Differentiation, Mied Eercise f() 0 f( + ) f( ) f () 0( + ) 0 0 + 0 + 0 0 0 + 0 (0+ 0 ) (0+ 0 ) As 0, 0 + 0 0 So f () 0 a A as coordinates (, ). Te y-coordinate of B is ( + δ) + ( + δ) + δ + (δ) + (δ) + + δ (δ) + (δ) + 6δ + Gradient of AB y y δ + δ + 6δ+ δ δ + δ + 6 δ δ δ + δ + ( ) ( ) ( ) ( ) ( ) 6 As δ 0, (δ) + δ + 6 6 Terefore, te gradient of te curve at point A is 6. y + + + + d 6 6 Wen, 6 d d 8 Wen, 6 d Wen, 6 8 7 5 7 7 Te gradients at points A, B and C are, 5 and 7, respectively. 7 y 7 d 6 wen d 6 + 6 0 ( 8)( ) 0 8 or 5 y + d d wen ± Wen, y () + Wen, y () () + 5 Te gradient is at te points (, ) and (, 5). 6 a f() + + f () Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free.
6 f () 0 wen ± 7 8 a a y + d + + ( ) y d 6 d ( ) Te gradient is zero wen d y d 0: ( ) 0 Wen, y 6 Te gradient is zero at te point wit coordinates (, 6). + + y + + + d c Wen, + d + + + 6 0 Let y + + + + + + + + + 6 + d 6 + Te point (, ) lies on te curve wit equation y a + + c, so a + + c () Te point (, ) also lies on te curve, so a + + c () () () gives: a + () d a + Te gradient of te curve is zero at (, ), so 0 a + () () () gives: a Sustituting a into () gives Sustituting a and into () gives c 5 Terefore, a,, c 5 a y 5 + 5 + d 0 + 5 i d 0 + 5 0 + 0 ( )( ) 0 or is te coordinate at P, so at Q. 6 Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free.
ii y 7 5 + 5 + So equation of te tangent is y + ( ) y 7 iii Wen 0, y 7 and wen y 0, 7 So points R and S are (0, 7) and ( 7, 0). 7 Lengt of RS ( ) ( 7) + 7 + 5 7 y 8 + 8 + d 8 8 + 6 + 6 Wen, y 8 + d 8 + Te equation of te tangent troug te point (, ) wit gradient is y ( ) y 8 + y Te normal at (, ) as gradient. So its equation is y ( ) y + 8 a y 7 + y 7 + d 7 + At (0, 0), 0, gradient of curve is 0 0 +. Gradient of normal at (0, 0) is. Te equation of te normal at (0, 0) is y. At (, 0),, gradient of curve is 7 +. Gradient of normal at (, 0) is. a Te equation of te normal at (, 0) is y ( ). Te normals meet wen y and y : 5 5 y 5 5 N as coordinates ( ) ceck in y,. 5 5 Area of OAN ase eigt Base () Heigt () 5 Area 5 5 5 y Wen 0, y so te point P is (0, ) For te gradient of line L: d At point P, wen 0, d y d Te y-intercept of line L is. Equation of L is y. Point Q is were te curve and line intersect: 0 Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free.
5 ( ) 0 0 or 0 at point P, so at point Q. Wen, y sustituting into te original equation Using Pytagoras teorem: distance PQ 68 ( 0) + ( ( )) 7 7 8 8 d Putting d y d 0: 8 5 6 a y + ( > 0) 8 y Sustituting into + gives: y 8 + 0 So and y 0 wen d y d d y 6 + d d y 6 5 Wen, d 8 + 6 8 > 0 minimum 7 y 5 + 7 d 0 + 7 Putting 0 + 7 0 ( 7)( ) 0 So 7 or Wen 7, 7 7 7 y 5 + 7 7 5 7 y 7 Wen, y 5() + 7() So ( 7 5, 7 ) and (, ) are stationary points. f () Wen, f () 8 6 7 8 a f () + ( > 0) For an increasing function, f () 0 + 0 0 Tis is true for all, ecept (were f () 0). So te function is an increasing function. y 6 + d + Putting + 0 ( + ) 0 ( )( ) 0 So or So tere are stationary points wen and. d y 6 d d y Wen, 6 6 < 0, so d maimum point d y Wen, 8 6 > 0, so d minimum point Wen, y 6 + So (, ) is a maimum point. Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free.
0 a f() 8 6 + + 0 f () + ( + ) ( )( ) ( )( )( + ) So, or f() 8 6 + + 0 f() () 8() 6() + () + 0 8 f() + 8 6 + 0 So (, ), (, 8) and (, ) are stationary points. f () 6 8 f () 6 8 < 0, so maimum f () 6() 8() 6 > 0, so minimum f (), y 6 + 8 7 > 0, so minimum So (, ) is a maimum point and (, 8) and (, ) are minimum points. Te coordinates of B are (5, 5). a P as coordinates m,,5. OP ( 0) + 5 0 + 5 5 + + 5 Given f() + 5 f () 8 Wen f () 0, 8 0 ( 8) 0 0 or 8 0 or ± c f () 8 Wen 0, f () 8 < 0, so maimum Wen 8, f () 8 8 6 > 0, so minimum Sustituting 8 into f(): OP 8 8 + 5 So OP wen ± 50 a f() 00 f () 50 a y + 5 + Let y 0, ten + 5 + 0 ( )( + + ) 0 ( )( + ) 0 or wen y 0 Te curve touces te -ais at (A) and cuts te ais at (C). C as coordinates (, 0) At te maimum point, B, f () 0 50 0 50 50 5 5 Wen 5, y f(5) 00 50 5 5 5 5 + d Putting d y d 0 5 + 0 (5 )( + ) 0 So 5 Wen 5, or Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free. 5
5 5 5 y + 5 + 7 5,. So B is ( ) 7 Wen, y 0 So A is (, 0). y f() y f () 0 < < 0.5 Positive gradient Aove -ais 0.5 Maimum Cuts -ais 0.5 < <. Negative gradient Below -ais. Minimum Cuts -ais. < <. Positive gradient Aove -ais. Maimum Cuts -ais >. Negative gradient Below -ais wit asymptote at y 0 6 A π + 000 π + 000 da d π 000 000 π Putting d A d 0 000 π 000 π 500 π 7 a Te total lengt of wire is π y+ + m As total lengt is m, π y+ + π y + Area, R y + π π Sustituting y + gives: π π R + 8 8 (8 π + π) 8 (8 π) 5 V π(0r r r ) dv 0π πr πr dr Putting d V dr 0 π(0 r r ) 0 ( + r)(0 r) 0 r 0 or r As r is positive, r 0 Sustituting into te given epression for V: 0 00 000 00 V π 0 π 7 7 c For maimum R, d R d 0 π R 8 dr d π Putting d R d 0 π + +π Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free. 6
7 c Sustituting into R: +π 6 π R 8 ( +π ) +π +π + 8π6π R ( +π) + π 6 + π +π + π ( ) ( +π) ( +π) +π 8 a Let te eigt of te tin e cm. Te area of te curved surface of te tin π cm Te area of te ase of te tin π cm Te area of te curved surface of te lid π cm Te area of te top of te lid π cm Total area of seet metal is 80π cm. So π + π + π 80π 0 Te volume, V, of te tin is given y V π π ( 0 ) π(0 ) dv d π(0 ) Putting d V d 0 0 0 (0 )( + ) 0 So 0 or But is positive, so 0 0 0 0 8 d V π 0 00 00 000 π 7 00 π 7 e Lid as surface area π + π Wen 0, 00 0 60 tis isπ + π Percentage of total surface area 60 π 00 00...% 80π a Let te equal sides of ADE e a metres. Using Pytagoras teorem, a + a a a Area of ADE ase eigt a a m c d V π( 6) d Wen 0, d V π( 0) < 0 d So V is a maimum. Area of two triangular ends Let te lengt AB CD y metres Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free. 7
Area of two rectangular sides ay ay So S + y y + y But capacity of storage tank y So y 000 6 000 y Sustituting for y in equation for S gives: S 6 000 + c ds d 6 000 Putting d S d 0 6 000 6 000 0 8.8 ( s.f.) Wen 0, S 00 + 800 00 d d S d 000 + d S d Wen 0, > 0, so value is a minimum. Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free. 8
Callenge a ( + ) 7 7 7 6 7 5 7 + + + + 7 + 7 6 + 5 + 5 +...... 7 d( ) f( + ) f( ) d 0 7 7 ( + ) 7 6 5 + 7+ + 5 6 5 7+ + 5 6 5 (7 + + 5 ) 6 5 ( 7 + + 5 ) As 0, 7 6 + 5 + 5 7 6, so 7 7 d( ) d 7 6 Pearson Education Ltd 07. Copying permitted for purcasing institution only. Tis material is not copyrigt free.