Calculus Assignment Solutions for BTEC Higher National Engineering

This document presents a solved assignment on engineering mathematics, focusing on calculus. It includes problems related to finding maxima and minima of functions, differentiation, and integration. The assignment demonstrates step-by-step solutions for various calculus tasks, such as finding turning points, determining concavity, and evaluating definite integrals. Specific problems involve finding derivatives of complex functions (e.g., v = (t^2 + 6)^2, v = loge(2t), v = sin(2t^3+4t-2)), evaluating definite integrals (e.g., ∫(1-e^(-t))dt, ∫e^(-t) dt), and solving practical problems related to electrical circuits involving integration. The solutions are detailed and include the application of calculus principles to solve engineering-related problems. The document also includes a reference list of relevant textbooks and articles.

Engineering Mathematics 1
ENGINEERING MATHEMATICS
Student’s Name
Professor
Course
Institution
City/State
Date

Engineering Mathematics 2
Engineering Mathematics
Task #4
Part #1
Solution
Y-axis
Max turning point
X-axis
Min turning point
y = 3x2-5x
Dy/dx = y’= d (3x2-5x)/dx
d (3x2-5x)/dx = 6x-5
y’= 6x-5
At the turning point the gradient is zero thus y’ = 0
6x-5 = 0
x = 5/6
But y = 3x2-5x therefore inserting for the values of x in the equation we get the value for y
y = 3(5/6)2-5(5/6)
= 25/12 – 25/6

Engineering Mathematics 3
= 25/4
Therefore the turning point is (5/6, 25/4)
To determine whether the curve is a maxima or a minima then we differentiate it second time,
second derivative.
y’ = 6x-5 = dy/dx
d2y/dx2 = y” = d2(6x-5)/dx2
d2 (6x-5)/dx2 = 6
but y”<0 means a negative concave and y”>0 positive concave
Since the answer after the second derivative is a positive value then it means the concave curve
faces up.
Therefore the turning point is (5/6, 25/4) and the curve is a Minima.
Part #2
Find the maxima and the minima values for the function;
y = x3- 4x + 6
dy/dx = y’= d (x3- 4x + 6)/dx
D (x3- 4x + 6)/dx = 3x2-4
Dy/dx = 3x2-4
At the turning point the gradient is always zero, therefore, y’= 0
Dx/dy = 3x2-4 = 0
x2 = 4/3
x = +2/ √ 3 or -2/ √ 3
To determine the maxi and minima then we go to the second derivative;
Dy/dx = 3x2-4

Engineering Mathematics 4
d2y/dx2 = y” = d2 (3x2-4) dx2
d2y/dx2 = 6x
Y” = 6x
When x is +2/ √ 3
y” = 6(+2/ √ 3 )
= 12/ √ 3 thus it is the value of the minima
When x -2/ √ 3
y” = 6(-2/ √ 3 )
= -12/ √ 3 thus it is the value of the maxima
Maxima, -12/ √ 3 ,
Minima, +12/ √ 3
Task #1
a) v = (t2 + 6)2
Solution
dv/dt = v’ =d(t2 + 6)2/dt
dv/dt = 2 x 2t (t2 + 6)
= 4t (t2 + 6)
But t = 5therefore,
4(5)*((5)2 + 6) = 620 seconds
= 620 seconds
b) v = (3t3-4t+6)3

Engineering Mathematics 5
Solution
dv/dt = y’ = d(3t3-4t+6)3/dt
dv/dt = (9t2-4)( 3t3-4t+6)2(3)
But t = 5, therefore,
= (225-4)*(375-20+6)2(3)
=221*3612 *3
c) v=loge(2t)
Solution
dv/dt = loge(2t)=ln(2t)
d(ln2t)/dt = 2/t
=2/t
But t=5
=2/5sec
d) v=4 e−0.5 t
Solution
Note that: dv/dt = f(x)y=f(x)y *y
dv/dt =4(−0.5) e−0.5 t
= (-2.0¿ e−0.5 t
But t = 5thus,
= -2.0(e−2.5)
= -0.16
=0.16 sec
e) v=sin(2t3+4t-2)
Solution
dv/dt =d sin(2t3+4t-2)dt

Engineering Mathematics 6
= (6t2+4)*Cos (2t3+4t-2)
t= 5
= (6(5)2+4)*Cos (2(5)3+4(5)-2)
= 154 Cos (127) seconds
=35.8 sec
f) v = Cos(3t4-5t+4)
Solution
dv/dt = d Cos(3t4-5t+4)/dt
dv/dt = -sin (3t4-5t+4)(12t3-5)
= sin (3t4-5t+4)(-12t3+5)
= sin (3(5)4-5(5) +4)*(-12(5)3+5)
=-1495sin (604)
= 1087sec
Task #3
Part #1
y =∫
2
4
(¿ 1−e−t )dt ¿
Solution
y =∫
2
4
(¿ 1) dt ¿- ∫
2
4
(¿ e−t ) dt ¿
For ∫
2
4
(¿ 1)dt ¿ = (t+ c)
For ∫ (e−t )dt = -e−t +c
Combining the two solution,
= 4 [t+e−t+c]
2

Engineering Mathematics 7
= [4+e−4+c] - [2+e−2 +c]
= 2+ e−4 -e−2
= 2 +e−2
=2.14unit sq.
Part #2
y =∫
2
4
(¿ e−t ) dt ¿
Solution
y =∫
2
4
(¿ e−t ) dt ¿ = (−e−t +c) dt 4
2
= [−e−t +c ¿ – [(−e−t +c)¿
= [(−e−4) + ¿)]
= 0.12 unit sq.
Task #2
Part #2
IL =1/ L∫ cos ( 100t ) dt
Solution
1/ L∫ cos ( 100t ) dt = Sin (100t)/100 + C
= Sin (100t)/100 + C
1/L {[Sin (100t)/100 + C] - [Sin (100t)/100 + C]} 0.9
0
= Sin (100*0.9)/100
= Sin (90)/100

Engineering Mathematics 8
0.009A
Part #1
I = E/R e−t /RC
Solution
E/R∫
0
0
(¿ e−t / RC)¿
E /R∫e−t / RC dt
But E /R= ε /R
Therefore
E /R∫e−t / RC dt= ε / R ∫ e−t /RC dt= - ε / R *RC [e−t / RC]
= - ε C [e−t / RC]
When t = 0 then,
e−t / RC] = 1
Therefore
- ε C [e−t / RC] = - ε C
= ε C
When t = 1 seconds
= - ε C [e−1 / RC]
C = 1, E = 10, R =1
=-10*0.36
= -3.6

Engineering Mathematics 9
References List
Brebbia, C.A. and Walker, S., 2016. Boundary element techniques in engineering. Elsevier.
Kapadia, A.S., Chan, W. and Moyé, L.A., 2017. Mathematical statistics with applications. CRC
Press.
Lu, Y. and Peng, Z., 2017. Maxima and minima of independent and non-identically distributed
bivariate Gaussian triangular arrays. Extremes, 20(1), pp.187-198.
Stroud, K.A. and Booth, D.J., 2013. Engineering mathematics. Macmillan International Higher
Education.

Engineering Mathematics 10

Engineering Mathematics 11