Solutioon to Topic 4 Applicattion of Inte gration
Q1
x = -/2
x = /2
y = tan
t x
x = /3
V =
3
0
=
( tan x )
( sec
3
0
2
2
dx
x 1) dx
= [ tan x x ]0 3
= 3 3 unnits3
Q2
Area =
0
= y dy
1
1
= tann ( 2 ) d
0
2
= 0.35113
Q3
y = ex
2
+1
When x = 0 , y = e and
Solution to Blk A Calculus Topic 1 Application of Differentiation I
1.
(i)
40 = 2 y + 3x
3
y = 20 x
2
1 2
x sin(60 )
2
1 3
= xy + x 2
2 2
Area, A = xy +
3 1 3
= x 20 x + x 2
2 2 2
3
3 2
= 20 x x 2 +
x
2
4
3 3 2
= 20 x +
x
4
2
(ii)
(Shown)
3 3
dA
=
Pioneer Junior College
2016 J2 H2 Mathematics Revision
Topic 8 Differential Equations
(To be completed by 13 May, Friday)
Tip 1 : (Time Management) 1 mark
1.5 min (Read+ Do) + 0.3 min (check)
Tip 2 : Always check your answers (using different approach)
:
Pioneer Junior College
H2 Mathematics
Session : Applications of differentiation (Maxima and minima)
At the end of the session, students should be able to understand the underlying concepts in solving
maxima and minima problems.
(1) Form equations for the
Name: _
Class: _
Cluster 6: Integration and its applications
XIX Integration techniques
Integration is the reverse process of differentiation. So the formulae of integration can be
deduced from formulae of differentiation with arbitrary constant being con
Solution to Topic 2
dx
1
dy
1. (i)
= 1+ ,
= 1,
dt
t
dt
dy
1
t
=
=
dx 1 + 1 t + 1
t
dy
t
=
> 0 for all t > 0
dx t + 1
Hence C does not have a stationary point
Since t > 0, t + 1 > 0,
(ii)
(0, 1.567)
y=1
When x = 0, t + ln t = 0 t = 0.5671432904 (by g.c.)
y
Pioneer Junior College
H2 Mathematics
Session: Vectors (Algebra and Lines)
At the end of the session, students should be able to:
(1)
(2)
(3)
(4)
(5)
Find equation of line OR direction vector of line
Find foot of perpendicular and point of intersection be
Name: _
Class: _
Cluster 4: Differentiation and its applications, Maclaurins series
TOPIC XIV Differentiation techniques and its applications (I)
(1) Techniques & implicit differentiation
xn
lnx ex sinx cosx tanx secx
y
nex cosx -sinx sec2x secxtanx
dy nx
Solution to Topic 5 DE
d2 y
1.
= x sin x
dx 2
dy
= x( cos x) cos x dx
dx
= x cos x + sin x + C
= x cos x + sin x + C
y = x sin x (1)(sin x)dx cos x + Cx + D
y = x sin x 2 cos x + Cx + D
y = f ( x ) passes through the origin f (0) = 0 D = 2
Tangent at the
Pioneer Junior College
H2 Mathematics
Session : Application of Differentiation (Parametric)
At the end of the session, students should be able to
(1) State the coordinates of a general point on curve given in parametric form.
dy
(2) Find
in terms of the p