Engineering Mathematics Assignment: Calculus Problems and Solutions

This document presents a complete solution to an Engineering Mathematics assignment. The assignment covers several key calculus concepts, including limits, differentiation, and integration. Question 1 focuses on evaluating limits, while Question 2 involves differentiating various functions using c...Read More

Engineering
Mathematics

Table of Contents
Question 1:.......................................................................................................................................3
Find the following equations.......................................................................................................3
Question 2:.......................................................................................................................................4
Differentiate................................................................................................................................4
Question 3: ......................................................................................................................................5
The energy E stored in a spring is given by the formula E = 200 x ^ 2 where x is the
compression. The rate of change of the energy with respect to the compression is called the
thrust T of the spring...................................................................................................................5
Question 4:.......................................................................................................................................6
Find dy/ dx at the point (0, 2) given that y ^ 3 = 8 + xe^ y ........................................................6
Question 5. ......................................................................................................................................6
The total force due to a distributed load acting on a beam from x = a to x = b is given by........6
∫a-b f(x)dx....................................................................................................................................6
Question 6:.......................................................................................................................................8
Evaluate the following integrals using integration by parts:.......................................................8

Question 1:
Find the following equations
(a) limx→−5 (25 – x^2 / x^2 + 2 x- 15)
Ans: => 25-x ^ 2 / x^2 + 2 x-15
= - x-5 / x-3
= 25- x^2 / x^2 + 2 x -15
Factors: -x^2 + 25 : (x+5)(x-5)
= - (x+5)(x-5) / x^2 + 2 x - 15
Factor x ^ 2 + 2 x – 15
=-(x+5)(x-5) / (x-3)(x+5)
cancel the common factor : x+5
= - x-5/ x-3
= lim x->-5 (-x-5 / x-3)
Putting value x= -5
= – -5-5 / -5-3
= - 10 / 8
cancel common factor:
= -5/4
(b) limx→∞ 4x^2 – 19 / 3x^2 + 11
Ans: limx→∞ (4 x^2-19 / 3 x^2+11)
Divide by highest denominator power:
4- 19/x^2 / 3+ 11/x^2
= limx→∞ (4 - 19/x^2 / 3+ 11/x^2)
= limx→∞ (4 - 19/x^2) / limx→∞ (3+ 11/x^2)
= limx→∞ (4 – 19/x^2) = 4 : simplify
= limx→∞ (3+ 11/x^2) = 3 : simplify
= 4 / 3

Question 2:
Differentiate
(a) e ^ sin(x 2 ) ;s
Ans : d / dx (e ^sin x^2)
= e ^sin (x^2) cos(x^2). 2X
Apply chain rules:
= d / du (e^u)d/dx(sin(x^2))
= d / du (e^u)= e^u
= e ^u
apply this common derivations:
d / d x (sin x^2) = cos (x^2).2x
= e ^u cos(x^2).2x
substitute back u= sin(x^2)
= e^sin(x^2) cos(x^2).2x
(b) ln sin−1 x 3 , where sin−1 x is an inverse function for sin x, i.e. sin−1 x = arcsin x
Ans: d/ dx (sin ^-1 x^3)
= d/dx( arcsin(x^3) = 3x^2 / √1-x^6
= d / dx (arcsin (x^3))
apply chain rule:
d/ du ( arcsin (u)) d/ dx(x^3)
= d/ dx (x^3) = 3x^2
= 1/ √1- u^2 . 3X^2
substitute back u = x^3
= 1/ √1- (x^3)^2 . 3X^2
Simplify :
= 1/ √1- (x^3)^2 . 3X^2 : 3x^2 / √1-x^6
= 3x^2 / √1-x^6

Question 3:
The energy E stored in a spring is given by the formula E = 200 x ^ 2 where x is the
compression. The rate of change of the energy with respect to the compression is
called the thrust T of the spring.
(a) Express the thrust T of the spring in terms of x;
Ans: R indirectly propositional to 1/E
T = c. E
T= thrust
C = Compression
E= energy
(b) Find the energy E at x = 1.5;
Ans: E= 200. (1.5) ^ 2
= 2.25 x 200
= 450 (energy)

Find the thrust T at x = 1.5;
Ans: T= 450 x 1.5 {thrust = Compression x energy}
= T= 675
Find the compression x when the energy E = 400;
Ans: E= 200(x)^2
= 400 = 200 X^2
x^2 = 400 / 200
x^ 2= 2
x= √ x
(e) Find the compression x when the thrust T = 100
Ans: Compression = Thrust / Energy
x= 100 / 450
x= 0.2
Question 4:
Find dy/ dx at the point (0, 2) given that y ^ 3 = 8 + xe^ y .
Ans: d / dx (y . 3) = d / dx (8+ xe^y)
= d/ dx (y . 3) = 3d/dx(y)
= d/dx(8+xe^y)= e^y+e^y x d/dx(y)
= 3 d/dx(y) = e^y+e^y x d/dx(y)
islolate in terms of y:
y' = e^y / -xe^y +3
d/dx(y)= e^y / -xe^y +3
Question 5.
The total force due to a distributed load acting on a beam from x = a to x = b is given by
∫a-b f(x)dx
where f(x) is the force at the point x.
Ans : ∫a->b f(x)dx
= ∫a->bfxdx
Compuet the integral : ∫ fx dx= fx^2/2 +c
= f.∫ xdx

= f x^1+1/ 1+1
Simplify:
=fx^2/2
(a) Find the indefinite integral of the force
f(x) = 6x − 2x 3 − x √ x + 3e ^x
i.e. ∫ f(x)dx
Ans: ∫ ( 6 x − 2x 3 − x √ x + 3e ^x)
= apply the sum rules:
= ∫ 6x dx- ∫ 2x^3dx - ∫ x √ x dx + ∫ 3e ^x dx
= ∫ 6x dx = 3x^2
= ∫ 2x^3 dx= x^4 / 2
=∫ x √ x dx = ½ x^2 √ x – 1/10 x^5/2
=∫ 3e^x dx= 3e^x
Simplify:
= 3x^2-x^4/2-1/2x^2 √ x – 1/10 x^5/2 + 3e^x
(b) Hence calculate the exact value of the total force if a = 0 and b = 1, i.e. evaluate the
integral
F =∫ 1 - 0 (6x − 2x ^3 − x √ x + 3e ^ x ) dx
Ans: ∫ 1 – 0 (6x − 2x ^3 − x √ x + 3e ^ x ) dx
Compute integral:
∫ 1 – 0 (6x − 2x ^3 − x √ x + 3e ^ x ) dx = 3x^2-x^4 / 2-1 / 2x^2 √ x +1/10 x^5/2 + 3e^x+c
Apply chain rules:
= ∫6xdx=3x^2
= ∫2x^3dx={x^4}/{2}
= ∫2x^3dx
taking constant out:
= 2 ∫ x^3dx
simplify by using power rule:
= 2. {x^{3+1} /{3+1}: {x^4}/ {2}
= {x^4} / {2}
= ∫x √ x dx={1}/{2}x^2{x}-{1}/ {10}x^{5}/ {2}}

Compute and simplify:
=3e+ {21}/ {10}-3
= =3e-[9} / {10}
Question 6:
Evaluate the following integrals using integration by parts:
(a) ∫ in x / x^5 dx;
Ans: ∫ x / x^5 dx = -{1} / {3x^3}+C
= ∫ x / x^5 dx
= x / x^5= x^{-4}
= ∫ x^{-4} dx
applying power rule:
= x^-4+1 / -4 +1
Simplify:
x^-4+1 / -4 +1= -1/3x^3
= -1/3x^3
= -1/3x^3 +c
(b) ∫ (2x + 5) cos 4x dx.
Ans: Apply integration u= (2x+5), v= cos(4x)
= {1} / {4} sin (4x)(2x+5)- ∫ ½ sin(4x) dx
= 1/ 2 sin (4x) (2x+5) - (-1/8 cos(4x))
Simplify:
= ¼ sin (4x) (2x+5)+ 1/8cos (4x)
= ¼ sin (4x) (2x+5) + 1/8 cos (4x) +c