Solutions to Maths Assignment

This document provides solutions to a Maths assignment, including integration and differential equations. It covers topics such as exponential decay, rate of change, and area calculations. Suitable for various courses and universities.

Maths Assignment
Student Name:
Instructor Name:
Course Number:
23 April 2019

SOLUTIONS TO THE MATHEMATICS
ASSIGNMENT
6 a)
Let t=3x2+1
dt =6x dx
dt
6 = xdx
Substituting dt
6 = xdx in ∫ x
√3 x2 +1 dx we get
1
6 ∫ dt
t0.5 = 1
6 ∫ t -0.5 dt
1
3t0.5 + c
= 1
3 √3 x2 +1 + c
b) Shaded area= ∫
0
π
(1+sin 2 x ¿) dx ¿ -∫
0
π
sec2 xdx
Working each integration separately)
∫ (1+sin 2x)dx=¿x- cos(2 x)
2 ¿+c
When x= π
4 then¿ x- cos(2 x)
2 ¿ = π
4 − 1
2 cos( π
2 ¿=0.7855-0.4998=0.2857
When x=0 then ¿ x- cos(2 x)
2 ¿=0- 1
2cos 0= 0-0.4970=-0.4970
0.2875- - 0.4970=0.7827 square units
∫ sec2x dx= tan x + c

When x= π
4 , thentan x=tan π
4 = 0.001371
When x=0 then tan x=tan 0=0
0.001371-0=0.001371
Shaded area= ∫
0
π
(1+sin 2 x ¿) dx ¿ -∫
0
π
sec2 xdx
=0.7855-0.0.001371
=0.784129 square units
7 i) The equation takes the form of an exponential decay where the height of water decreases
with increase in time. The decrease in the height of water is in the form of rate of change which
leads to a differential equation.
The differential equation is;
−dh
dt =k √ h
Rate of height of water remaining is proportional to square root of height.
Because the height decreases we introduce a minus sign.
Thus
Rate of height means i.e.
Because of the decrease we shall have -
Therefore - = k .
K (constant of proportionality) has been introduced because of the existence of direct variation
between rate of change in height and square root of height.
ii) –dh= k √h dt
dh= - k √h dt
∫dh= ∫- k√h dt

∫ dh= - k √h ∫ dt
h=- k √h t + c
iii) h= - k √h t + c
when t=0 ,h=2
2= (−k √2 )0 +c
2=c .
Hence
h= - k √h t + 2
when t=10, h=1.5
1.5= -10k √1.5 + 2
- 0.5 = -10k √1.5
−0.5
−10 = k √1.5
0.05
√1.5 = k
0.0408 =k
Particular equation is h= -0.0408 √h t + 2
iv) when h=1 we have
1=-0.0408 √1 t + 2
0.0408t = 1
t= 1
0.0408 = 24.51
=25 minutes (to nearest whole number)

8 . i) dA
dt = n √2 A−5
Where n is a positive integer
ii) dA= n √ 2 A−5 dt
∫dA = ∫ n √ 2 A−5 dt
∫dA = n √ 2 A−5 ∫dt
A= n √2 A−5 t + c
When t=0, A=7
7= ( n √ 2× 7−5 )0 + c
7=c
When t=10, A=27
27= 10n ( √ 2× 27−5 ) + 7
27-7=10n √49
20=70n
2
7 =n
Hence A= 2
7 √2 A−5 t + 7
iii) A= 2
7 ( √ 2 A−5 )t + 7
when t=20

A= 2
7 ( √ 2 A−5 )20 + 7
A-7= 40
7 ( √ 2 A−5 )
7
40 ( A−7 )=√2 A−5
7
40 A - 49
40 =√ 2 A−5
Squaring both sides we get
49
1600A2 - 686
1600A+ 2401
1600 =2A-5
Multiplying both sides by 1600
49A2-686A+2401=3200A-8000
49A2 -686A-3200A+2401+8000=0
49A2-3886A+10401=0
Using Quadratic formula
A=3886 ± √15100996−4 × 49 ×10401
2 ×49
A= 3886 ± √ 13062400
98
A= 3886 ±3614.2
98 = 3886+3614.2
98 =76.53
OR
A=3886−3614.2
98 =2.773
In conclusion the area covered after 20 weeks is 76.53 m2.