# Chapter 6 Applications of differentiations

Chapter 6 : Applications of differentiation (Part 1) (A) Small approximations

By definitions of first principle :

dy ?y ? lim ? x ? 0 dx ?x Here ? x and ? y are small increments in x and y respectively. Hence , if ? x is very small , ? y dy ? then ? x dx ? Percentage change in y: ?y ?100% y

Example 1 : (a) If the radius of a circle increases from 5cm to 5.01 cm , (b) If the radius of a sphere decreases from 5cm to find the approximate increase in the area ? 4.98cm , find the approximate decrease in [solution:] the volume ? Let A be the area of the circle and r be the radius . = 2 ? r= 5.01 – 5 = 0.01 dA ? 2? r dr ? A dA ? ? r dr dA since ? A ? ?? r dr ? 2? (5) ? 0.01

? 0.1? Hence , the approximate increase in the area is 0.1 ? cm2 .

Example 2: (a) Given that y=x2 – 6x , use differentiation to find (b)Given that y=x4-2x3 . use calculus to find , in terms the small change in y when x increases from 2 to 2.01 .of p , the approximate increase in y when x ? x= [solution:] y = x2 – 6x increases from 2 to 2+p, where p is small . dy ? dx ? y dy since ? ? x dx dy ? y ? ?? x dx

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Example 3: Given y = x3 , find the approximate change in x if y decreases from 27 to 26.73 . [solution:] when y = 27 , x3 = 27 x=3. ? y = 26.73 – 27 = - 0.27 y = x3 dy = 3x2 dx ? y dy Then ? ? x dx ? 0.27 ? 27? x ?0.27 ?x? ? ?0.01 27 Hence , the approximate change in x is -0.01 . Example 4: (a) If y = a 4 x5 , find the approximate percentage change in x (b) If y=3x-2x2 , find the approximate percentage when there is a 5% change of y .change in y when there is a 4% change of x . ?y ?100% ? 5% [Solution :] Given that y

?y

y

?

5 ?? ? y ? 0.05 y 100

for y = a 4 x5 =ax 5/4

dy 5 1 ? ax 4 dx 4 ? y dy ? since ? x dx dy ? y ? ?? x dx 5 1 0.05 y ? ax 4 ? ? x 4

?x?

0.05 y ? 4 5ax

1 4

?

0.04(ax )

1 4

5 4

? 0.04 x

ax ?x 0.04 x ? ?100% ? ?100% x x ? 4% hence , there is a 4% change in x corresponding to a 5% change in y .

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Example 5: A closed cylinder has base radius r units and constant height h units .If the radius increases by a p% , find the approximate % increase in the surface area A .

Example 6: Given that y = [Solution : ]

243 243 , use the calculus to estimate the value of . 5 x 2.985 243 y = 5 =243x -5 ? x ? 2.98 ? 3 ? ?0.02 x 243 when x = 3 , y = 5 =1 3

找一个靠近 2.98 的 数

dy ?1215 ? -1215x -6 = dx x6 dy ?1215 5 ? ?? when x = 3 , 6 dx 3 3 243 so ? y ?? y 2.985 dy ? y ? ?? x dx ?5 ? 1 ? ( )( ?0.02) 3 1 ? 1? 30 31 ? 30

Note : The approximate value of y is given by y new = y initial + ? y where ? y ?

dy ?? x dx

Example 7: Use the method of differentiation to find the approximate value of ( both correct to 4 decimal places) (a)

9.1

(b) 3 123 [Ans : (a)3.0167 ; (b)4.9733 ]

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Misc.Problem: 1. The area of a circle increases from 25 ? to 25.5 ? . Calculate the approximate increase in the radius . 2. An error of 0.2% is made in measuring the radius of sphere . Find the corresponding percentage error in the volume of the sphere . ( 0.6% ) 3. The height of a cone is 20cm but the radius of its circular base is increased from 10cm to 10.01 cm . Find the approximate change in the volume of the cone in terms of ? . ( 4/3 ? ) 4. Given that y =

x , use the calculus to obtain an approximate value for 16.08 . ( 4.01 )

5. Each side of a cube is increased by p% when p is small . What is approximate percentage increase in the volume of the cube in terms of p ? ( 3p ) 6. Given that y = 8x2 –x3 , find the approximate change in y as x increases from 2 to 2.1 . Stating whether this is an increase or a decrease . ( 2 ; increase ) 7. Given that y = 15x +

24 dy , calculate the value of when x = 2 . Hence find an expression for the 3 x dx approximate value of y when x = 2+p , where p is small . ( 33+10.5p)

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(B) Rate of change

Example : Let the equation of the free fall object s(t) = velocity of object when t = 3 . t ( seconds) s(m)

?t ?s

Average velocity

1 2 gt where g is gravitational acceleration of object . Find the 2

3.1 3.01 3.001 3.0001 3.00001

47.089

0.1

2.989

?s ?t 29.89

=

From the above table , we found that the instantaneous velocity of object at t=3 is ____________. Differentiation is all about finding rates of change of one quantity compared to another . We need differentiation when the rate of change is not constant .

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Instantaneous Rate of Change

The instantaneous rate of change of f at a , also called the rate of change of f at a , is defined to be the limit of the average rates of change of f over shorter and shorter intervals around a .

An object is said to be moving with variable speed or non-uniform speed if it covers equal distances in unequal intervals of time or vice-versa. Example: A rubber ball dropped from a certain height (h1) on reaching the ground bounces up to a height less than the initial one (h2). It continues to bounce but the height to which it raises keeps decreasing (h3,h4). The distance covered by the ball in unit time decreases. The speed of the ball varies from point to point. Such a speed is called a variable speed.

Average Speed and Instantaneous Speed

When we travel in a vehicle the speed of the vehicle changes from time to time depending upon the conditions existing on the road. In such a situation, the speed is calculated by taking the ratio of the total distance travelled by the vehicle to the total time taken for the journey. This is called the average speed. If an object covers a distance S1 in time t1, distance S2 in time t2 and distance Sn in time tn then the average speed is given by, Average speed =

S1 ? S2 ? S3 ? ...... ? Sn total distance travelled ? t1 ? t2 ? t3 ? .... ? tn total time taken

When we say that the car travels at an average speed of 60 km/h it does not mean that the car would be moving with the speed of 60 km/h throughout the journey. The actual speed of the car may be less than or greater than the average speed at a particular instant of time. The speed of a moving body at any particular instant of time is called instantaneous speed. 5

Example 1: A particle moves in a straight line and its distance , s m , from a fixed point O is given by s=t 3-6t2+9t+56 , where t is the time in seconds after passing O. Find (a) its velocity after 4 seconds ; (b) its acceleration after 4 seconds ; (c) the time when it is momentarily at rest ; (d) the total distance moved in the first 3 seconds of its motion. [Solution : ] dv (a) s = t3 – 6t2 +9t +56 (b) a = = 6t – 12 dt ds v? ? 3t2 – 12t + 9 when t = 4 , a = 6(4) – 12 = 12 (m/s2) dt when t=4 , v = 3(4)2-12(4)+9 = 9 (m/s) ? its acceleration after 4 seconds is 12 m/s2 . ? its velocity after 4 seconds is 9m/s . (c) When the body is momentarily at rest , its velocity is zero . i.e. 3t2 – 12t + 9 = 0 t2 – 4t + 3 = 0 (t-3) ( t-1) = 0 t = 3 or t = 1 ? the body is momentarily at rest when t = 1 or t = 3 . (d) When t = 0 , s=03 – 6(0)2 + 9(0) + 56 = 56 When t =1 , s = (1)3 – 6(1)2 + 9(1) + 56 = 60 When t = 3 , s = 3(3)3 – 6(3)2 + 9(3) + 56 = 56 ? total distance moved in the first 3 seconds = (60-56) + ( 60 -56 ) = 4 + 4 = 8(m)

Figure 1 : the diagram shows the motion of the particle in the first 3 seconds.

0

56

60

t=0 t=3

t=1

Example 2: A particle move on a straight line pass through a fixed point O with velocity V=17+t 2 m/s , where t is time in second after pass through O . Find a)the initial velocity (b) velocity when acceleration is 2m/s2 .

Example 3: The distance s m which a body had travelled in t s , is given by s = 4t 3-15t2-18t+2 . Find when the body is at rest , the distance from the origin and acceleration at this time.

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Example 4 : A particle O moves in a straight line so that its distance , sm from a fixed point O on the line is given by s=6t2 – t3 , where t is the time in seconds after passing O . Find (a) the velocity after 5 seconds . (b) the distance of Q from O when it is instantaneously at rest after passing through O . ( 15m/s ; 32m)

Example 5: A particle starts from a fixed point O and moves along a straight line . Its displacement , s m , from O , t seconds after leaving O is given by s = t3 – 6t2 . Find the acceleration (a) its velocity is 15m/s ; (b) when it returns to O . 3 2 [solution] s = t – 6t ds v? ? 3t2 – 12t dt dv a? ? 6t-12 dt (a) when v = 15 , i. e. 3t2 -12t = 15 (b) when it returns to O , s = 0 2 3t – 12t – 15 = 0 t3 – 6t2 = 0 t2 – 4t – 5 = 0 t 2 (t – 6 ) = 0 (t-5)(t+1) = 0 t = 0 or t = 6 t = 5 or t = -1 (rejected) hence, when it returns to O when t=6. 2 when t = 5 , a = 6(5)-12 = 18 (m/s ) When t = 6 , a = 6(6) – 12 = 24(m/s2 ) ? the acceleration is 18m/s2 . ? the acceleration is 24 m/s2 . Example 6 : A particle O moves in a straight line so that its distance , sm from a fixed point O on the line is given by s=3t – t3 , where t is the time in seconds after passing O. Find the velocity and accelerations of the point when t = ? s . At what time is the point at rest ? ( 2009 fykulai mathematics unified test )

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