Chapter 7 Applications of differentiations 2

APPLICATIONS OF DIFFERENTIATION(2)
1. What is Higher Derivatives ?
In mathematics we use the higher derivatives in many applications . The higher derivatives are used in different fields . They are graphing , differentiation , linear equations , and partial differentiation . Higher derivatives are also used in limits . Higher derivatives are also used for finding the rate of change of acceleration of the vehicle . If we repeat the differentiation process to find the differential coefficient of the differential coefficient of y d2y d dy ( )= with respect to x and is written dx dx dx 2 d2y dy 2 3 ? ( )2 Note : d d y d y 2 dx dx ( )= 3 . third derivative is written as dx dx 2 dx Example 1 2 dy 2 2 d y x ? ( x ? 1) ? y ? 8 .(x=3/5 , -1 ) Given that y = x + 2x + 7 , find the values of x if 2 dx dx [Solution:]y = x2 + 2x + 7 ---------- (1) dy ? ----------- (2) dx d2y ? ----------- (3) dx 2 Substitute (1) , (2) , (3) into the equation d2y dy x 2 2 ? ( x ? 1) ? y ? 8 dx dx

Example 2
d 2v d 2v dv dv If v = t +2t + 3t + 1 , find and 2 . Hence calculate the value of and 2 when t = 1 . dt dt dt dt [Solution :] v = t3 + 2t2 + 3t + 1 dv = 3t2 + 4t + 3 dt d 2v = 6t2 + 4 2 dt d 2v dv When t = 1 , = 3(1)2 + 4(1) + 3 = 10 , = 6(1) +4 = 10 . dt dt 2
3 2

Example 3 Given y = a + bx-2 + cx2 , where a,b,c are constant , show that x
d2y dy ? 3 ? 8cx . (2004 Unified Test(2)) 2 dx dx

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2. Sign Of Derivative (Graphical interpretation of Derivative)
dy =rate of change of y with respect to x dx = gradient of the curve at the point P(x,y) = gradient of the tangent line to the curve at the point P(x,y) ? A function f(x) is said to be a strictly increasing function on (a,b) if x1< x2?f(x1) < f(x2) for all x1,x2 ? ( a,b )

?

How to Test y is increasing function ? ? If y = f (x) is an increasing function , dy ? > 0 for all x ? (a , b ) . dx ? y increases (?) as x increases (?) .
?graph is sloping upwards from a to b .

Hence , f is strictly increasing in the interval a<x<b. ? A function f(x) is said to be a strictly decreasing function on (a,b) if x1< x2? f(x1) > f(x2 ) for all x1 , x2? ( a,b ) How to Test y is decreasing function ? ? If y = f(x) is an decreasing function , dy ? < 0 for all x ? ( a , b ) . dx ?y decreases (?) as x increases (?) . ?graph is sloping downwards from a to b . Hence , f is strictly decreasing in the interval a<x<b . Example 4 (i) For what values of x when the function y = 10 – 6x – 2x2 is increasing as x increases ? [Solution :] y = 10 – 6x – 2x2 ,
dy = -6 – 4x dx

? a b

(ii) If y = 4x3 – 3x2 – 6x + 1 , find the ranges of values d2y of x for which decreases as x increases . dx 2

Since y is increasing function ,

dy >0 dx -6 – 4x > 0 - 4x > 6 ?3 x< 2

Hence , the function y is increasing when x <

?3 . 2

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Example 5 For what values of x when the function y = x3 – 3x2 – 9x + 4 is decreasing . ( 2005 unified test2 , -1 < x < 3 )

3. Stationary Points
Stationary points

Turning points

Points of inflexion

Maximum Point

Minimum Point

A turning point is a stationary point which is either a maximum or minimum point . Not all stationary point are turning point .
Consider the graph below in Fig 13.2 . A point on a curve , y = f(x) is a stationary point if
dy = 0 at that point . They include turning points and dx

stationary points of inflexion . On the graph , points A , B and C are stationary points . ? Point A is called minimum point because y has a minimum value at A as compared to the neighboring points around A . ? Point A and Point C are called local minimums . ? Point B is only local maximum and does not necessarily represent the maximum value of the whole curve. ? In Fig 13.3 and Fig 13.4 , the point dy dy where = 0 and does not dx dx change sign in the neighborhood of the point of inflexion.
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3.1 How to distinguish between the maximum and minimum ?
(A) First derivative Test Let P be a stationary point when x = a . a - : refers to a number which is slightly < a . a+ : refers to a number which is slightly > a . (a) If P is minimum point : x aa dy -ve 0 dx Slope a+ +ve
P

dy is changes dx from negative to positive in passing through point P.

As x increases ,

? P is minimum point .
aaa+

(b) If P is maximum point : x dy dx Slope a+ve a 0 a+ -ve
P

dy is changes dx from positive to negative in passing through point P.

As x increases ,

? P is maximum point .
aaa+

(c)(i) If P is neither maximum nor minimum point : x dy dx Slope a+ve a 0 a+ +ve As x increases , the sign of does not change before and after the point PP . ? P is stationary point of
aaa+

dy dx

Inflexion .

(ii) x dy dx Slope a-ve a 0 a+ -ve
P

dy dx does not change before and after the point P .

As x increases , the sign of

? P is stationary point of
aaa+

Inflexion .

*Note : At a point of inflexion ,

d2y dy = 0 , but is not necessarily zero . 2 dx dx
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Example 6 : Find the coordinates of the stationary points on the curve y = x3 – 12x – 5 and determine the nature of these points . Hence sketch the graph of y = x3 – 12 x – 5 . y (-2,11) [Solution : ] y = x3 – 12 x – 5 dy = 3x2 – 12 dx x dy For the stationary points , =0 dx 3x2 – 12 = 0 3 (x2 – 4 ) = 0 y=x3-12x-5 3(x +2 ) (x -2 ) = 0 x = 2 or x = -2 3 When x = 2 , y = 2 – 12(2) – 5 = -21 (2,-21) When x = -2 , y = (-2)3 – 12(-2) – 5 = 11 ? the stationary points are (-2, 11) and (2 , -21) dy To determine the nature of the stationary points , we consider the sign of : dx X Sign of
dy dx

Slightly < -2 +ve

x= - 2 0

Slightly > -2 -ve

x Sign of
dy dx

Slightly < 2 -ve

x= 2 0

Slightly > 2 +ve

Sketch of tangent

Sketch of tangent

Hence , maximum point is (-2,11 ) , minimum point is (2,-21) . The sketch of graph is shown as above . Example 7: Determine the nature of turning point(s) on the curve y = 27x +
4 = 27x + 4x -2 2 x dy 8 = 27 – 8x -3 = 27 - 3 dx x dy For stationary points , =0 dx 8 27 - 3 = 0 x 8 2 x3 = ,x= 27 3 2 2 4 when x = , y = 27 ( ) + = 27 . 2 2 3 3 ( ) 3 2 ? the stationary point is ( , 27) . 3 To determine the nature of the stationary point , 2 Hence , ( , 27 ) is minimum point . 3
4 . Draw a rough sketch of the curve. x2
250 200 150 100 50

[Solution:] y = 27x +

? 3

? 2

? 1 ? 50

1

2

3

4

X Sign of
dy dx

Slightly < 2/3 -ve

x= 2/3 0

Slightly > 2/3 +ve

Sketch of tangent

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Example 8: Find the turning points on the curve y = 4x3 – 3x4 and determine the nature of the stationary points . Hence sketch it . ( max (1,1) , pt of inflexion (0,0) )

Example 9: The curve y = ax3 + bx + c has two turning points , (1,1) and (-1,9) . (a) Find the values of a , b and c . (b) Determine whether each of the turning points is a maximum or a minimum point .

Review Question :
dy is increasing . (Ans : x > 2 ) dx 2. Find the points on the curve y = x4 – 8x2 + 2 where the gradient is zero and determine the nature of the points . Ans: (0,2) max. pt. , (2,-14) min.pt. , (-2,-14) min.pt.

1. Given that y = x 3 – 6x2 + 9x +5 , find the range of the values of x for which

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d2y ?0 2 3.2 Second Test ( Note : Only valid when dx at the stationary point ) Consider the maximum point A :

dy dy is change from positive to zero , then to negative , in other words , is decreasing function , dx dx ? dy ?0 2 ? d y ? dx i.e. < 0 . Thus , a point is maximum when ? 2 . dx 2 d y ? ?0 ? ? dx 2
From the graph , Consider the minimum point at B :

dy dy is change from negative to zero , then to positive , in other words , is increasing function , dx dx ? dy ?0 2 ? d y ? dx i.e. > 0 . Thus , a point is minimum when ? 2 . dx 2 ?d y ? 0 ? ? dx 2
From the graph , Note : The second derivative is simple and effective BUT

? It’s sometimes inconclusively ? It can involvedunnecessary calculation d2y ? When = 0 ( which is points of inflexion ) dx 2

Be ready switch to First derivative test Page 7 of 14

Example 10: Find the coordinates of the turning points of the curve y = (x2-1)2 and distinguish between the maximum and minimum points . [Solution:] y = (x2 – 1)2 = x4 – 2x2 + 1 dy = 4x3 – 4x = 4x ( x2 – 1) = 4x ( x+1) (x-1) dx dy For turning points , =0 dx 4x (x+1)(x-1) = 0 X =0 , x = 1 , x = -1 When x = 0 , y = (02 – 1)2 = 1 When x = 1 , y = ( 12 -1)2 = 0 When x = -1 , y = [(-1)2 – 1]2 = 0 Thus , the coordinates of turning points are (0,1),(-1,0),(1,0) . 2 d y ? 12x2 – 4 2 dx d2y ? 12(02)-4 = -4 < 0 When x = 0 , 2 dx d2y ? 8>0 When x = 1 , dx 2 d2y ?8>0 When x = -1 , dx 2 ? (0,1) is maximum point , (1,0) and (-1,0) are minimum points . Example 11 Determine the maximum , minimum and the inflexion points of the curve y = x 4 – 4x3 . Use the result to give a sketch of the curve . Ans : (3,-27),min.pt. , (0,0),pt. of inflexion .

Review Question 1. Determine the maximum and minimum points on the curve y = 4x3-48x. Ans: min(2,-64),max(-2,64) 2. The derivative of the function y = ax2 + bx + c is 6x – 12 . If the curve has a minimum value of -2 , find the values of a,b and c . Ans : a=3 , b= - 12 , c=10 3. Find the coordinates of the stationary points on the curve y=x 3-2x2-4x and ?2 13 distinguish between these points . Hence sketch the curve. Ans : (2,-8) min.pt. , ( ,1 ) max.pt. 3 27
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4. Problems of Maxima and Minima
Steps (1) Find a relation between the quantity to be maximized and minimized and the variable involved. dy (2) Find the value(s) of x when =0. dx (3) Test the nature of the stationary point(s) and choose the required one . (4) substitute the relevant value of x into equation and find the required maximum or minimum value of y . Example 12: Three variables x , y and A are related by the equation A=2x2 y . Given that x + y = 9 , find the values of A at the stationary points and determine the nature of these values of A . [Solution : ] x + y = 9 ? y = 9 – x ? A = 2x2 y = 2x2 ( 9 - x ) = 18x2 – 2x3 d2A dA = 36x – 6x2 , = 36 – 12x dx dx 2 dA For stationary value of A , =0 dx 36x – 6x2 = 0 6x ( 6 – x ) = 0 x = 0 or x = 6 2 d A When x = 0 , = 36 – 12(0) = 36 > 0 (give min value of A) dx 2 d2A When x = 6 , = 36 – 12(6) = -36 < 0 (give max value of A) dx 2 ? maximum value of A = 18(0)2 – 2(0)3 = 0 Minimum value of A = 18(6)2 – 2(6)3 = 216 . Example 13: A company that manufactures dog food wishes to pack the food in closed cylindrical tins . What should be the dimensions of each tin if each is to have a volume of 128 ? cm3 and the minimum possible surface area? [Solution:]Let r be the radius of tin , h be the height of tin the required V be the volume . V = ? r2 h 128 ? r2 h = 128 ?? h = 2 ---- (1) h r S be the surface area of the tin S = 2 ? r2 + 2 ? r h r 256? = 2 ? r2 + (from (1) ) r = 4 (cm) r dS 256? 128 = 4? r h = 2 = 8(cm) 2 dr 4 r d 2S 512? dS 256? ? 4? ? 3 When = 0 , then 4?r = 0 2 2 dr r dr r 2 d S 512? ? 4? ? 3 = 12?> 0 (min) 4?r3 - 256? = 0 when r = 4 , 2 dr 4 Hence , the surface area S has a minimum value when r = 4cm and h=8cm .

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Example 14: Find the maximum volume of a cylinder that can be inscribed in a cone of height 18cm and base radius 8cm , if the axes of the cylinder and the cone coincide . [Solution : ]Let the cylinder inscribed in the cone have a radius r , height h and volume V . AO OC 18 8 A ? From the diagram , ? ? BD DC h 8?r 18 9 (8 ? r ) = (8 ? r ) h= 8 4 2 V=?r h r B 2 9 = ? r [ (8 – r)] 18 cm 4 9? h cm (8r 2 ? r 3 ) = 4 dV 9 9 C ? ? (16 ? 3r 2 ) ? ? r (16 ? 3r ) O D dr 4 4 dV 9 When = 0 , ? r (16 ? 3r ) = 0 8 cm 4 dr 16 r = 0 (rejected) or r = 3 2 d V 9? ? (16 ? 6r ) dr 2 4 16 d 2V 9? ? (16 ? 32) = -36?< 0 When r = , 3 dr 2 4 16 Hence , the volume is maximum when r = . 3 9 16 16 9 256 8 512? )( ) ? V max = ? ( ) 2 (8 ? ) ? ? ( (cm2). 4 3 3 4 9 3 3 Example 15 A right circular cylinder is to be made so that the sum of its base radius and its height is 6cm . Find the maximum volume of the cylinder . Ans : V max = 32? cm3

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Assignment : Problems of maxima and minima (2) 1. A farmer wishes to fence up a rectangular enclosure of area 128m2 . One side of the enclosed is formed by part of a river. Find the least possible length of fencing required for the other three sides . 2. Find the volume of the largest box that can be made by cutting equal squares out of the corners of a piece of cardboard of dimensions 15 inches by 24 inches , and then turning up the sides . [486 inch 3]

3. An open paper box with a square base of side x cm is made from a vanguard sheet of area 75cm2 . Show that its volume , V cm3 , is given by V = 1 (75 x ? x 3 ) . Find the value of x for which V is a 4 maximum and find the maximum volume .

4. Which point on the curve y=x2 is nearest to the point (3,0)?

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5. A 100 cm length of wire is bent to form two squares , one of side x , and the other of side y . Prove that the area enclosed by the squares is least when x = y .

6. A solid circular cylinder has a radius r cm , height h cm . It has a fixed volume of 400 cm3 . Show that the 800 total surface area , A cm2 , of the cylinder is given by A = 2? r 2 ? . Find , to 3 significant figures , the r value of r that gives the cylinder its minimum surface area .

7. The curve whose equation is y=x3 – ax2 + bx + c passes through the point (2,7) , has a local maximum when x = 1 , a point of inflexion when x = 3 . Find the values of a , b and c . Hence find the minimum point on the curve .

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Assignment : Problems of maxima and minima (3) 1.
A

S y cm B P x cm Q

R

2. An iron block is used to make a cylindrical can which volume is 250 ? cm3 . Find the dimension of the can when the material used is a minimum . [ r = 5 , h = 10]

C

In the diagram , ABC is an isosceles triangle with AB=AC=13cm and BC = 10cm . A rectangle QRS is drawn in the triangle such that PQ= x cm , QR= ycm and the vertices R and S are on the sides AC and AB respectively . (a) Show that the area , A cm2 of the shaded region 6 is given by A = 60 – 12x + x2 . 5 (b) Hence , find the minimum area of the shaded region .

3. The three sides of a trapezium are 10cm , find the other side if the area obtained is maximum ? [20cm]

[Ans : 30cm2]
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4. . A man rows 30 metres out to sea from a point P on a straight coast . He reaches a point M such that MP is perpendicular to the coast . He then wishes to get as quickly as possible to a point Q on the coast 400 metres from P . If he row at 40m/min and cycle at 50m/min , how far from P should he land ? [40m from P]

5. Divide 48 into two parts , such that the sum of the square of one part and the cube of the another part is 1 2 least . [ 5 , 42 ] 3 3

6. An open-topped cylindrical container has a capacity of 24? cm3 . The cost of the material used for the base of container is three times that of the curved surface . Find the radius of the base so that the manufacturing cost is minimized . [2cm]

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