MATH130 Assignment 2: Detailed Algebra Problem Solutions

This document contains the complete solutions for MATH130 Assignment 2, focusing on algebra concepts. The assignment includes problems related to exponential decay, where the quantity of medicine in a patient's bloodstream decreases over time, requiring the formulation of equations and determination of the halving time. Trigonometry problems involve the unit circle and trigonometric identities, specifically dealing with cosine and sine functions. Calculus problems involve finding marginal revenue, area calculations using upper sums, and derivative applications to determine the rate of change and maximum/minimum values of a function. The solutions include step-by-step explanations, graphical representations, and detailed calculations to demonstrate the problem-solving process.

Algebra
1)
a) Given,
Q=f(t)=300 ∗ (0.75)
𝑡
40
a)
To write the formula for Q=𝐴𝑒−𝑘𝑡, let g(t) =𝐴𝑒−𝑘𝑡
0.75-1=-0.25
Which is 25% decreasing
We need to find the points, calculate f(0), f(1)
f(0)=300*0.75=225; (0,300)
f(1)=300*(0.75)
1
40=297.85; (1,297.85)
g(t)= =𝐴𝑒−𝑘𝑡
g(1)= =𝐴𝑒−𝑘=297.85
here A=300(from Q=f(t))
so, 297.85=300=𝑒−𝑘
=𝑒−𝑘 = 297.85
300
k=7.192*10−3
b)
finding t half
given Q=300
half of Q=150
plugging this into Q=𝐴𝑒−𝑘𝑡,
150=300𝑒−7,192∗10−3𝑡
Solving for t,
T=96.377minutes

c)
here,
red curve is the graph, violet line is the halving time.
2)
Starting from (1,0), move along the unit circle until the angle θ is formed with x axis.
cosθ is the x coordinate and sinθ is the y coordinate.
cos(90+θ) lies in the second coordinate. Therefore the x coordinate is negative and y coordinate is
positive.
Cos(90+θ)=cos90cosθ-sin90sinθ
=0-sinθ
=-sinθ

So cos(90+θ)=-sinθ
That is, cos (
𝜋
2+θ)=-sinθ
3)
Amplitude is the half range.
That is A=1
2 (35 −(−5)) = 20
Mean value is the average of the two extremes.
That is,
𝑇𝑚 = 35−5
2 =15
Time period for one complete cycle 𝑡𝑝= 12+12=24
So the period = 2𝜋
𝑡𝑝
= 𝜋
12
So putting it all together,
𝐻 = 𝑇𝑚 + 𝐴𝑠𝑖𝑛 (𝑡𝑝 ∗ 𝑡 −
𝜋
2)
That is , 𝐻 = 15 + 20sin ( 𝜋
12𝑡 −𝜋
2)
a)
b)
given,
𝐻 = 𝐴 + 𝐵𝑠𝑖𝑛(𝐶𝑡 + 𝐷)
Comparing with 𝐻 = 15 + 20sin (
𝜋
12𝑡 −𝜋
2)
A=15

B=20
C= 𝜋
12, D=-𝜋
2
4)given f(x)=2𝑥3 − 7𝑥 + 2
a)
f(0) is found by putting x=0
that is f(0)=2
f(1)=-3
f(2)=4
f(-1)=-2+7+2=7
f(-2)=-16+14+2=0
b)
f(x)=2𝑥3 − 7𝑥 + 2=0
x=-2 is a solution.
So x+2=0
Dividing f(x) by x+2
That is, 2𝑥3−7𝑥+2
(𝑥+2) = 2𝑥2 − 4𝑥 + 1
Putting 2𝑥2 − 4𝑥 + 1 = 0
𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐
2𝑎
There fore ,x =1 ± 1
√2
c)

d)
x<-2,
𝑥 < 1 +1
√2,
x< 1 − 1
√2
the region shaded is the required answer.
5)
Given,
𝑅(𝑞) = ln(1 + 1000𝑞2)
Marginal revenue is fond by differentiating with respect to q.
That is, 𝑅′(𝑞)= 2000𝑞
1+1000𝑞2
Marginal revenue when q= 10 is,
𝑅′(10)= 20000
1+100000= 0.199998 0.2⩪
That is R’(x) is the slope of the curve which is=0.2
6)
Given, f(x)=𝑥3 + 2
a)
Graph of y=f(x) is given by,
The interval given is [-1,2]
So the coordinates are (-1,1); (2,10)

b) the lines are x=-1 and x=2.
Subdivision, n =3
Therefore ∆x=𝑏−𝑎
𝑛 = 2−(−1)
3 = 1
So the intervals are, [-1,0]; [0,1]; [1,2]
Area by upper sum, A=∑ 𝑓(𝑥𝑖−1)∆x = f(𝑥0)∆x + f(𝑥1)∆x + f(𝑥2) ∆x3
𝑖=1
Here, f(𝑥0) = −13 + 2 = 1; f(𝑥1) = 0 + 2 = 2;f(𝑥2) = 1 + 2 = 3
So the area=1+2+3=6 sq units
c)
the area can be improved by increasing the number of sub division.
That is value of n can be increased.
7)
Given, ℎ(𝑥) = 50 + 45𝑠𝑖𝑛
𝜋𝑥
50
a)h’(x) is found by differentiating h(x) with respect to x.
that is ,h’(x)=0 + 45 cos (
𝜋𝑥
50) ∗𝜋
50
=45 cos (
𝜋𝑥
50) ∗𝜋
50
b)
h’(40) is found by substituting x=40
that is,
h’(40)= 45 cos (
𝜋∗40
50 ) ∗ 𝜋
50
=45 cos (
4𝜋
5 ) ∗𝜋
50
=-2.2874

This means that slope at the point 40 is -2.2874
c)
maximum is found by putting derivative=0
that is h’(x)=0
implies, 45 cos (
𝜋𝑥
50) ∗𝜋
50 = 0
=cos (
𝜋𝑥
50) = 0
=𝜋𝑥
50 = 𝜋
2,3𝜋
2
x=25,75
that is maximum is at x=25
that is maximum=h(25)
=50 + 45𝑠𝑖𝑛
𝜋∗25
50
=95m
The minimum is at x=75
That is h(75)=50-45
=5m
d)
given, h’’(x)=-45( 𝜋
50)2
sin (
𝜋𝑥
50)
rate of change is greatest at h’(x)=0
that is, at x=25
that is upper limit =25
lower limit is 0,
distance is calculated by integrating h’(x)
that is,
= ∫ 45𝜋
50
25
0
cos (
𝜋𝑥
50)
=9 ∗ 5𝑠𝑖𝑛
𝜋𝑥
50,0,25
=9*5sin𝜋
2
=45m
Hence the answer.