Unit 4 Maths: Calculus Techniques for Engineering Technicians - BTEC

This assignment focuses on applying calculus techniques to solve engineering problems. It covers differentiation of various functions, including polynomial, exponential, and trigonometric functions. Integration techniques are applied to solve problems involving summation, and graphical methods are used to analyze exponential growth and decay. The assignment also includes error analysis to compare graphical and analytical solutions. Specific tasks involve finding derivatives and integrals of given functions, determining gradients, calculating temperature changes, and finding mean voltage and charge. The student work provides detailed step-by-step solutions for each problem.

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