UNIVERSITY OF RIZAL SYSTEM Morong, Rizal College of Engineering
Assignment in CE-22: CE Professional Prociency Development Course
DIFFERENTIAL AND INTEGRAL CALCULUS ENGR. ROMMEL BARAQUIEL nstructor
JAIRELL JANE JANE . . BARRINUEVO BARRINUEVO !-"#CE
6 ¿ ¿ ¿3
I. ro!le"# $%& ro!le" $. $in% t&e volume of t&e soli% generate% 'y rotating t&e curve ()2 * +y2 . along t&e line +) * y 2/0 Sol'(ion)
ro!le" *. $in% t&e volume generate% 'y rotating t&e area in t&e t&ir% an% fourt& 1ua%rants 'oun%e% 'y t&e curve 2!)2 * .y2 (// an% t&e )-a)is, a'out t&e )-a)is0 Sol'(ion) 2!)2 * .y2 (// 2 2 x y 36
+
25
=1
a ., ' ! 2!)2 * .y2 (// 2
y =25 −
25 36
x
2
(
6
25 36
2
)
x dx
∫ π ( 25− 36 x )dx 25
V =
¿ ¿ ¿3 ¿ 25 ¿ −¿ 25 ¿−¿
25 ( 6 )−¿ ¿ V =¿
V + *,,- '. 'ni(# ro!le" /. $in% t&e area enclose% in t&e curve r2 +cos40 Sol'(ion) r2 +cos4 E1uating r /, / +cos4 π θ=± = 90 ° 2
% 3y2%)
dV = π 25−
25 (−6 ) −6
2
$or r to 'e /: π θ =± 2
A =
(
V = π 25 x −
)
∫ r dθ 2 2
π
3 6
108 −6
β
α
−6
25 x
1
A =
1
2
∫ 4cos θ dθ
2 −π 2
π 2
A = 2
∫ cos θ dθ
− π 2
π
A = 2 ( sin θ ) −2 π 2
A = 4 sq.units
ro!le" 0. Determine t&e surface area generate% 'y revolving a'out t&e y-a)is t&e portion of t&e curve ) 2 + * y2 from y/ to y+0 Sol'(ion) 2 2 x =4 + y
x
2
4
−
y
2
4
=1
ro!le" 1. A tan5 is full of oil 6%ensity !/ pcf70 t &as a %iameter of 8/ ft0 n% t&e 9or5 %one in ft-l' in pumping all t&e li1ui% out of t&e top of t&e tan50 a0 f t&e tan5 is a &emisp&erical tan50 '0 f t&e tan5 is a conical tan5 9it& a &eig&t of 2/ ft0 c0 f t&e tan5 is cylin%rical tan5 9it& a &eig&t of 2/ ft0 Sol'(ion)
V ( 0,0 ) a=b =2 For y = 0 : 2
x =4 2
x −4 =0
( x + 2 ) ( x −2 ) =0 x =∓ 2
ro!le" 2. &e velocity of a particle along a straig&t pat& is + * t * t2, 9&ere t is time of travel in secon%s an% v is in m;s0 $rom t&e start it travels < m in 8 secon%0 $in% t&e total %istance it travels in 2 secon%s0 Sol'(ion) V = 4 + t + 3 t 2
For y = 4 : x
2
4
−
16 4
=1
2
x −20= 0 x =2 √ 5 x =√ 4 + y
2
v%t 6+ * t * t 27%t 2 s = ( 4 + t + 3 t ) dt
∫
2
dV = π x dy dV = π ( 4 + y 2 ) dy
2
s = 4 t +
4
∫ ( 4 + y ) dy 2
V =
0
( ) = [ ( )+ ]
V = π 4 y +
V π 4 4 V =
112 3
y
3 4
3
4
0
π cu. units
2
+t 3 +C ; @ t = 1 sec,
s =8 m 8 =4 + 1 / 2+ 1 + C
C =
5 2
3
3
t
2
s = 4 t +
t
2
+ t 3 +
5 2
S + *,.1 " ro!le" &. Determine t&e perimeter of t&e curve r +68 = sin470
Sol'(ion) r = 4 ( 1 – sin θ )
%ubstitutin! t&e 'a(ue o) # :
quatin! r = 1,
s =2 √ 32+ 32sin θ
0 =4 ( 1 – sin θ )
π −π θ= ∨ 2
'a(uatin! ¿−
2
2
r =16 ( 1− sin θ )
π π 2
¿
2
2
√
s =2 32+ 32sin
r =16 ( 1 −2sin θ + sin θ ) 2
2
π 2
−2
√
32 + 32sin
−π 2
=16
2
θ +¿ 16sin θ
*u(ti+(yin!by 2 ( symmetrica( ) :
2
r =16 −32 sin ¿
r = 4 – 4sin θ
# + *3$24 + /* 'ni(#
dr =−4cos θ dθ
( )
2
dr = 16cos2 θ dθ
∫ √ 16 −32sin θ +( sin θ + cos θ ) 16 d s =∫ √ 16 −32sin θ + 16 dθ s =∫ √ 32−32sin θ dθ 2
s=
2
r = 4 ( 1 – sin θ ) is a cardioid on y − axis
∫ √ 32−32sin θ dθ "et # =sin θ
II. ro!le"# $%$, ro!le" $. f # t = t2, 9&ere s is %istance in meters an% t is time in secon%s, n% t&e velocity 9&en t&e rate of c&ange of velocity is zero0 Sol'(ion) 3 2 s =t −t
$=sin −1 #
dθ =
ds d t
d#
√ 1− # 2
∫ √ 32 √ 1−# d# =√ 32∫ √ 1√ +1−# √ #1d# − # √ 1−# 2
s =2 √ 32+ 32 #
='
2
' =3 t −2 t ------ > d' dt
=a
a =6 t −2 ; a=0 0 =6 t −2
" A -= 27.60 m
1
t = sec ------ ? 3
[] [] 2
1
' =3
−2
3
'=
−1 3
1
5¿ 2 5
+
x
2=0.0191 ( 2 )( 27.60 )
m/ s
()
2
5
dV =2 dt
3
ro!le" *. @ater o9s on a conical tan5 at t&e rate of 2 m ;s0 &e conical tan5 is + m across t&e top an% . m %eep0 @&en t&e %ept& of t&e 9ater is ! m, &o9 fast is t&e 9ette% surface area of t&e tan5 c&angingB Sol'(ion) 2
2
3
¿
( )
d"A =1.90 m2 / s dt
ro!le" /. @ater is o9ing into a sp&erical tan5 2+ m in %iameter at a rate of 2 m;s0 $in% t&e rate of 9ater rise after +/ minutes0 Sol'(ion) 3 dV 2 m = dt s 4
= ; , =√ ¿
V )u(( = π r
5
5
5 √ 10
3
3
1
3
12 ¿ 4
V )u(( = π ¿
2
V = π r &
3
3
r=
1 2
3
V )u(( =7238.23 m
( 2 πr ) s &
V @ 40=2 ( 4 ) ( 60 )
"A πs& 1
V =
1 3
V @ 40= 4800 m
( )
2
"A (& ) πs&
V = π 3
3
3
x = ; , =
"A =
( ) d"A dt
π 2 V one = y ( 3 − y )
[ ( )]
3
2
"A ( 5)
π
5 √ 10
2
( )
dV π dy = ( 3 y −3 y 2 ) −① dt 3 dt
3
V =0.0191 " A
2
( )
dV d"A =0.0191 (2 )( "A ) dt dt
( )( )
" A -= π
3
5
5 √ 10
3
3
y =/ 7238.23 − 4800=
π 3
2
y ( 3 − y )
=1 2
dV d, d, dx = xy + x, + y, ; dt dt dt dt
y =9.35 m− ②
dV : dt
①∧②∈
dV =138 )t 3 / s dt
2
9.35 ¿
6 ( 12 ) ( 9.35 ) −3 ¿ 2=
π 3
dV =4 ( 5 ) ( 3 ) + 4 ( 6 ) ( 2 ) + 5 (6 )( 1 ) dt
( ) dy dt
¿
( )
dy = 4.63 x 1 0−3 m/ s dt
ro!le" 1. 9o si%es of a triangle measure 8! m an% 2/ m0 @&en t&e inclu%e% angle is ./, it increases at t&e rate of 80!;sec0 &o9 fast is t&e area of t&e triangle c&angingB Sol'(ion) 1
A = ab%inθ 2
a =15 m , b =20 m ro!le" 0. At a certain instant, t&e %imensions of a rectangular parallelepipe% are +, ! an% . feet, an% t&ey are eac& increasing respectively at t&e rates of 8, 2 an% feet per secon%0 At 9&at rate is t&e volume increasingB Sol'(ion) x =4 )t
y =5 )t =6 )t
' = xy, dx =1 )t / s dt dy =2 )t / s dt d, =3 )t / s dt
θ= 60 ° ,
dθ 1.5 ° = dt s
dA 1 dθ = abCosθ dt 2 dt dA 1 = ( 15 ) ( 20 ) cos60 ° ( 1.5 ° ) dt 2 dA 2 =112.5 m / s dt
ro!le" 2. &e &an%s of t&e cloc5 are 8! cm an% 2/ cm long0 o9 fast are t&e en%s of t&e &an%s approac&ing at +:// PMB Sol'(ion)
ro!le" &. A man is 9al5ing at 2 m;s to9ar%s a 'uil%ing 8! m &ig&0 A 'ill'oar% + m is locate% on top of t&e 'uil%ing0 o9 fast is t&e su'ten%e% angle of t&e 'ill'oar% c&anging 9&en
t&e man is 2/ m from t&e 'uil%ingB &e eye level of t&e man is 80! m a'ove t&e groun%0 Sol'(ion) θ= α − β - > tan ∝=
17.5
tan β =
13.5
x
x
−1 17.5
; ∝= tan
; β = tan
- ?
x
−1 13.5
-
x
ro!le" 5. An airplane is traveling nort& at !./ 5p& is %irectly a'ove a certain to9n at 2 PM0 A secon% airplane traveling east at .// 5p& is %irectly a'ove t&e same to9n / minutes after0 o9 fast is t&e %istance 'et9een t&e t9o c&anging apart at + PMB Sol'(ion) V A =560 #+ &
V - =600 #+ &
②∧ ③ ¿ ① : θ= tan
−
dθ = dt
−1
17.5
x
−1
− tan
x 1+
2
x
s =( d a ) + ( d b 2
( ) ( )
17.5 dx
dt
306.25
13.5 dx
+
x
2
1+
2
x
x =
d = 't
13.5
20∧dx
dt
dt
182.25
√
2
)
[ ( )]
s = ( 560 t ) + 600 t −
2
x
2
√
2
2
(
2
1
2
2
s = 313600 t + 360000 t − t +
=2
1 4
)
s = √ 673600 t −360000 t + 90000 2
dθ = dt
−
17.5
1+
2
20
13.5
( 2)
306.25 2
20
2
+
20 1+
( 2)
182.25 2
20
dθ −3 =−3.19 x 10 rad / s dt dθ =−0.18 ° / s dt
1347200 t −360000 ds = dt 2 √ 673600 t 2−360000 t + 90000 1347200 ( 2 )−360000 ds = dt 2 √ 673600 ( 2 )2−360000 ( 2)+ 90000
ds =812.36 #m / &r dt
ro!le" 6. $in% t&e area of t&e largest rectangle t&at can 'e inscri'e% in t&e ellipse .) 2 * 2!)2 (//0 Sol'(ion) 2 2 36 x + 25 x =90 0
2
x
25
+
y
2
36
2
0 = x
=1
a =6, b=5
x =±
A = 2 x (2 y )= 4 xy - >
5
√ 2
−
25 2
5 2 =± √ 2
0a#in! Abso(ute Va(ue :
¿ ((i+se quation : 2
x
25 36 x
2
+
y
x =
2
36
=1
+ 25 x 2=90 0
√
y = 36 −
36 25
√
2
x ②
√
A = 4 x 36 −
y = 36 −
[√
y =3 √ 2 36 25
−36 x
25 36 −
0 =−144 x
2
36 25
+100
0 =−144 x
2
2
③
to ?:
② ¿①:
A = 4 x
5 √ 2
2
x
(
]
( )
36 5 √ 2 25
2
2
F
2
x an% F to >:
+4
36 −
√
36−
36 25
2
x
A = 4
( )( 5 √ 2 2
3 √ 2 )
A = 60 sq.units 36 25
2
x
+ 3600−144 x 2
)
ro!le" $,. $in% t&e volume of t&e smallest cone t&at can 'e circumscri'e% a'out a sp&ere of ra%ius < cm0 Sol'(ion)
