3 −4 x2
2
¿ 2 x4 + 5
3 x3 −2 x2 +C
Question e
∫(¿ e2 t−3 e6 t ) d t =∫e2 t dt−3∫e6 t dt ¿
In evaluating ∫ e2 t dt ,let 2 t=u , dt=1
2 du
∫ e2 t dt= 1
2 ∫eu du= 1
2 eu= 1
2 e2 t

CALCULUS TECHNIQUES 4
In evaluating 3∫ e6 t dt ,let 6 t=u , dt= 1
6 du
3∫ e6 t dt=3 × 1
6 ∫eu du= 1
2 eu= 1
2 e6 t
Therefore ,∫(¿ e2 t−3 e6 t )d t= 1
2 e2 t −1
2 e6 t +C ¿
Task 2(a)
Part a
v=60 ( 1−e−0.2 t )
t 0 2 4 6 8 10 12 14
v 0 19.78 33.04 41.93 47.89 51.88 54,57 56.35

CALCULUS TECHNIQUES 5
Gradient at t=9 seconds =56−42
12−5 = 14
7 =2 v s−1
Part b
Gradient= d v
dt = d
dt ( 60 ( 1−e−0.2t ) )= d
dt (60 )− d
dt (60 e−0.2 t )
¿ 0−60(−0.2)(e−0.2t )
¿ 0−60 (−0.2 ) ( e−0.2t ) =12 e−0.2 t
Gradient ( t=9 seconds )=12 e−0.2 (9 )=1.9836 v s−1
Part c

CALCULUS TECHNIQUES 6
% error= Graph value− ActualValue
Actual Value × 100 %
¿ 2−1.9836
1.9836 × 100 %=0.8275 %
Task 2(b)
Part d
θ( temperature)=25+ 80 e−t
t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
θ 10
5
73.52 54.43 42.85 35.8
3
31.57 29.9
8
27.42 26.4
7
25.89 25.53 25.3
3
25.2

CALCULUS TECHNIQUES 7
Part e
I nitial temperature ( when t=0 )=105 ℃
Temperature ( as t → ∞ )=25 ℃
Part f
Slope ( at t=3 seconds ) = 20−35
5.2−1.4 =−3.9474
Slope= d θ
dt ( 25+80 e−t )=0+80 (−1 ) e−t =−80 e−t
Slope ( t=3 ) =−80 e− ( 3 ) =−3.9830
% error=−3.9474−(−3.9830)
−3.9830 =0.8938 %
The slopes using the two methods is similar but with a percentage error of 0.8938 %
Task 3
Part a
voltage , v=220 sin (140 πt)
mean voltage= total voltage
time =
∫
t =0
3.6
220 sin (140 πt )
3.6 ms
∫
t =0
3.6
220 sin ( 140 πt ) =220 ( 1
140 π ) cos ( 140 π t ) ∨¿t =0
t =3.6¿

CALCULUS TECHNIQUES 8
¿ 220
140 π ( cos ( 140 π ×3.6 )−cos (140 π ×0 ) ) = 220
140 π ( 1−1 )=0
mean voltage= totalvoltage
time = 0
3.6 =0
Part b
Charge , I = ∫
t =0
t =7
3 t2 dt =3 ∫
t=0
t=7
t2 dt
¿ 3 ( t2 +1
2+1 ) ∨¿t =0
t =7 ¿
3 ( t3
3 )∨¿t=0
t=7= ( t3 ) ∨¿t =0
t =7 =73−03=343 Coulombs ¿ ¿
Charge=343 Coulombs