MATH130 Assignment 2: Solutions to Algebra Problems with Graphs

This document contains the complete solutions to MATH130 Assignment 2, covering various algebra concepts. The first problem involves modeling exponential decay of a medicine in a patient's bloodstream, including writing a formula in the form Q = A * e^(-kt), calculating the half-life, and sketching a graph. The second problem focuses on trigonometric identities and the unit circle to determine the equivalent of cos(θ + π/2). Further solutions include problems on amplitude, period, and general forms of trigonometric functions, finding the zeros of a polynomial function, and solving inequalities. The assignment also covers marginal revenue, the calculation of area using upper sums, and applications of derivatives to find maximums, minimums, and rates of change. The final problems involve finding the area under a curve and applying calculus concepts to analyze a function's behavior.

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∗10−3 t
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 ¿+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,
T m= 35−5
2 =15
Time period for one complete cycle t p= 12+12=24
So the period = 2 π
t p
= π
12
So putting it all together,
H=T m + Asin (tp∗t− π
2 )
That is , H=15+20 sin ( π
12 t − π
2 )
a)
b)
given,
H= A +Bsin (Ct + D )

Comparing with H=15+ 20 sin ( π
12 t− π
2 )
A=15
B=20
C= π
12, D=- π
2
4)given f(x)=2 x3 −7 x+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)=2 x3 −7 x+2=0
x=-2 is a solution.
So x+2=0
Dividing f(x) by x+2
That is, 2 x3−7 x +2
( x+ 2 ) =2 x2 −4 x +1
Putting 2 x2 −4 x +1=0
x=−b ± √ b2−4 ac
2 a
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+1000 q2 )
Marginal revenue is fond by differentiating with respect to q.

That is, R ' ( q )= 2000 q
1+ 1000 q2
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=∑
i=1
3
f ( xi−1 ) ∆ x=f ( x0 ) ∆ x+ f ( x1 ) ∆ x+ f (x2)∆ x
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+ 45 sin 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 ( 4 π
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 , 3 π
2
x=25,75
that is maximum is at x=25
that is maximum=h(25)

=50+45 sin 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,
¿∫
0
25
45 π
50 cos ( πx
50 )
=9∗5 sin πx
50 ,0,25
=9*5sin π
2
=45m
Hence the answer.