Mathematical Calculations for Science

This document contains solved problems for logarithms, algebra, differentiation, integration and more for Mathematical Calculations for Science. It includes tables, graphs and equations for better understanding.

APPLIED SCIENCE
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Mathematical Calculations for Science

APPLIED SCIENCE
Contents
Logarithms and Indices –.................................................................................................................3
Task-1 Solution............................................................................................................................3
Task-2 Solution............................................................................................................................5
Logarithms and Algebra-.................................................................................................................6
Task-1, Solution...........................................................................................................................6
Task-2, Solution...........................................................................................................................9
Differentiation and integration-.....................................................................................................10
Task-1, Solution.........................................................................................................................10
Task-2, Solution.........................................................................................................................13
Circular measure and trigonometry –............................................................................................15
Task-1 Solution..........................................................................................................................15
Task-2 Solution..........................................................................................................................16
Task-3- Solution........................................................................................................................17
Task 4- Solution.........................................................................................................................18
References......................................................................................................................................20
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Logarithms and Indices –
Task-1 Solution
1a) C2 x C3 (On multiplication power are added for same base)
⇒ C2+3=C5 Ans
1b) b3 x b4 xb5 (On multiplication power are added for same base)
b3+4 +5=b9 Ans
1c) ( a2 )3 (On power of the power is multiplied)
( a2 x 3 )=a5 Ans
1d) m7
m3 (Power is subtracted on division)
( m7−3 ) =m4 Ans
1e) ( 5 a3 b c4 )2 (On power of the power is multiplied)
25 a3 x2 b1 x2 c4 x2 =25 a6 b2 c8 Ans
1f) p−3 q2 (The negative power goes to the denominator)
q2
p3 Ans
1g) ( b2
c4 )
3
(On power of the power is multiplied)
b2 x3
c4 x3 = b6
c12 =b6 .c−12 Ans
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2. considering base 10 for given logarithm
2a) log8 + log6
Log is multiplied when it is in addition form
Log8 +log6 = log48 = 1.68124. Ans
2b) log8 -log6
Log is divided when it is in subtraction form.
Log8 – log6 ¿ log ( 8
6 ) = log1.33 = 0.12483 Ans
2c) log (8)c =
Log is multiplied when it is in power form.
log (8)c = c.log (8) = 0.90308C Ans
3) Solution
3a) 102x = 320.2 (Taking log on both side)
log102x = log320.2
then, 2x = 2.5054, then x = 1.2527 Ans
3b) b.3.1x = 80 where b = 8
Taking log from both side
Logb.3.1x = log80
Logb + xlog3.1 = log 80 (putting the value of b and taking log value of 8 and 80)
Log8 + xlog3.1 = log80
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X x 0.491 = 1.9030 – 0.9030
X = 2.036 Ans
3c) logx3 = 9
As per logarithm rule,
Logx3 = 9
⟹ 3 logx=9
⟹ logx=3
⟹ x=103 or x = 1000 Ans
Task-2 Solution
1) As given in question,
T = 27.3 days, G = 6.67 x 10-11 Nm2kg-2, M = 5.975 x 1024 kg
R =?
As per given formula,
T = √ 4 π2 r3
GM (Giving power on both side by 2
T 2= 4 π2 r3
GM
⇒ T2 GM
4 π2 =r3 or, ⇒ r3 = ( 27.3 x 24 ) 2 x 6.67 x 10−11 x 50975 x 1024
4 x 3.14 x 3.14
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⇒ r3 = ( 27.3∗24∗60∗60 )2 x 6.67 x 10−11 x 50975 x 1024
4 x 3.14 x 3.14
⇒ r3 =5.622 x 1025 , ⇒ r =3
√5.622 x 1025= 383088467.7 m = 3.83 x 108 m Ans
2) As given in question,
ρ=1 x 105, V = 7x10-5 m3, m = 6.6 x 10-27 and N = 2 x 1022 atoms
a) As per given equation,
pV =⅓ Nm c2
3 pV
Nm =c2
Putting the value in above equation
c2= 3 x 105 x 7 x 10−5
2 x 1022 x 6.6 x 10−27
⇒ c2= 21
0.00132 =159090.909 m2 s−2
Mean square speed = 159090.909 m2 s−2
b) As given in question,
r . m. s . speed= √c2 Putting the value from previous solution
r . m. s . speed= √159090.909=398.862m/s Ans
Logarithms and Algebra-
Task-1, Solution
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1. The factorization is as follows
a) 4x -6xy
= 2(2x-3xy) = 2x(2-3y) Ans
b) 2abc-6abd
= 2(abc-3abd) = 2ab(c-d) Ans
c) 4(a-b) – c(a-b), Taking common (a-b) from both equation
= (a-b) (4-c) Ans
d) 2pr-4ps+qr-2qs
= 2pr+qr -4ps-2qs Arranging
= r(2p+q) -2s(2p+q) Taking common (p+q)
= 2(2p+q) (r-2s) Ans
2. The factorization is as follows
a) x2 -5x+4
= x2 -4x – x +4 (splitting the middle part in suitable addition)
= x(x-4) -1(x-4) (Taking common as (x-4)
= (x-1) (x-4) Ans
b) 4x2 – 36 (We know that, a2 – b2 =(a-b) (a+b))
= (2x-6) (2x+6) Ans
c) X2 -4x +4
= x2 -2x-2x + 4
= x(x-2) -2(x-2) = (x-2) (x-2) = (x-2)2 Ans
d) 3x2 – 5x-28 (splitting the middle part in suitable addition)
= 3x2 – 12x+7x-28
= 3x(x-4) 7(x-4) = (3x-7) (x-4) Ans
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3. The The solution is as follows
a) 5x -7y=1 ………(i)
2x+5y = 21 ……. (ii)
Multiplying equation (i) x 5 and (ii) with 7
25x – 35y = 5
14x + 35y = 147 By adding both the equation
39x = 152, x = 152/39 = 3.8974
Putting the value of x and solving 7y = 18.48717, y = 2.641 Ans
b) Y = 5x +4 ……….(i)
Y = 4x+6 ……….(ii)
Equating both the equation,
5x+4 = 4x+6
= 5x -4x = 6-4
x = 2
Putting the value of x in equation (i)
Y = 5*2 +4 = 14
Therebefore, x = 2 and y = 14
4. The solution of quadratic equation is as follows
a) 3x2 -7x +2 = 0
= 3x2 -6x -x +2
= 3x(x-2) -1(x-2) = (3x-1) (x-2) Ans
b) 2x2 -9x +7
= 2x2 – 7x-2x+7
= x(2x-7)-1(x-7)
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= (x-1) (2x-7) Ans
c) 10x2 +5x-15
= 10x2 + 5x -15
= 10x2 +15x -10x -15
= 5x(2x+3) -5(2x+3)
=(5x-5) (2x+3) Ans
Task-2, Solution
1) For the given problem, we have to construct table for value of y at a given time t, we
have to draw graph and find out the solution.
As given in question,
Y = 20t – 5t2 . Putting the different value of t we will get different value of y
t y t y t t y
0.1 1.95 1.1 15.95 2.1 19.95 3.1 13.95
0.2 3.8 1.2 16.8 2.2 19.8 3.2 12.8
0.3 5.55 1.3 17.55 2.3 19.55 3.3 11.55
0.4 7.2 1.4 18.2 2.4 19.2 3.4 10.2
0.5 8.75 1.5 18.75 2.5 18.75 3.5 8.75
0.6 10.2 1.6 19.2 2.6 18.2 3.6 7.2
0.7 11.55 1.7 19.55 2.7 17.55 3.7 5.55
0.8 12.8 1.8 19.8 2.8 16.8 3.8 3.8
0.9 13.95 1.9 19.95 2.9 15.95 3.9 1.95
1 15 2 20 3 15 4 0
If we draw the graph for t verses y we will get like this
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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
0
5
10
15
20
25
y Height
As per table at 0.6 seconds the ball goes above 10 m and remain 10 m up while 3.4 seconds.
The time for ball up to height 10 m = 3.4 – 0.6 = 2.8 seconds.
Differentiation and integration-
Task-1, Solution
1. Differentiation is as follows
a) X5
d
dx x5=5 x4Ans
b) Sin.x
d
dx Sinx=Cosx Ans
c) e3 x
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d
dx e3 x= d
dx e3 x . d
dx 3 x=e3 x .3=3 e3 x Ans
d) Lnx
d
dx lnx= 1
x Ans
e) X4
d
dx x−4 =−4 x−5 Ans
f) Cosx
d
dx Cosx=−Sinx Ans
2 Integration is as follows
a) e5x
∫ e5 x=1
5 e5 x+ C Ans
b) Sinx
∫ Sinx=−Cosx+C Ans
c) x7
∫ X7= X7+ 1
7 +1 =¿ x8
8 +C ¿ Ans
d) √ x
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∫ √ x= x
1
2 +1
1
2 +1
= x3 /2
3
2
= 2
3 x3 /2 +C Ans
e) x-3
∫ X−3= X−3 +1
−3+1 =¿ x−2
−2 = −1
2 x2 +C ¿ Ans
h) 5
x
∫ 5
x =5∫ 1
x =¿ 5 ln |x|+C Ans
3. To find the gradient first I must do derivative of the equation
Y = x3
dy
dx =3 x2 (Putting the value x =2)
dy
dx (At x=2)=3. 22=12
The gradient of the curve = 12 Ans
4. as given equation x = t3 -2t2 +3t -4 where x is the distance and t is the time
The velocity can be given by differentiating the equation
dx
dt =3 t2−4 t+ 3
Putting the value t =4 sec
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dx
dt =3 x 16−4 x 4+3 = 29 m/sec
The acceleration of the equation is given by differentiation of velocity equation
d2 y
d x2 =3 x 2. t−4 (putting the value t=4)
Acceleration = 12 x4 – 4 = 8m/s2 Ans
5 The integral is as follows
a) ∫
0
5
7. dx
∫
0
5
7. dx=7 x +c=|7 x+c|0
5
=(7 x 5−7 x 0)=35 Ans
b) ∫
2
4
( 3 x +2 ) dx
∫ ( 3 x +2 ) dx= 3 x2
2 + 2 x +C
∫
2
4
( 3 x +2 ) dx=|3 x2
2 +2 x|2
4
=|3 x 42
2 +2 x 4− 3 x 22
2 −2 x 2|
¿|24 +8−6−4|=32−10=22 Ans
6) Area under the curve is integral of equation with given boundaries of limit
Therefore,
Y = x3, integral of ∫
1
2
x3 dx=| x4
4 |1
2
=( 24−14)=32−1=31 unit2
Ans
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Task-2, Solution
7) The distance covered by particle P is given by the distance by the equation
x=8 t− 1
2 t2
, to calculate the velocity, we must differentiate the given equation.
x=8 t− 1
2 t2
dx
dt =8− 1
2 x 2 xt=8−t
a) Putting the value of t =3 sec
Velocity we get = 8-3 = 5 unit/sec
b) To calculate the distance, we must calculate the acceleration by double differentiating
the velocity function.
d2 x
d t2 =0−1=−1
As we can see the acceleration is negative, -1 unit/sec2, therefore velocity at t = 0 is 8
unit/sec,
Therefore, using the formula
v2=u2−2 as (Putting the value v,u,a )
0=82−2 s
Or s = 64/2 = 32 unit.
Therefore, distance covered when its velocity becomes 0 = 32 unit Ans
8) As given in question, the upward speed of stone is 24 m/s which is reducing by time.
To calculate the velocity, we must differentiate the given equation,
y=24 t− 4
5 t2
(Differentiating the given equation)
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dy
dt =24− 4 x 2
5 t=24−8
5 t …………(i)
a) At the time t when velocity will be zero, in this condition dy
dt = 0, Putting the value of in
equation (i)
dy
dt =24− 8
5 t
0=24− 8
5 t
8
5 t=24
t= 24 x 5
8
t=15 sec Ans
b) The distance covered in 15 secs is given by the distance equation
y=24 t− 4
5 t2 (Putting the value of t =15)
Y = 24*15 – 4 x 15x15/5
Y = 360 – 180 = 180 m Ans
c) The velocity after time t is given by the equation (i) which is
dy
dt =24− 8
5 t
To calculate the acceleration, we must again differentiate the velocity equation.
d2 y
d t2 =0− 8
5 =−8
5 m/s2
The acceleration of thrown object = -8/5 m/sec2 Ans
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Circular measure and trigonometry –
Task-1 Solution
1a) 135o = 135o x3.14/180o = 2.3569 radian
1b) 150o = 150o x3.14/180o = 2.6179 radian
1c) 70o = 70o x3.14/180o = 1.22173 radian
1d) 75o = 75o x3.14/180o = 1.309 radian
2a) π
2 = π
2 x 180
π = 90o
2b) π
3 = π
3 x 180
π = 60o
2c) 5 π
2 = 5 π
2 x 180
π = 450o
2d) 11 π
6 = 11 π
6 x 180
π = 330o
3) radius = 8 cm, then arc ¿ 2 πr
360 xgiven angle
¿ 2 x πx 8
360 x 46=6.42 cm Ans
4) radius = 7 cm, then arc ¿ 2 πr
4 π x 1.4=4.9 cm
¿ 2 x πx 7
4 π x 1.4=4.9 cm Ans
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Task-2 Solution
1a) Sin230o = -0.76604
1b) Tan 230o = 1.1917
1c) Tan(-60o) = -1.7320
1d) Cos 130o = -0.6427
1e) Sin30o = 0.5000
1d) tan30o = 0.5773
2a) As given in question,
c = 6 cm, b = 8 cm, <BAC = 90 o
Using, Pythagoras’s theorem
BC= √ a2+ b2= √ 62 +82= √ 36+64= √ 100=10 cm Ans
2b) Using, Pythagoras’s theorem
x= √32 +32= √32 +32 = √9+9=3 √2=4.2426 cm
3) From Cosine rule, we know that
p2=q2 +r2−2 qrCos 42o Putting the value from question
p2=102 + 92−2 x 10 x 9 x 0.7431
p2=100+ 81−133.7660
p=6.87 cm Ans
4) Frome Sine rule we know that
t
SinT = w
SinW = r
SinR =d (Putting the value from question)
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3
sin 82o = r
sin39o
r = 3 xSin 39o
sin 82o = 3 x 0.6293
0.9902 =1.9065 cm Ans
5) From Cosine rule, we know that, we have to find angle between 6 cm and 9 cm
p2=q2 +r2−2 qrCos θo Putting the value from question
82 =62 +92−108 cos θo
108 cos θo=36+81−64
cos θo=53 /108
cos θo=0.4907
θo=60.6133o Ans
Task-3- Solution
1) As given in question, the length of piece contains twice radius + some circumferential
part
R = 4cm,
Circumference of piece= 60
360∗2∗3.14∗4=4.1866 cm
Now Total length of piece = 2R+4.1866 = 8+4.1866 = 12.1866 cm Ans
2) As given in question,
AB = 110 km
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<AOB = 1o
Then Circumference of Earth = 360 x110/1
= 39600 km,
a) Since we know that
Circumference = 2*Π*R
Then R = Circumference / 2 Π
(Putting the value in above formula
R = 39600/2*3.14 = 6304 km
Then radius of the earth is = 6304 km
b) If the Second place is 200 km north of first one,
Then total distance from latitude = 200 +110 = 310 km
Angle from latitude = 310/110 = 2.88o
Angle between two places = 2.88 – 1 = 1.88o Ans
Task 4- Solution
1a) As given in question,
The weight acting downward on rough
surface = 5.4 N, the force due to friction of rough
surface is hold due to Sin component of force.
Therefore, Force of friction = 5.4Sin30o = 2.7 N
Ans
1b) The support force is second component of force on the object, which Cosθ which acting
normal to the slope surface.
Support force = 5.4Cos30o = 5.4* 0.86602 = 4.6765 N
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2a) As given in question,
L = 4.5 m, distance from wall to foot of ladder = 1.15, < = 75.2o
Then, Sin75.2o = p/h (Where, p is perpendicular height and h is hypotenuse = 4.5
Then, Sin75.2o = p/4.5
Then, p = 0.9668*4.5 = 4.3507 m Ans
2b) From the triangle, Angle between wall and ladder = 180 – (90+75.2) = 14.8o Ans
3) As given in figure,
As per given question,
AC = 71 m, <ACB = 28o
Using Sin for this triangle,
Sin28o = AB/AC
Then. AB = Sin28o x AC
= 0.4694 x 71 = 33.332 m
The true thickness of stratum = 33.332 m Ans
References
Gelfand, I, M, & Saul, M, 2015, Trigonometry, 4th ed, Cambridge: Springer.
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Higgins, P, M, 2015, Algebra: A Very Short Introduction, 1st ed. Oxford: Oxford University
Press.
Mackenzie, A, 2005, Mathematics and Statistics for Life Scientists, 1st ed. Newyork: Taylor and
francis.
Stewart, J, 2015, Calculus, Boston: Cengage Learning.
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