1.

For a certain curve, the point of contact of each tangent to it bisects the part of the
tangent termination on the coordinate axes. Find the equation of the curve.

Figure:

(0 , y)

(x , 0)

Solution:

𝑦 −𝑦 𝑑𝑦⁄
Slope = 𝑥1 − 𝑥2 = 𝑑𝑥
1 𝑧

𝑑𝑦⁄ 𝑦−0
𝑑𝑥 = 0−𝑥
𝑑𝑦⁄ 𝑦
𝑑𝑥 = −
𝑥
𝑑𝑦 𝑑𝑥
= −
𝑦 𝑥

𝑑𝑦 𝑑𝑥
+ =0
𝑦 𝑥

𝑑𝑦 𝑑𝑥
∫ +∫ =0
𝑦 𝑥

ln 𝑦 + ln 𝑥 = ln 𝐶

xy = c

2. Find the equation of the curve so drawn that every point on it is equidistant from the
origin and the intersection of the x-axis with the normal to the curve at the point.

1|Page
Figure:

(x , y)

Y (0 , 2x)

Solution:

y−0
mn =
0−x
y
mn =
−x
dx y
− = −
dy x

∫ xdx = ∫ ydy

y2 = x2 + C

3. The area bounded by a curve, the x-axis, a fixed ordinate, and a variable ordinate is
proportional to the difference between the ordinates. Find the equation of the curve.

2|Page
Figure:

(x,y)
dx

Solution:

A = ∫ 𝑦𝑑𝑥

A = ∫ 𝑘𝑑𝑦

𝑘𝑑𝑦
∫ 𝑑𝑥 = ∫
𝑦

x = klny + C

𝑒 (𝑘𝑙𝑛𝑐+𝑥) = 𝑘𝑙𝑛𝑦 (Multiply by 1⁄𝑘 )

𝑥⁄ )
𝑒 (𝑙𝑛𝑐+ 𝑘 = 𝑙𝑛𝑦
𝑥⁄
y = C𝑒 𝑘

4. Find the orthogonal trajectories of the ellipses 4x 2 + y 2 = cx.

Solution:

3|Page
4x 2 + y 2 = cx

4x 2 + y 2
=c
x
8x + 2𝑦𝑦1 = 𝑐

𝑐 − 8𝑥 𝑑𝑦
𝑦1 = = 𝑚𝑡 =
2𝑦 𝑑𝑥
2𝑥𝑦
= − 4𝑥 2 + 𝑦2 − 8𝑥 2

𝑑𝑦 2𝑥𝑦 2𝑥𝑦
= − =
𝑑𝑥 −(4𝑥 2 − 𝑦 2 ) (4𝑥 2 − 𝑦 2 )

dy(4𝑥 2 − 𝑦 2 ) = 𝑑𝑥 (2𝑥𝑦)

Let x = vy ; dx = vdy + ydv ; v = 𝑥⁄𝑦

(4𝑣 2 𝑦 2 − 𝑦 2 )𝑑𝑦 = 2𝑣𝑦 2 (𝑣𝑑𝑦 + 𝑦𝑑𝑣)

(4𝑣 2 𝑦 2 − 𝑦 2 )𝑑𝑦 = 2𝑣 2 𝑦 2 𝑑𝑦 + 2𝑣𝑦 3 𝑑𝑣 (Divide both sides with 𝑦 2

(4𝑣 2 − 1 )𝑑𝑦 = 2𝑣 2 𝑑𝑦 + 2𝑣𝑦𝑑𝑣

(2𝑣 2 − 1)𝑑𝑦 = 2𝑣𝑦𝑑𝑣

𝑑𝑦⁄ 2𝑣𝑑𝑣
𝑦 = (2𝑣 2 − 1)

1
lny = 2 ln(2𝑣 2 − 1) + 𝑙𝑛𝑐

𝑥2
𝑦 2 = (2 𝑦 2 − 1) 𝑐 (Multiply both sides with 𝑦 2

𝑦 4 = (2𝑥 2 − 𝑦 2 )𝑐

5. Determine the isogonal trajectories of the circles x 2 + y 2 = c it the angles of

intersection are to be 45º.

Solution:

4|Page
 x2 + y2 = C
dy
2x + 2y dx = 0
dy −x
( )g =
dx y

dy x
( )+
dx y
Arctan 45º = x dy
1− ( )
y dx

xdy dy x y
1 - ydx = + y = xv and v = ⁄x
dx y
dy = xdv + vdx
xdy dy x
- − = –1
ydx dx y (x + xv)(xdv + vdx) – (xv – x)dx = 0
x (1 + v) (xdv + vdx) – x (v – 1) dx = 0
x dy x
(y + 1) =1-y xdv + vdx + xvdv +v 2 dx - vdx + dx = 0
dx
x (1 + v)dv + (v 2 + 1)dx
=0
(
x+y
) dy = (
y− x
) dx (v 2 + 1)x
y y (1 + v)dv dx
+ =0
(x + y) dy – (y –x) dx = 0 (v 2 + 1) x

Integration:
dv vdv dx
∫ (v2 + 1) + ∫ (v2 + 1) + ∫ x
=0

1
Arctan v + 2 ln(v 2 + 1) + 𝑙𝑛𝑥 + 𝑐 = 0

ln(v2 + 1)+ 2𝑙𝑛𝑥

Arctan v + =0
2

1⁄
𝑦2 2
Arctan v + 𝑙𝑛 [(𝑥 2 + 1) (𝑥 2 )] + 𝑐=0

𝑦 1⁄
Arctan 𝑥 + ln (𝑥 2 + 𝑦 2 ) 2 =𝑐

6. A weight of 450 N falls from rest. If the resistance of the air is proportional to the
speed, and if the limiting speed is 52.5 m⁄s, find the speed after 5 sec.

Figure:
W = 450 N

5|Page
 Air resistance = KV

(Note: Taking downward direction as positive.)

ƩF = W - KV
W dv
W – KV = ⁄dt
g

When dv⁄dt = 0 ; v = 52.5

W – 52.5K = 0
W
K = 52.5

W dv
W – KV = ⁄dt
g
450450
450 – V 52.5 = 9.81 dv⁄dt
450 – 8.5714V = 45.8716 dv⁄dt
(450 – 8.5714V) dt = 45.8716 dv

45.8716
∫ dt = ∫ dv
450 − 8.5714v
45.8716
t = −8.5714 ∫ du⁄u
t = -5.3517 ln|450 – 8.5714v| + C

When t = 0 ; v = 0 (Starts from rest)

C = 5.3517 ln|450|

When t = 5 seconds ; find v

t = -5.3517 ln|450 – 8.5714v| + 5.3517 ln|450|
t - 5.3517 ln|450| = -5.3517 ln|450 – 8.5714v|
-27.6949 = -5.3517 ln|450 – 8.5714v)
−27.6949
e −5.3517 = 450 – 8.5714v

V = 31.8744 m⁄s or 32 m⁄s

7. A 45 N weight is projected downward with an initial speed of 4.6 m⁄s. If the air
resistance is proportional to the speed and if the limiting speed is double the initial
speed, how far has the body traveled when it has reached a speed of 6 m⁄s?

Figure:

6|Page
 45 N

Solution:

ƩF =
W dv
⁄dt 9.81 4.6
g 𝑡 = 𝑙𝑛 ( )
9.2 9.2 − 𝑣
W dv
W–R= ⁄dt but dv⁄dt = 0 𝑒
9.81𝑡⁄
9.2 = 4.6⁄9.2 − 𝑣
g
W=R
W = KV −9.81𝑡⁄
9.2 – v = 4.6𝑒 9.2
K = 𝑊⁄2(4.6)
−9.81𝑡⁄
v = 9.2 - 4.6𝑒 9.2
𝑊𝑉 W
W - 9.2 = g dv⁄dt
s = ∫ 𝑣𝑑𝑡
9.81
(9.2 − 𝑉) = dv⁄dt 0.34 −9.81𝑡⁄
9.2 s = ∫0 [9.2 − 4.6𝑒 9.2 ]dt

9.2 𝑑𝑣
∫ 𝑑𝑡 = ∫ ⁄(9.2 − 𝑣) s = 1.81 m
9.81
9.2
t = − 9.81 ln (9.2 – v) + c

When t = 0; v = 4.6
9.2
C= ln (4.6)
9.81

When v = 6𝑚⁄𝑠
9.2
t = 9.81 [– ln(9.2 − 𝑣) + ln(4.6)]

t = 0.3403395048

8. A body falls from rest against a resistance proportional to the cube of the speed at any
instant. If the limiting speed is 3 m⁄s, find the time required to attain a speed of 2
m⁄ .
s
Figure:

𝑉𝑜 = 0

𝑉𝑓 = 2 m⁄s

7|Page
Solution:
W dv
W-R= ⁄dt but dv⁄dt = 0
g
W=R
W = K𝑣 3 K = 𝑤⁄27
𝑊 W
W - 27 𝑉 = g dv⁄dt
9.8
(27 − 𝑣) = dv⁄dt
27
27
∫ 𝑑𝑡 = ∫ 𝑑𝑣⁄(27 − 𝑣)
9.81
27
t = − 9.81 ln(27 − 𝑣) + 𝑐

When t = 0 ; v = 0
27
C = 9.81 ln(27)

When v = 2 𝑚⁄𝑠
27
t = 9.81 [− ln(27 − 𝑣) + ln(27)]

27
t = 9.81 𝑙𝑛 27⁄27 − 𝑣

t = 0.21

9. A body above the surface of the earth is pulled toward the center of the earth with a
force proportional to the reciprocal of the square of the distance of the body from the
center. If the radius of the earth is 6375 km, find the velocity of the body as it strikes
the surface of the earth if it falls from rest at a distance of four times the earth’s radius
measured form the center of the earth. What is the velocity if it falls from an infinite
distance?

r
Figure:
R

8|Page
 Let r be the variable distance
Let R be the radius of the earth
𝑑𝑟 𝐾
a = 𝑑𝑡 = 𝑟2

The acceleration is negative because the velocity is decreasing. Hence, the k is negative when
r = R, then a = -g, thus;

𝐾 𝑔𝑅 2
−𝑔 = 2 − − − − − − − 𝑎 = − 2
𝑟 𝑟
𝑣𝑑𝑣 𝑔𝑅 2
= − ------- Equation 1
𝑑𝑟 𝑟2

2𝑔𝑅 2
𝑣2 = − + 𝐶 ------ General Equation
𝑟

When R = 6375 km or 6, 375, 000 meters

2𝑔𝑅 2
(a) 𝑣 2 = + 2𝑔𝑅
4𝑅
−2(9.81)(6,375,000)
= + 2(9.81)(6375000)
4
𝑚
v = 9685.459 𝑠
v = 9.7 𝑘𝑚⁄𝑠

(b) As R approaches infinity

2𝑔𝑅 2
𝑣2 = + 2𝑔𝑅
4𝑅
𝑣 2 = 2𝑔𝑅
𝑣 2 = 2(9.81)(6375000)
v = 11183.805 𝑚⁄𝑠
v = 11.18 𝑘𝑚⁄𝑠
10. A boat with its load weighs 1430 N. if the force exerted upon the boat by the motor in
the direction of motion is equivalent to a constant force of 67 N, and if the resistance
to motion is equal numerically to twice the speed, find the speed after 10 sec. assume
the boat starts from rest.

W = 1430 N
Figure:

2V Fa = 67 N
9|Page
 Solution:

ƩF = (W⁄g)(dv⁄dt)
67 – 2v = (1430⁄9.81)(dv⁄dt)

∫ dt = ∫ 145.77 dv⁄(67 − 2v)

t = -72.88 ln |67 – 2v| + C

When t = 0 ; v = 0 (Starts from rest)

0 = -72.88 ln |67| + C
C = 306.438

When t = 10 seconds ; find v

t = -72.88 ln |67 – 2v| + 306.438
10 = -72.88 ln |67 – 2v| + 306.438
4.067 = ln |67-2v|
e4.067 = 67 − 2v
V = 4.3 m⁄s

11. A certain radioactive material follows the law if exponential change and has a half life
of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.

Solution:

S = Cekt

When t = 0 ; S = So
0
So = Ce
So = C

10 | P a g e
S = So ekt
1
When t = 38 hrs. ; S = 2 So
1
S = So ek(38)
2 o
1
= ek(38)
2
1
1 38 1
( ) = [ek(38) ]38
2
t
1 38
S = So ( )
2

S = (.10) (So )
t
1 38
So ( ) = (.10) (So )
2
t
1 38
(2) = (.10)
t
ln (.10) = 38 ln 0.5

t = 126 hours

12. A bacterial population follows the law of exponential growth. If between noon and
2p.m. the population triples, at what time should the population become 100 times
what it was at noon? At 10 a.m. what percentage was present?

Solution:

P = Cekt

(a) When t = 0, P = P0

P0 = C
P = P0 ekt

When t = 2, P = 3P0

11 | P a g e
3P0 = P0 ek(2)
3 = ek(2)
√3 = √ek(2)

√3 = ek

t
P = P0 (3) ⁄2

When P = 100P0
t
100P0 = P0 (3) ⁄2
t
100 = (3) ⁄2
t
ln 100 = 2 ln 3

t = 8.38 hours

Time = 8:22 P.M.

(b) When time is 10 a.m.

P = xPekt
1⁄ = e0.55 (2)
x

x = 33.29%

13. A thermometer reading 75ºF is taken out where the temperature is 20ºF. The reading
is 30ºF 4 minutes later. Find (a) the thermometer reading 7 minutes after the
thermometer was brought outside, and (b) the time taken for the reading to drop from
75ºF to within a half degree of the air temperature or 20.5ºF.

Solution:

Let u = Tb

du⁄ = k (u − 20)
dt

∫ du⁄(u − 20) = ∫ kdt

12 | P a g e
ln |u-20| = kt + lnC

ln |u -20| - lnC = kt

(u – 20) = ekt + 𝑒 𝑐

Let 𝑒 𝑐 = 𝑎

T = aekt + 20

When t = 0 ; u = 75ºF

75 = aekt + 20

a = 55

When t = 4 ; k = -0.4262

u(t) = 55e−0.4262t + 20

(a) u(7) = 55e−0.4262(7) + 20

= 22.78ºF or 23ºF

(b) 20.5 = 55e−0.4262t + 20

t = 11.0236 minutes

14. At a certain time, a thermometer reading 70ºF is taken outdoors where the
temperature is 15ºF. Five minutes later, the thermometer reading is 45ºF. After
another five minutes, the thermometer is brought back indoors where the temperature
is fixed at 70ºF. What is the thermometer reading after ten minutes after the
thermometer is brought back indoors? When will the reading, to the nearest degree,
return back to its original reading of 70ºF?

Solution:
du⁄ = K (u − 15) (u − 15)⁄
du
dt C = Kt
∫ ⁄(u − 15) = ∫ Kdt (u-15) = CKt
ln(u-15) = kt + lnC
ln(u-15) – lnC = Kt (u-15) = Cekt

13 | P a g e
When t = 0, u = 70ºF
70 -15 = Ce0 (b) When u = 58.5ºF
55= C
t
u-15 = 55ekt
58.5 – 70 = -38.65 (30⁄55)
5
When t = 5, u = 45ºF
t

-11.5 = -38.65 (30⁄55)

5
45-15 = 55 ek(5)
30⁄ = ek(5)
55 t
1 −11.5
= (30⁄55)
5
ln
[30⁄55 = e k(5) 5
] −38.65
1
(30⁄55) = ek
5 t = 46 minutes

u-15 = 55 (30⁄55)
5

(a) When t= 10, u= 31.36ºF

31.36-70 = Ce0
C = -38.64
10

u-70 = -38.74 (30⁄55)

5

u= 58.5ºF

15. A tank contains 200 litres of fresh water. Brine containing 2.5 N⁄litre of dissolved salt
runs into the tank at the rate of 8 litres⁄min and the mixture kept uniform by stirring
runs out at 4 litres⁄min. Find the amount of salt when the tank contains 240 litres of
brine. The concentration of salt in the tank after 25 minutes amounts to how much?

Solution:

ds⁄ = ds⁄ en − ds⁄ es S

C = 200+4t
dt dt dt

ds⁄ = (2.5)(8) − ( S S
) (4) ds⁄ = 20 −
dt 200 + 4t dt 50 + t

14 | P a g e
ds⁄ + S C = -25,000
dt 50 + t = 20
When t = 10 minutes ; find S
P = 1⁄50 + t
S = 10 (50 + t) - [25000⁄(50 + 10)]
Q = 20
dt
Ø = e∫50+t S = 183.33 N
= eln(50+t) S
C = 200+4t
= 50 + t
When t = 25 seconds ; S = 0
S (50 + t) = ∫(50 + t)(20)dt
S = 10 (50 + 25) -[25000⁄(50 + 25)]
20
S (50 + t) = (50 + t)2 + C
2
S = 416.6667 N
S = 10 (50 + t) + C⁄(50 + t) 416.6667
C = 200+4(25)
When S = 0 ; t = 0
C = 1.4 N⁄Lit

16. A tank contains 400 litres of brine. Twelve litres of brine, each containing 2.5 N of
dissolved salt, enter the tank per minute, and the mixture, assumed uniform, leaves at
the rate of 8 litres per min. if the concentration is to be 2 N⁄litre at the end of one
hour, how many newtons of salt should there be present in the tank originally?

Figure:
𝑅1

400 L

𝑅2

15 | P a g e
R1 = 12 Lit⁄min (2.5 N⁄Lit)

A N⁄ (8 Lit⁄ 2A N
R2 = Lit min ) = 100+t ⁄min
400+4t

2
dA⁄ = R − R P = 100+𝑡
dt 1 2
Q = 30
dA⁄ = 30 − 2A 2𝑑𝑡
Ø = 𝑒 ∫ ⁄100+𝑡
dt 100 + t
dA⁄ + 2A = 30 = 𝑒 2 ln(100+𝑡)
dt 100+t = (100 + 𝑡)2

A Ø = ∫ ØQ dt

A(100 + t)2 = ∫(100 + t)2 (30)dt

30(100+t)3
A(100 + t)2 = 3
A(100 + t)2 = 10(100 + t)2 + C
C
A = 10 (100 + t) + (100+t)2

When t = 60 minutes
A = 2[400 + 4(60)]
A = 1280
C
1280 = 10 (100 + 60) + (100+60)2
C = -8192000

When t = 0 ; find A
−8192000
A = 10 (100) + (100)2
A = 180.8 N
17. Tank A initially contains 200 litres of brine containing 225 N of salt. Eight litres of
fresh water per minute enter A and the mixture, assumed uniform, passes from A to B,
initially containing 200 litres of fresh water, at 8 litres per minute. The resulting
mixture, also kept uniform, leaves B at the rate of 8 litres per minute. Find the amount
of salt in tank B after one hour.

Figure:

Tank A Tank B
𝑥 𝑦
200 200

16 | P a g e
Solution:

When t = 0 ; x = 225 N

TANK A:

dx⁄ = (0 N ) (8 lit ) − (8 lit ) ( xN )

dt lit min min 200 lit
dx⁄ = − 8x N xN
dt or −
200 min 25 min

dx⁄ = − x
dt 25

∫ dx⁄x = ∫ − dt⁄25

t
ln|x| = ln | − 25 + C|

t
x = ln |e−25 + eC |
t
x = Ce−25
225 = ce0
C = 225
t
x = 225e−25

TANK B:

dy⁄ xN lit lit yN

dt = ( ) (8 ) − (8 ) ( )
200 lit min min 200 lit

dy⁄ x⁄ − y
dt = 25 ⁄25
dy⁄ y x⁄
dt + ⁄25 = 25
t
(Substitute x = 225e−25 in the equation)
t 1
dy⁄ y − P = 25
dt + ⁄25 = 225e 25⁄25
t
dy⁄ y 𝑑𝑡 𝑡

dt + ⁄25 = 9e−25 Ø = 𝑒 ∫ 25 = 𝑒 25
y Ø = ∫ Ø Qdt
t

t t t
Q = 9e−25
−
y e25 = ∫ e25 (9e 25 ) dt

t
y e25 = ∫ 9dt
t
y e25 = 9t + C

17 | P a g e
When t = 0, y = 0
t
y e25 = 9t + C
t
(0) e25 = 9 (0) + C

C=0
t
y e25 = 9t

When t = 60 ; find y
60
y e25 = 9 (60)

y = 48.99 N or 49 N

18. Tank A contains 400 litres of brine holding 225 N of dissolved salt, and tank B
contains 400 litres of fresh water. If brine runs out of A into B at the rate of 12
litres⁄
min while the mixture, kept thoroughly stirred, is pumped back from B to A at
the same rate, when will A contain twice as much salt as B?

Figure:

Tank A Tank B

Solution:

Find t if 𝑆𝐴 = 2𝑆𝐵

𝑑𝑠⁄ = 𝑆𝐵 (12) − 𝑆𝐴 (12)

𝑑𝑡 400 400
Since 𝑆𝐴 + 𝑆𝐵 = 225 ; 𝑆𝐵 = 225 - 𝑆𝐴

18 | P a g e
𝑑𝑠⁄ = (225− 𝑆𝐴 )3 3𝑆
𝑑𝑡 - 100𝐴
100

𝑑𝑠⁄ = 675 3𝑆 3𝑆
𝑑𝑡 - 100𝐴 - 100𝐴
100

𝑑𝑠⁄ = 675 6𝑆
𝑑𝑡 - 100𝐴
100

100𝑑𝑠
∫ 𝑑𝑡 =
675 − 6𝑆
−100𝑑𝑠
∫ 𝑑𝑡 =
6( 𝑆 − 112.5)
100
t=- ln(𝑆 − 112.5) + 𝑐
6

100
When t = 0 ; s = 225 c = ln(112.5)
6

100
t= [− ln(150 − 112.5) + ln(112.5)]
6

t = 18.3 minutes

19. A tank with a horizontal sectional area constant at 10m2 and 4 m high contains water
to a depth of 3.5 m. the tank has a circular orifice 5 cm in diameter and located at its
side 0.5m above the bottom. If the coefficient of discharge of the orifice is 0.60, find
the duration of flow through the orifice.

10 𝑚2
Figure:

3.5 m
4m
0.5 m

Solution:

A = 10 m2
Ao = Π⁄4 (0.05)2 = 1.9635 x 10−3
h1 = 3.5 m − 0.5 m = 3m
h2 = 0m
C = 0.6

19 | P a g e
 3 Adh
t = ∫0
CAo √2gh

3 10dh
t = ∫0
0.6 (1.96 x 10−3 )(4.43)√h

3 10dh
t = ∫0
5.22 x 10−3 √h

3 dh
t = 1916.33 ∫0
√h

t = [1916.33 (2)(√3) − 1916.33 (2)(√0)]

t = 6638.347473 seconds

t = 1 hour, 50 minutes and 38.35 seconds

20. A tank in the shape of an inverted cone has a base diameter of 1.5 m and a height of 2
m. a 4-cm orifice (C=0.60) is located at the apex. Determine the time required to
empty the tank assuming that it is initially full of water.

Figure:

Solution:

Πd2 Π (1.5)2
A= = = 1.767
4 4
Ao = (L)(W) = (0.04)(0.04)

h1
Adh
t=∫
h2 CAo √2gh
2 1.767 dh
= ∫0 (0.60)(.04)(√(2)(9.81)h

2 dh
= 415.5772 ∫0
√h

20 | P a g e
 2
t = 415.5772 [2√h]0

t = 1175 seconds

t = 19 minutes and 35 seconds

21. A tank in the shape of an inverted frustum of a cone has an upper base diameter of 2m
and a lower base diameter of 1m. A square orifice 5cm x 5cm (C=0.61) is located at
the lower base. If the tank is 3m high and initially full of water, find the time required
for the water surface to drop 2m.

Solution:

1 2
=
𝑧 3+𝑧
3 + z = 2z

z = 3m

𝑠 2
=
ℎ+3 6
ℎ+3
s= 3

h1
Adh
t=∫
h2 CAo √2gh

21 | P a g e
 ℎ+3 2 Π
( 3 ) ( ) dh
t=∫ 3 4
2
1 0.61(0.05) √2(9.81)h

3(
ℎ+3 2
t = 116.27 ∫ 3 ) dh
1 √h

116.27 3 (h2 + 6h + 9)dh

t= ∫
9 1 √h
3 3 1 1
t = 12.92 ∫ h2 + 6h2 + 9h−2 dh
1

5 3 1 3
t = 12.92 [45ℎ2 + 4ℎ2 + 18ℎ2 ]
1

t = 751.9 – 289.108

t = 462.492 seconds

22. A tank in the shape of a hemispherical shell has a diameter of 2m. Water escapes
through a circular orifice 10 cm in diameter (C=0.60) and located at the lowest point
on the tank. Determine the time required to empty the tank assuming that it contains
water amounting to half the volume of the tank.

Figure:
2 meters

1 meter Height of the water =

Half the volume of the
tank
10 cm.

Solution:

Volume of hemisphere = 2⁄3 (Πr 3 )

Compute for the radius o the circle:
= 2⁄3 [Π(1)3 ]
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
2𝛱
= 3
(𝑥 − 0)2 + (𝑦 − 1)2 = (1)2
1 2𝛱 𝛱
𝑉𝑜 = ( )=
2 3 3 𝑥 2 + (𝑦 − 1)2 = 1

22 | P a g e
 ℎ
Let r be the radius of the sphere 1
= ∫ 1 − ℎ2 + 2ℎ − 1 𝑑ℎ
3 0
Let h be the height of the water
ℎ
1
𝑟 2 = 1 − (ℎ − 1)2 = ∫ − ℎ2 + 2ℎ 𝑑ℎ
3 0
𝑟 = √1 − (ℎ − 1)2 ℎ
1 ℎ3
= [− + ℎ2 ]
𝑉𝑜 = ∫ 𝐴𝑑ℎ 3 3 0

ℎ
h = 0.6527m
𝛱
= ∫ 𝛱𝑟 2 𝑑ℎ 0.653 𝛱(1− (ℎ−1)2 )
3 0 t = ∫0 𝑑ℎ
(0.60)(𝛱⁄4)(0.01)2 √(2)(9.81)ℎ
ℎ
1
= ∫ 𝑟 2 𝑑ℎ t = 150.51 ∫0
0.653 2ℎ− ℎ2
3 0 √ℎ

ℎ 1 3
1 t = 150.51 ∫0
0.653
[2ℎ2 − ℎ2 ]
= ∫ 1 − (ℎ − 1)2 𝑑ℎ
3 0
t = 84.7 seconds
23. A swimming pool has a gradually sloping bottom. It has a depth of 1m at one end and
a depth of 3m at the other end. The horizontal sectional shape of the pool is a
rectangle 50m x 20m. Two short tubes each 15 cm in diameter (C=0.80) are located at
the lowest level of the pool. If the pool is initially full of water, find the time to empty
the pool through the tubes.

Figure: 𝑣1 = 500
𝑣1 = 500 1m 𝑣2 = 1000

𝑣2
2m

Solution:
v2 v1
=
t2 t1

1000 500
=
t2 t1

t2
t1 =
2

23 | P a g e
 h Adh
t = ∫h 1
2 CAo √2gh

3 (50)(2)dh
t = ∫2 9Π
(0.80)( ⁄1600)(2)√2(9.81)h

t 2 = 1.40990678

1.40990678
t1 =
2
t1 = 0.70495339

t T = t1 + t 2

t T = 0.70495339 + 1.40990678

t T = 2.11486017 hours

t T = 2 hours, 6 minutes and 53.28 seconds

24. A coil of inductance 1 henry and a resistance 10 ohms is connected in series with an
e.m.f. of Eosin10t volts. When t=0, the current is zero. If I= 5 Amp. when t= 0.1 sec,
what must be the value of Eo?

Solution:

di
L + RI = Eo sin10t
dt
Give L = 1 Henry and R = 10 ohms
di P = 10
+ 10I = Eo sin10t Q = 𝐸𝑜 𝑠𝑖𝑛10𝑡
dt
Ø = 𝑒 10𝑡

I Ø = ∫ ØQ dt

If e∫ 10dt = E0 sin10t
emt
∫ emt sinntdt = 2 (msinnt − ncosnt)
m + n2
So;
10t
e10t (I) = Eo [e ⁄200 (10sin10t − 10cos10t) + c]

t=0;I=0
1
0 = Eo [200 (−10) + c]

24 | P a g e
 E
0 = − 20o + c

E
C = 20o

If I = 5 amp. ; t = 0.1 sec. ; Eo =?

e10(0.1) E
5e10t = Eo { [10sin10(0.1) − 10cos10(0.1)]} + o
200 20

Eo = 149.47 volts

25. An inductance of 1 henry and a resistance of 2 ohms are connected in series with an
e.m.f. of 𝐸𝑒 −𝑡 volts. No current is flowing initially. (a) If the current I= 10 Amp.
when t= 1 sec, how much must E be? (b) If E= 50 volts, when will the current be 5
Amp.?

Solution:
𝑑𝑖
P=2
𝐿 𝑑𝑡 + 𝑅𝐼 = 𝐸 Q = Ee−t
Ø = 𝑒 2𝑡
𝑑𝑖
(1) + 2𝐼 = 𝐸𝑒 −𝑡
𝑑𝑡

I𝑒 2𝑡 = ∫ 𝑒 2𝑡 (𝐸𝑒 −𝑡 ) dt
I𝑒 2𝑡 = 𝐸𝑒 𝑡
I𝑒 2𝑡 = 𝐸𝑒 𝑡 + C

When I = 0 ; t = 0 and C = -E
I𝑒 2𝑡 = 𝐸𝑒 𝑡 − 𝐸

(a) If I = 10 Amp. ; t = 1 sec.

10 (𝑒 2 ) = 𝐸 (𝑒 − 1)
E = 43.003 V

(b) If E = 50V ; I = 5A
𝑒 2𝑡 𝐸
=
(𝑒 𝑡 − 1) 𝐼

𝑒 2𝑡 50
𝑡
=
(𝑒 − 1) 5

t = 0.119574012 or 0.12 second

25 | P a g e
 26. An inductance L henries and a resistance of R ohms are connected in series with an
e.m.f. of 𝐸𝑒 −𝑎𝑡 volts where E and a are positive constants. Initially, the current I is
zero. Find an expression for I as a function of t, and determine at what time the
current reaches its maximum value.

Solution:
𝑅
P=𝐿
𝑑𝑖
𝐿 + 𝑅𝐼 = 𝐸 𝐸𝑒 −𝑎𝑡
𝑑𝑡 Q=
𝑑𝑖 𝐿
𝐿 + 𝑅𝐼 = 𝐸𝑒 −𝑎𝑡
𝑑𝑡 𝑅𝑡
𝑑𝑖 𝑅𝐼 𝐸𝑒 −𝑎𝑡 Ø= 𝑒𝐿
+ =
𝑑𝑡 𝐿 𝐿

I Ø = ∫ Ø (𝑄)𝑑𝑡
𝑅𝑡 𝑅𝑡
𝐸𝑒 −𝑎𝑡
I𝑒 𝐿 = ∫ (𝑒 𝐿 ) ( )dt
𝐿
𝑅𝑡 𝑅𝑡
𝐸 −𝑎𝑡 1
I𝑒 𝐿 = 𝐿 ∫𝑒 𝐿 (𝑅⁄ )
𝐿 −𝑎
𝑅𝑡 𝑅𝑡
𝐸 𝐿
I𝑒 𝐿 = 𝐿 (𝑅−𝑎𝐿) (𝑒 𝐿 −𝑎𝑡 ) + c

−𝐸
When t = 0 ; I = 0 ; c = 𝑅−𝑎𝐿
𝑅𝑡 𝑅𝑡
𝐸 𝐸
I𝑒 𝐿 = ( ) (𝑒 𝐿 −𝑎𝑡 ) -
𝑅−𝑎𝐿 𝑅−𝑎𝐿
−𝑅𝑡
𝐸 𝐸𝑒 𝐿
I = (𝑅−𝑎𝐿) (𝑒 −𝑎𝑡 )- 𝑅−𝑎𝐿

−𝑅𝑡
𝐸
I = (𝑅−𝑎𝐿) (𝑒 −𝑎𝑡 − 𝑒 𝐿 )

Current at its maximum value:

26 | P a g e
 −𝑅𝑡
𝑑𝐼 𝑑 𝐸 𝑑𝐼
= ( ) (𝑒 −𝑎𝑡 − 𝑒 𝐿 ) but =0
𝑑𝑡 𝑑𝑡 𝑅−𝑎𝐿 𝑑𝑡

𝑅 −𝑅𝑡 𝑅 𝑅𝑡
0 = -a𝑒 −𝑎𝑡 + 𝐿 𝑒 𝐿 -at = ln 𝑎𝐿 − 𝐿

−𝑅𝑡 𝑅 𝑅
a𝑒 −𝑎𝑡 =
𝑅
𝑒 𝐿 t ( 𝐿 − 𝑎) = ln 𝑎𝐿
𝐿
𝐿 𝑅
−𝑎𝑡
𝑅 −𝑅𝑡 t = 𝑅−𝑎𝐿 𝑙𝑛 𝑎𝐿
𝑒 = 𝑒 𝐿
𝑎𝐿
−𝑅𝑡
𝑅
ln𝑒 −𝑎𝑡 = ln 𝑎𝐿 + ln 𝑒 𝐿

27. A pipe 20 cm in diameter contains steam at 100ºC and is covered with a certain
insulation 5 cm thick. The outside temperature is kept at 40ºC. By how much should
the thickness of insulation be increased in order that the rate of heat loss shall be
decreased by 20%?

Figure:

𝑥2 = 15 𝑐𝑚

20 cm

𝑥1 = 10 𝑐𝑚

Solution:
2𝛱𝐾𝐿 (𝑢1 − 𝑢2)
q= 𝑥
𝑙𝑛 2
𝑥1

Let y be the additional thickness of insulation

2𝛱𝐾𝐿 (𝑢1 − 𝑢2)
0.8q = 𝑥2+𝑦
𝑙𝑛
𝑥1

2𝛱 (100−40) 2𝛱 (100−40)
0.8 [ 15 ]=[ 15+ 𝑦 ]
𝑙𝑛 𝑙𝑛
10 10

120𝛱
743.8196005 = 15+𝑦
𝑙𝑛
10

15+𝑦
𝑙𝑛 = 0.5068313852
10

27 | P a g e
y = 1.600228796 cm. or 1.60 cm.

28. A steam pipe of radius 3 cm and at 100ºC is wrapped with a 1 cm layer of insulation
of thermal conductivity 0.0003 cal/cm.deg.sec and then that layer is wrapped with a 2
cm layer of insulation of thermal conductivity 0.0002 cal/cm.deg.sec. At what
temperature must the outside surface be maintained in order that 0.008 𝑐𝑎𝑙⁄𝑠𝑒𝑐 will
flow from each square cm of pipe surface?

Figure:

3 cm
1 cm

2 cm

Solution:

q = 𝑞1 + 𝑞2

A = 1𝑐𝑚2

1 = 2ΠrL
1
L = 2𝛱(3)

1
L = 6𝛱

2𝛱𝐾𝐿 (𝑢1 − 𝑢2) 2𝛱𝐾𝐿 (𝑢1 − 𝑢2 )

0.008 = 𝑥 + 𝑥
𝑙𝑛 2 𝑙𝑛 3
𝑥1 𝑥1

1 1
2𝛱(0.0003)( ) 2𝛱(0.0002)( )
6𝛱 6𝛱
0.008 = (100 - 𝑢2 ) [ + ]
ln4⁄3 ln6⁄4

28 | P a g e
𝑢2 = 28.33ºC

29. A pipe 10 cm in diameter, contains steam at 100ºC. it is to be covered with two coats
of insulating material each 2.5cm thick; the inner with K= 0.0006 cal/cm.deg.sec and
the outer with K= 0.00017 cal/cm.deg.sec. If the outside surface temperature is 30ºC,
find the heat loss per hour from a meter length of pipe.

Solution:
(𝑢 − 𝑢2 )
q = ln𝑟2⁄ 1 𝑟3
𝑟1 ln ⁄𝑟1
+
2𝛱𝐾1 𝐿 2𝛱𝐾2 𝐿

𝑢1 = 100ºC

𝑢2 = 30ºC

𝑟1 = 5 cm

𝑟2 = 7.5 cm

𝑟3 = 10 cm

L = 1 meter or 100cm

100−70
q=
ln7.5⁄5 ln10⁄5
+
2𝛱(0.0006)(100) 2𝛱(0.00017)(100)

q = 18.57342744 cal/sec

q = 66,864.33879 cal/hour

29 | P a g e
 30. A beam L m long is simply supported at its end and carries a uniform load of w 𝑁⁄𝑚
along its entire length. Determine the equation of the elastic curve and the maximum
deflection of the beam.

Figure:

x L -x

Solution:
𝐿−𝑥
mA = -w (L – x) ( )
2

−𝑤
= (𝐿2 − 2𝐿𝑥 + 𝑥 2 )𝑑𝑥
2

𝑑2 𝑦⁄
EI 𝑑𝑥 2 = 𝑚𝐴
2
𝑑 𝑦⁄ 𝑤
(𝐿2 − 2𝐿𝑥 + 𝑥 2 )𝑑𝑥
∫ 𝐸𝐼 𝑑𝑥 2 = ∫ − 2
𝑑𝑦 𝑤 𝑥3
EI ⁄𝑑𝑥 = − 2 (𝐿2 𝑥 − 𝐿𝑥 2 + 3 + 𝑐)

𝑑𝑦⁄
When x = 0 ; 𝑑𝑥 = 0 ; 𝑐 = 0
𝑤 2
𝐿 𝑥 2 𝑥3𝐿 𝑥4
EIy = − 2 ( 2 − 3 + 12)
𝑤 𝐿2 𝑥2 𝑥3𝐿 𝑥4
y = − 2𝐸𝐼 ( − + ) ------- Equation of the elastic curve
2 3 12

𝐿
When x =
2

30 | P a g e
 2
2 L L 3 L 4
w L ( ) ( ) L ( )
𝑦𝑚𝑎𝑥 = − ( 2 − 2 + 2 )
2 2 3 12

5wL4
ymax = − 384EI --------- Maximum deflection of the beam

31. A beam L ft long is fixed at both ends and carries (a) a concentrated load of P lb at the
midspan, (b) a uniform load of w lb⁄ft along its entire length. Determine the equation
of the elastic curve and the maximum deflection of the beam for each type of loading.

HINT: Since both ends of the beam are fixed, there is a moment of couple C
exerted by the masonry at each end.
x P

Solution: 𝐴
𝑃𝐿⁄
(a) 8

𝐿⁄ 𝐿⁄
2 2

1
(0.5𝐿)(𝑃𝐿⁄4) − 𝐿⁄2 (𝐴) = 0
2

But A = 𝑃𝐿⁄8

𝑑2 𝑦 𝑃 𝑃𝐿⁄
EI𝑑𝑥 2 = (𝑥) − 8
2

𝑑𝑦 𝑃𝑥 2 𝑃𝐿𝑥
EI𝑑𝑥 = 4
− 8
+ 𝑐

𝐿 𝑑𝑦⁄
When x = ; 𝑑𝑥 = 0 ; 𝑐 = 0
2

𝑃𝑥 3 𝑃𝐿𝑥 2
EIY = 12
− 16
+ 𝑐1

𝐿
When x = 2 ; y = 0 ; c= 0

𝑃𝐿3
𝑦𝑚𝑎𝑥 = − 𝑓𝑡
192𝐸𝐼
1 𝐿 𝐿 1 1 𝑊𝐿2
(b) (2) (2) 1⁄4 𝑊𝐿2 − 2
𝐴 − (
3 2
𝐿) ( 8
) =0

𝑊𝐿2 𝑑2𝑦 𝑊𝐿 𝑊𝐿2

But A=
12
EI𝑑𝑥 2 = 12
(𝑥) − 𝑊𝑥 (𝑥⁄2) − 12

31 | P a g e
 𝑑𝑦 𝑊𝐿𝑥 2 𝑊𝑥 3 𝑊𝐿2 𝑥 𝐿
EI𝑑𝑥 = − − + 𝑐 When x = 2 ; y = 0 ; c= 0
4 6 12

𝑊𝐿4 𝑊𝐿4 𝑊𝐿4

When x = 2; 𝑑𝑦⁄𝑑𝑥 = 0 ; 𝑐 = 0
𝐿
EIY = 96
− 384
− 96

𝑊𝐿𝑥 3 𝑊𝑥 4 𝑊𝐿2 𝑥 2 𝑊𝐿4

EIY = − − + 𝑐 Y = − 384𝐸𝐼
12 24 24

32. A cable weighing 58.4 N⁄m is suspended between two supports on the same level and
305 apart. Compute the maximum tension (at the supports) and the length of the cable
if the sag is 60m.

Figure:

305 m

60 m = d

d = 60m
W = 58.4 N⁄m

Solution:

T = wy = w (a + d)

L
d = a (cosh 2a - 1)
305
d = a (cosh – 1)
2a
60 305
= cosh –1
a 2a
60 305
+ 1= cosh 2a
a
60 1 152.5 −152.5
+ 1 = (ρ + ρ )
a 2 4 a
a = 203.082 m

T = 58.4 (203.082 + 60)

T = 15.300 N

L
S = 2a(sinh 2a)
305
S = 2 (203.082) sinh 2 (203.082)

32 | P a g e
S = 335.4 m

33. Compute the span (L) and sag (dip d) of a cable 183 m long and weighing 14.60 N⁄m
and supported at points on the same level with a maximum tension at the supports of
3560 N.

Figure: L

d=?

Solution:

T = W (a + d)

3560 = 14.60 (a + d) ----------- Equation 1

L
S = 2asinh 2a
L
183 = 2asinh 2a
183
L = 2asinh−1 [ ] ----------- Equation 2
2a

L
d = a (cosh 2a - 1) ------------- Equation 3

(Equation 1 and 3)
L
3560 = 14.60 [a + a (cosh 2a − 1)]
L
243.8356 = a + a(cosh 2a − 1)

243.8356
L = 2acosh [ ] ---------- Equation 4
a

(Equation 2 and 4)

183 243.8356
2asinh−1 ( ) = 2 acosh ( )
2a a
a = 226.02 m
3560
d = 14.60 − 226.02 = 18m

33 | P a g e
 243.8356
L = 2(226.02) cosh ( ) = 178 m
226.02

34. A cable weighing 87.5 N⁄m is suspended between points 244 m apart on the same
level. The sag or dip is 61 m. compute the maximum tension at the support and the
length of the cable.

Figure:
244 m

d = 61m

T = W (a + d)
T = 87.5 (a + 61) -------- Equation 1

L
d = acosh (2a − 1)
244
61 = acosh ( 2a − 1)
a = 131.1 m

T = 87.5 (131.1 + 61)

T = 16810 N

L
S = 2asinh 2a
244
S = 2 (131.1) sinh 2(131.1)
S = 280 m.

34 | P a g e
 35. A cable weighing 29.2 N⁄m is to be suspended between supports on the same level
and 183 m apart. If the tension at the supports is 5340 N, determine the minimum sag
or dip to which the cable may be drawn up, and the length of the cable.

Figure:
183 m

T= 5340 N

W= 29.2 N⁄m

Solution:

T = W (a + d)
5340 = 29.2 (a + d)
182.88 = (a + d) ------ (Eq. 1)

L
d = a (cosh 2a - 1)
183
d = a (cosh - 1) ----- (Eq. 2)
2a

183
182.88 = a + a (cosh − 1)
2a
a = 155.1 m
183
d = a (cosh - 1)
2a
183
d = 155.1 (cosh2(155.1) – 1 )
d = 28 m

L
S = 2a sinh 2a
183
S = 2 (155.1) sinh 2 (155.1)
S = 194 m

35 | P a g e