differentiation

302 DIFFERENTIAL CALCULUS
height reached, (d) the velocity with which
the missile strikes the ground.
[ ]
(a) 100 m/s (b) 4s
(c) 200 m
(d) −100 m/s
2. The distance s metres travelled by a car in t
seconds after the brakes are applied is given
by s = 25t − 2.5t 2 . Find (a) the speed of the
car (in km/h) when the brakes are applied,
(b) the distance the car travels before it stops.
[(a) 90 km/h (b) 62.5 m]
3. The equation θ = 10π + 24t − 3t 2 gives the
angle θ, in radians, through which a wheel
turns in t seconds. Determine (a) the time
the wheel takes to come to rest, (b) the
angle turned through in the last second of
movement.
[(a) 4 s (b) 3 rads]
4. At any time t seconds the distance x metres
of a particle moving in a straight line from
a fixed point is given by x = 4t + ln(1 − t).
Determine (a) the initial velocity and
acceleration (b) the velocity and acceleration
after 1.5 s (c) the time when the velocity is
zero.
⎡
(a) 3 m/s; −1m/s 2 ⎤
⎢
⎣
(b) 6 m/s; −4m/s 2
⎥
⎦
(c) 3 4 s
5. The angular displacement θ of a rotating disc
is given by θ = 6 sin t , where t is the time in
4
seconds. Determine (a) the angular velocity
of the disc when t is 1.5 s, (b) the angular
acceleration when t is 5.5 s, and (c) the first
time when the angular velocity is zero.
⎡
(a) ω = 1.40 rad/s
⎤
⎢
⎣(b) α =−0.37 rad/s 2 ⎥
⎦
(c) t = 6.28 s
6. x = 20t3
3 − 23t2 + 6t + 5 represents the distance,
x metres, moved by a body in t seconds.
2
Determine (a) the velocity and acceleration
at the start, (b) the velocity and acceleration
when t = 3 s, (c) the values of t when the
body is at rest, (d) the value of t when the
acceleration is 37 m/s 2 and (e) the distance
travelled in the third second.
⎡
(a) 6 m/s; −23 m/s 2 ⎤
(b) 117 m/s; 97 m/s 2
3
(c)
⎢ 4 sor 2 5 s
⎣(d) 1
2 1 s
⎥
⎦
(e) 75 1 6 m
28.3 Turning points
In Fig. 28.4, the gradient (or rate of change) of the
curve changes from positive between O and P to
negative between P and Q, and then positive again
between Q and R. At point P, the gradient is zero
and, as x increases, the gradient of the curve changes
from positive just before P to negative just after. Such
a point is called a maximum point and appears as
the ‘crest of a wave’. At point Q, the gradient is also
zero and, as x increases, the gradient of the curve
changes from negative just before Q to positive just
after. Such a point is called a minimum point, and
appears as the ‘bottom of a valley’. Points such as P
and Q are given the general name of turning points.
Figure 28.4
It is possible to have a turning point, the gradient
on either side of which is the same. Such a point is
given the special name of a point of inflexion, and
examples are shown in Fig. 28.5.
Maximum and minimum points and points of
inflexion are given the general term of stationary
points.
Procedure for finding and distinguishing
between stationary points:
(i) Given y = f (x), determine dy
dx (i.e. f ′ (x))