306 Higher Engineering Mathematics

Hence (−2, 9) is a maximum point.

Whenx=3,

d^2 y

dx^2

= 2 ( 3 )− 1 = 5 ,

which is positive.

Hence

(

3,− (^1156)
)
is a minimum point.
Knowing (−2, 9) is a maximum point (i.e. crest of
awave),and
(
3 ,− (^1156)
)
is a minimum point (i.e.
bottom of a valley) and that whenx=0,y=^53 ,a
sketch may be drawn as shown in Fig. 28.6.
x
y
22 21 210
212
(^21156)
28
24
4
89
12
x^3
3
x^2
2
5
y 5 2 26 x (^13)
3
Figure 28.6
Problem 14. Determine the turning points on the
curvey=4sinx−3cosxin the rangex=0to
x= 2 πradians, and distinguish between them.
Sketch the curve over one cycle.
Sincey=4sinx−3cosx
then
dy
dx
=4cosx+3sinx= 0 ,
for a turning point, from which,
4cosx=−3sinxand
− 4
3

sinx
cosx
=tanx
Hencex=tan−^1
(
− 4
3
)
= 126. 87 ◦or 306. 87 ◦,since
tangent is negative in the second and fourth quadrants.
Whenx= 126. 87 ◦,
y=4sin126. 87 ◦−3cos126. 87 ◦= 5
Whenx= 306. 87 ◦,
y=4sin306. 87 ◦−3cos306. 87 ◦=− 5
126. 87 ◦=
(
126. 87 ◦×
π
180
)
radians
= 2 .214rad
306. 87 ◦=
(
306. 87 ◦×
π
180
)
radians
= 5 .356rad
Hence (2.214, 5) and (5.356, −5) are the
co-ordinates of the turning points.
d^2 y
dx^2
=−4sinx+3cosx
Whenx= 2 .214rad,
d^2 y
dx^2
=−4sin2. 214 +3cos2. 214 ,
which is negative.
Hence (2.214, 5) is a maximum point.
Whenx= 5 .356rad,
d^2 y
dx^2
=−4sin5. 356 +3cos5. 356 ,
which is positive.
Hence (5.356,−5) is a minimum point.
Asketchofy=4sinx−3cosxis shown in Fig. 28.7.
 3
 5
5
0 2.214
5.356
/2  3 /2 2 
y4 sin x3 cos x
y
x(rads)
Figure 28.7