Download Application of Derivatives

Lesson – 3
Introduction to Calculus.
Application
of
Derivatives
1.
The slope of a curve.
Example 1.
Determine the equation of the tangent line to the function of
y = 3x2 – 2x – 5 at point x = 2
Solution:
The slope of the tangent line at point x to a function is
determined as the derivative of this function at this point.
We find of the derivative of a function y = 3x2 - 2x -5, first.
dy
= 6x – 2
dx
and the gradient (slope) is equal to a value this function at point
x=2
so m = 6(2) – 2 = 12 – 2 = 10
To find the value of y when x = 2, we substitute x = 2 to the
primary function.
3(2)2 – 2(2) – 5 = 12 – 4 – 5 = 3
Therefore, equation of the tangent to a function at point x, is equal
to the equation if a lie p[assign through point (2,3) and having the
gradient m.
y – y1 = m(x – x1) where x1 = 2 and y1 = 3 (from the equation of the
line)
y – 3 = 10(x – 2) or if we remove brackets, y = 10x – 17.
We point out not every curve has the tangent line at of its
points. For example, a curve y = |x| does not the tangent line at
point (0,0), as you can see in figure 3.1. At (0,0) a secant line from
the right hand must always be line x = y. From the left hand it is line
y = - x. Since there is no common limiting position, there is no the
tangent line. Thus, is no slope at point (0,0).
Fig.3.1.
Example 2.
The gradient of a function y = 5x2 – 6x + 3 at certain point is
equal to 4. Find of coordinates of this point on a curve.
Solution:
We must find first the derivative of a function y = 5x2 – 6x + 3
dy
= 10x – 6
dx
This function represent the slope at point x to a curve above.
Thus,
m = 4 = 10x – 6
Solving for x we have:
4 = 10x – 6 and: x = 4  6 =  (4  6) = 1
10
10
and co-ordinate y we determine by substitution x = 1 to the primary
function.
Y = 5(1)2 – 6(1) + 3 = 5 – 6 + 3 = 2
Therefore, these co-ordinates are (1,2)
1
The derivative as a rate of change.
If a body has a constant velocity V. The velocity can be
expressed as:
V= s
t
Where s is the distance travelled in t seconds by a body. The
velocity is a gradient of the distance over time taken as shown in
figure 3.2 (A).
Fig.3.2
If velocity is not constant, the distance-time graph is a curve
as seen in the same figure but (B). Average velocity over a short
period of time t and distance s is given by the gradient of the
chord AB, that is, the average velocity over the small interval of
time. This is:
V = s
t
As t → 0 the chord becomes a the tangent line, so that at
the point A, the velocity is given by the derivative of a distance in
respect of time at this point.
The same applied to the acceleration, which is defined as
change in velocity over the interval of time. Both velocity and
acceleration may be written as follows:
V = ds
a = dV
and
dt
dt
(3.1)
Example 3.
A body moves according to the equation s = 25t - t2 where
s is the distance in meters from a reference point A.
Find:
(i)
The velocity of the body at t = 3 seconds
(ii)
The acceleration
(ii)
The time when the velocity becomes 0
Solution:
As s 25t - t
(i)
Velocity V = ds = 25 - 2t
dt
when t = 3
V = 25 - 2(3) = 19 ms-1
(ii)
(iii)
Acceleration a = dV = d (25  2t ) = - 2 ms-2
dt
Since V
t = 12.5 s.
dt
25 - 2t, then t = 25/2 if V= 0
Example 4
A spherical balloon is being filled with air. Find the rate of
change of its volume with respect to its radius. Evaluate this rate
of change when the radius is 1 m.
Solution:
The volume U of a sphere of radius r is U = 4/3r3 .To find
the rate of change of a function we take its derivative. Thus, the
rate of change U with respect of r is:
3
dU
= d (4 / 3r ) = 3(4/3r2) = 4r2
dr
dr
When r = 1m, the rate of change is
dU
= 4r2 = 4 m3/m
dr
(because radius is 1 m long).
This means that when the radius is 1 m, changing the radius by 1
m will change the volume by 4.
Example 5
The rheostat frequency of a series a.c. (alternative current) circuit
is given by  or f if you wish.
f=
1
2 LC
Where L and C are the inductance and capacitance in the circuit,
respectively. Find the rate of change f, with respect to C, assuming
L remains constant.
Solution:
The ratio of change f in respect to C is equal to:
d(
1
2 LC
dc
)
1 / 2
1 / 2


= d  1 C  = - C
dc  2 L 
4 L
EXERCISES:
If a particle obeys the equation s = 4t – 64t2.
1.
Find:
(i)
expression for its velocity and acceleration
(ii)
its velocity when t = 2 seconds
2.
A car travels along a road so that its distance s in meters
from 0 is given by s = 2t + 6t2
Find:
(i)
its distance from o when t = 4 s
(ii)
the time at which velocity is 62 ms-1
3.
When viewed through a microscope a bacterium is seen to
move in accordance with the equation s = (4t + 6t2 )10-6
Find its velocity after 2 seconds.
4
The distance s meters moved by a body in t seconds is given by
s = 2 t3 - 13t2 + 24t +10
Find.
(i)
the velocity when t = 4 s
(ii)
the value of t when the body comes to rest
(iii)
the value of t when the acceleration is 10 ms-2
5.
In the following problems, you are given an equation of
motion for a particle travelling in a horizontal line. For the given
value of t, find the position, velocity acceleration and direction of
motion. Assume that t is in seconds and s is in meters and that the
positive direction is to the right.
a)
s = t2 - 3t, t = 4
b)
s = 1/2t + 1, t = 2
c)
s = 2t3 +6, t = 1
d)
s = -3t2 + 2t + 1, t = 1
f)
s = t4 –2t3 + t, t = 1
g)
s = 4t3 -3t + 2, t = 0
h)
6.
s = t4 - t2 + t, t = 0
i)
s = t4 - t5/2, t = 1
If w = i2 - 2i + 4, find the rate of change of w with respect to i
when, i = 3
7.
8.
If y = p + p-1 find the rate of change y with respect to p when p =1
Find the rate of change of the volume V of a sphere with
respect to its diameter d if, V = 1/6d3.
9.
Find the rate of change of the area A of a circle with respect
to its radius r if A = r2. Evaluate when, r = 3 cm.
1O.
The volume of a certain gas varies with pressure p according
to the equation:
p = 150
V
Find the rate of change of p with respect to V when V = 5.
11.
The current i in a certain resistor as a function of the power P
developed in the resistor is given by i =
P . Find the rate of
change of i with respect to P
when,
12.
P=4
The induced electromotive force , in volts, developed in a certain
coil is:
 = 0.05 di
dt
where the current i in the coil is a function of time. If i = 2t 2 + 3
amperes, where t is in seconds, find  at t = 2sek.
13.
Dalton's law of partial pressures for a mixture of gases is:
p = (n1 n2 ) RT
V
where p is the pressure, R the universal gas constant, T the
absolute temperature, V the volume and n1 and n2 the number of
moles of each gas. Find the rate of change of the pressure with
respect to the volume, assuming all other quantities remain
constant.
Answers:
1.
3.
(4 - 128t), -128, -252
2.
28 x 10-6
4.
(i)
3 sec
5.
a)
16m/s
3 s or 4/3s
f)
h)
0 m, 1m/s, -2 m/s2,
to right
4
7.
2p - p- 2, 1
8.
1/2d2
10.
–6
1/4
12.
(n1  n2 ) RT
V2
+5 sec
(iii)
11.
c)
0 m, - 1 m/s, 0 m/s2 , to
left
6 cm2/cm
13.
(ii)
(ii)
4 m, 5 m/s, 2 m/s2, to right
8m, 6m/s, 12m/ S2, to right
6.
(i) 104m
9.
0.4 V
2r,