Semester 1 Engineering Mathematics Problems and Solutions

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ENGINEERING MATHS

Contents
LO1.......................................................................................................................................................3
a.........................................................................................................................................................3
b.........................................................................................................................................................4
c.........................................................................................................................................................4
d.........................................................................................................................................................6
e.........................................................................................................................................................7
f.........................................................................................................................................................8
Lo2......................................................................................................................................................10
a.......................................................................................................................................................10
b.......................................................................................................................................................10
c.......................................................................................................................................................11
d.......................................................................................................................................................12
e.......................................................................................................................................................13
LO3.....................................................................................................................................................14
a.......................................................................................................................................................14
b.......................................................................................................................................................16
c.......................................................................................................................................................16
d.......................................................................................................................................................16
e.......................................................................................................................................................17
f.......................................................................................................................................................17
g.......................................................................................................................................................18
Lo4......................................................................................................................................................19
a.......................................................................................................................................................19
b.......................................................................................................................................................19
c.......................................................................................................................................................19
d.......................................................................................................................................................20
References...........................................................................................................................................22
List of figures
Figure 1: Part D.....................................................................................................................................6
Figure 2: Curve of radioactive...............................................................................................................7
Figure 3: curve of normal distribution.................................................................................................11

Figure 4: Curve of P(t).........................................................................................................................15
Figure 5: values for test.......................................................................................................................18

LO1
a.
To find the variables x, & the variable y, & the variable z required to utilising the
formula provided.
As, to get the values of the dimension of each unit in every part must be same
M constitute mass, T constitute time & L constitute the length, so to quantify the
power of the M, power of the T & power of the L, so the quantities unit will be on the
side of right & dimension of the velocity is on the other side & power need to be
mentioned.
The each dimension power needs to be balance to construct the equation.
As for M-
Where, X=-y (Eq. 1)
2x= 1 (Eq. 2)
On balancing the above equation, x will be ½ & y will be -1/2.
As for L-
Where, x+3y=3z-1 (Eq. 3)
Now, put the value of x & value of yin the eq. 3 & solve the equation the z’s value
will be comes zero.
(-1/2 + 3*(1/2) + 3* (z-1) as z= 0/3, & z = zero)
To make the eq. x & y & z are again put in the equation & the output equation will
be :
 = (C (P/))1/2

b.
To get equation’s in sequence to correspond with quantities of force like viscosity,
radius & velocity. The force is directly proportional to velocity, radius & viscosity
raised to the some power that provides the eq.:
F ∝ ηp * rq * vr
For removing the proportional replace it with the constant R for making the correct
eq.
F = R* ηp * rq * vr
To place the values, the dimension of the unit must be similar, as
[M1L1T-2] = [M1L-1T-1]r * [L1]q [L1T-1]p]]
[M1L1T-2] = [(Mp) (Lr+q-p )(T-r-p)]
To get the value of the p, q & r respectively, need to balance the dimension power.
- 1= p
- -r-1 = -2 as r= 1
- 1-1 +q = 1 as q = 1
By putting the value of p, q & r values respectively, the output comes to be:
F is C v1 r1 η1
c.
The above formula is utilise to get the total of the first N no. of the arithmetic
progression:
a constitutes the first, n will constitutes the total no. of terms are present in an
arithmetic progression & d constitutes the difference that is between the term of two
consecutive.
To compute the eight term sum, n is 8 & a is 3 as given.
S8 = (8 divides 2 * ((2 multiply 3) + (8-1) * d)) gives
S8 = (4 multiply (6 + 7 * d))

To compute the five term sum, n is 5 & a is 3 as given.
S5 = (5 divides 2 * ((2 multiply 3) + (4) * d))
S5 = (5/2 multiply (6 + 4 * d))
Contrast the above eq. to find the d value,
4 multiply (6 + 7 *d) = 2 multiply (5/2 multiply (6 + 4 *d)) (2 will be multiply as
eight term sum will be double that of the five term)
24 + 28 *d = 2 multiply (15 + 10 *d) will give 8 *d = 6, on solving further, d is 6 on 8
that provides 0.75.
As, the difference between an arithmetic progression will be 0.75
This formula will require to get the first sum of the geometric progression, a
constitutes the term first, r will constitute the ratio among the two term of consecutive
& n will constitute the no. of the term in the GP.
To compute the, five term sum, now put a, n & r value in the equation provided
above.
S5 = (8 multiply ((1-(-1 divides 2)5) divides with (1- (-1 divides 2))))
S5 = (8 multiply ((1+ 1 divides 32) divides with (1+1 divides 2)))
S5 = (8 multiply ((33 divides 32) divides with (3 divides 2)))
S5 = (8 * 33 * 2) divides (32 * 3))
S5 = 11 divides 2 = 5.5

d.
Figure 1: Part D
As plane altitude at B point make the angle 200 with respect to the ground & the C
point altitude make the angle 600 with ground.
To get h value, the triangles CAC & BAB need to be examine.
Utilising the angle of tan 200 as,
tan 200 is h divides (d + x)
To abolish the d value, it only left with the two variables as the d value will be
compute utilising the formula of distance.
(Distance = time * speed )
Time is 1/60 hrs.
Speed is 600 m/hr.
Therefore d will be [1 * 600]/ 60 that will provide the ‘d’ value will be 10 miles.
Now converting the x in the h,
In the triangle BAB,
Tan 200 is h divides (10 + x) as,
x = (h – 3.639)/ 0.3639 ( put tan 200 value in eq. & solve )

In the triangle CAC,
tan 600 = h divides x
(3)1/3 = (h / ((h – 3.639) divides 0.3639))
(3)1/3 = ((h × 0.3639) divides (h- 3.639))
(1.7320* h – 0.3693* h) is equal to 5.8352 (put value √3)
1.3627* h = 5.8352 & solving it more,
h is 4.28 (in miles)
The final plane altitude is 4.28 miles.
e.
1.
Figure 2: Curve of radioactive
2. The formula will be indicate the decayed amount material in the that time,
A is A0 (1/2)t/t ½ as, t1/2 constitute the time when its amount get reduced to its half
called as half life, & A0 constitutes the amount at the initial time.
R is R0 e−λt, is utilise to get the rate on which the element of radioactive gets decay
in the provided time period, R0 constitute the rate of decay at the initial time , t
constitutes the time & λ constitutes, constant of the radioactive decay.

3. To get, rate of the decay on the completion of the three weeks, t1/2 will be one week
as given & t is 3.
R is 20 * e−λ×3
For compute , the equation is  = L*n*2 (divides) t ½ that provide  = 0.693 / 1 is
0.693.
R is 20 * e−0.693×3
R is 2.50 counts per second (Yubo,, 2015).
f.
1.
The equation of the exponential in this question Y= c*ax, as Y constitutes the
difference among in the given year & 200.
X constitutes the people count with the phones in the year given.
For compute the constant value a,
a = √ (375 / 275), a’s value is 1.29.
To abolish the variable of c, the value will be compute by utilising the above
computed.
Y= c ×(1.29)x, to compute y, put the value of y is 1725.
Then x will come as to be 8.
For solving further, c will be 225.
Therefore the output will be, Y= (1.29x) * (225).
2. To compute the phone cells that will be bought in 2015 year, x is 2025-2000 that
is 25.
After applying the constant & x’s values,
Y= (225) multiply to (1.2925) is equal to 130895.6685 .
3. To compute the occurred change, x’s value will be 6 for the 2006 year & 10 for
the 2010.
Y2 is equal to (225) * (1.296) =2971.307
Y1 is equal to (225) * (1.2910) =1036.86
For compute the Y change w.r.t X,
Y2 – Y1 divides X2 – X1 implies (2971.307-1036.86) divides to (10-6) is equal to
483.61.

4. As, compute the change of instantaneous rate of 2006 year, take the approx. value
that is about six but not actually six.
Therefore x will be 5.999 & x will be 6.001 may be possible values.
As, compute the occurred change in the y w.r.t the x in 2006 year, put the Y & X
respectively in the eq.:
Y2 – Y1 divides X2 – X1 is equal to (1037.125636 – 1036.597579) divides (6.001-
5.999) = 264.0285.

Lo2
a.
Formula to estimate the mean value is:
As , xconstitutes value of mean, n constitutes the no. of the sample data entry & x
constitutes the value of middle.
provide the 49.2 the answer.
The value of mean will be 49.2.
As to compute, standard deviation the formula is:
S= √ [ 1
1−n ]∗∑
i=1
n
(x−x)2
Put the n & mean value of the n in the formula, S is 17 on solving. As, standard
deviations will be 17 (Braun, 2017).
b.
The distribution as normal will easy to compute the data as the records will be present
in the simple way.

Figure 3: curve of normal distribution
N is 20,
P is 25/100,
To compute the μ ‘s value the formula will be:
μ = n*p,
μ is equal to (20) * (0.25)
it provides, μ is 5
formula to evaluate σ, will be σ2 is equal to n*p (1-p) as provided,
σ2 = 3.75 (n & p will be put)
Will taking the sq. root, σ2, σ will be 1.94 (Zhang, 2018).
c.
Life is 10 years as given
The standard deviation will be 2.
To evaluate the Z, will be Z is equal to X−μ
σ .
Put values of σ, Z,& μ is Z is equal to X−μ
σ .