Solved problems in Applied Mathematics

This text provides solutions to problems in Applied Mathematics including solving equations, evaluating derivatives, and atmospheric pressure. It also includes a summary of the steps taken to solve each problem.

Applied Mathematics
Institution Name
Student Name
Date of Submission

1. Solve the equations
a) ( 1
5 ) x
= 1
25
xln ( 1
5 )=ln ( 1
25 )
x=
ln ( 1
25 )
ln (1
5 )
=2
b) 92 x ¿ 27x2
= 1
3
34 x∗33 x2
=3−1
4 x+3 x2=−1
3 x2+4 x=−1
3 x2+4 x +1=0
Obtaining the value of x
x=−b ± √ b2−4 ac
2 a
¿ −4 ± √42−4∗3∗1
2∗3 =−1∨−1
3
c) e0.134 t =7
0.134 t =ln7
t= ln7
0.134 =14.5217
d) −logc 729=6
log c 729−1=6
logc( 1
729 )=6
From the laws of logarithm this can be transformed to
c6= 1
729
6 lnc=ln ( 1
729 )
lnc=
ln ( 1
729 )
6

lnc=−1.0986
c= 1
3
2.
a) 2 logb 6−4 logb 2+ 1
3 logb 64=2
logb 36−logb 16+ logb 4=2
logb (36
16 ∗4 )=2
logb 9=2
This can be transformed to
b2=9
2 logb=log 9
logb= log 9
2
logb=0.4771
b=3
b) log ( x−2 ) −log ( x−7 ) =log (2−x)
log ( x−2
x−7 )=log (2−x)
x−2
x−7 =2−x
x−2=(2−x)( x−7)
x−2=2 x−14−x2 +7 x
2 x−14−x2+7 x−x +2=0
−x2+ 8 x−12=0
x2−8 x +12=0
x=−b ± √ b2−4 ac
2 a
8 ± √64−48
2
8 ± √16
2 = 8 ± 4
2

This gives 2 or 6
3. Waterfront murder
a) The T (0) indicates the value of the body temperature at time 0. This significantly
indicates the body temperature at the time the murder took place
b) Using the equation T ( t ) =10+ 27 ekt , given that k=−0.0953108 .
This modifies our equation to T ( t )=10+ 27 e−0.0953108t
At the time of death, the body temperature was
T ( 0 )=10+27 e−0.0953108∗0=37
The degree difference between the time of the murder and the discovery of the
body is 37−32=5
Using the temperature, we can calculate the time taken in between which is.
5=10+27 e−0.0953108 t
Ext we now solve for t.
−5
27 =e−0.0953108 t
ln ( 1
27 )=0.0953108t
t=
ln ( 1
27 )
−0.0953108 =17.69 hrs
The body was discovered at 3:15 am going 17 hrs 41 minutes behind gives
9:34 am
The victim dies at 9:34 am of the previous day.
4. Evaluation of derivatives
a) H ( x )= S ( x )∗t (x )
P(x)
Using the quotient and product rules of differentiation
Product rule
d ( uv )=u1 v +v1 u
Quotient rule
d ( u
v )= u' v +v ' u
v2

H' ( x )= ( s ( x )∗t ( x ) ) p ( x )−P ' ( x )¿ ¿
But ( s ( x )∗t ( x ) ) ' =S' ( x ) t ( x ) +t' ( x ) S (x)
This gives
H' ( x ) = ( S' ( x ) t ( x ) +t' ( x ) S ( x )) p ( x ) −P '(x )¿ ¿
¿ [ ( 3∗6 )+ ( 5∗−1 ) ]∗4− (−2 )∗[ (−1 )∗6]
42 = 5
2
b) J ( x )=S ( P ( x ) ) + P( s ( x ) )
J' ( x ) =S' ( P ( x ) )∗p' ( x ) +P' ( S ( x ) )∗S ' ( x)
Replacing the values gives
J' (2 )=S '( P ( 2 ))∗P' ( 2 ) +P '(S ( 2 ) )∗s ' (2)
S' ( 4 )∗(−2 ) +( P' (−1 )∗3)
¿ ( 1∗−2 )+ ( 10∗3 )=28
5. Differentiation
a) y=x2 ln ( 4 x )
Using the product rule
y' =2 xln4 x+ ( 1
4 x ∗4 )x2
¿ 2 xln 4 x + 1
x ∗x2
2 xln 4 x+ x
b) h ( t )= a e3t
t2
Applying the quotient rule
d ( u
v )= u' v−v ' u
v2
3 a e3 t∗t2−(2t∗a e3 t )
t4
3 a t2 e3 t −2 tae3 t
t4
c) f ( t )= 1
2 π cos ( π θ3 ) sin (2 πθ)

1
2 π d ¿
1
2 π ¿
¿−3
2 ∗π 2 sin ( π θ3 ) +cos (2 π θ)
d) g ( x )=2 ( x2 −b2 )
( 1
3 )
2 ¿
¿ 4
3 x ( x2−b2 )
−2
3
¿
4
3 x
( x2−b2 )
2
3
6. Atmospheric pressure
a)
p ( h )=760 e−0.145 h
p ( 10 )=760 e−0.145∗10
¿ 178.2734 mmHg
On the ground
p ( 0 )=760 e−0.145∗0 =760 mmHg
The atmospheric pressure is higher in the ground than in high altitude areas. The
atmospheric pressure on the earth’s surface is generated by the layer of air above
the ground. As one goes higher the layer becomes thinner hence reduction on the
atmospheric pressure.
b) Rate of change
p' ( h )=760 ¿
−110.2 e−0.145h
p ( 10 )=−110.2 e−0.145∗10
¿−¿25.8496mmHg/Km

As theplane goes higher by 1 km the atmospheric pressure reduces by 25.8496
mmHg.
7.
a) ∑
i=0
n
( ai−2 )
The ∑ of the firts n terms∈the arithmetic series is obtained by the formula
Sn= n
2 ¿ ¿
Since we have
∑
i=0
n
( ai−2 )=−2+ ( a−2 )+ (2 a−2 ) +…+n
Sn= n
2 [2 ( −2 ) − ( n−1 ) ( a−4 ) ]
a=−2 while d=(a−4)
¿ n
2 [−na+4 n+a−4 ]
¿ n
2 [−na+4 n+a−4 ]
¿ n
2 [ a−na+4 n−4 ]= n [ a ( n+1 )−4 ]−4
2
b) ∑
i=1
5
(3 i−2 )
a=3−2=1
Now ∑
i=1
5 5 [ 1 ( 5+1 ) −4 ] −4
2
5 [ 1 ( 6 ) −4 ] −4
2
¿ 5 ( 6−4 ) −4
2 =3
c) r =
4
9
2
3
= 2
3
∑
i=1
5
a rn−1=¿ ∑
i=1
5
( 2
3 )i−1
¿

d) Sn= a ( 1−r ) n
1−r
Replacing the value of r and n we have
Sn=
a (1− 2
3 )6
1−2
3
= 1
243
e) At the first one million values we can use the sum to infinity which is given by
the formula ¿ a
1−r = 1
1− 2
3
=3

References
Thomas, G. B., W., M. D., Joel, H., & Frank, R. G. (2008). Calculus. Addison-Wesley.
Zill, D. G., Wright, S., & S., W. (2009). Calculus: Early Transcendentals. Jones & Bartlett
Learning.