Apply differentiation to connected rates of — Paper 2 · IGCSE Mathematics - Additional 0606

12 In this question all lengths are in metres. x A B 30° h 5 C A water container is in the shape of a triangular prism. The diagrams show the container and its cross-section. The cross-section of the water in the container is an isosceles triangle ABC, with angle ABC = angle BAC = 30°. The length of AB is x and the depth of water is h. The length of the container is 5. (i) Show that x = 2 3 h and hence find the volume of water in the container in terms of h. [3] (ii) The container is filled at a rate of 0.5 m3 per minute. At the instant when h is 0.25 m, find (a) the rate at which h is increasing, [4] (b) the rate at which x is increasing. [2]

9 marks

Mark scheme: 12(i) h M1 tan30 = oe x 2 Correct completion to given answer A1 V = 5 3 h 2 isw B1 12(ii)(a) dV 5 3 B1 FT theirV = k h 2 = their10 3 h or dh 2 dh dh dV M1 = × soi dt dV dt dh 1 M1 = × 0.5 dt  d V  their    dh  0.115 or 0.11547 to 0.1155 oe A1 12(ii)(b)  dx dx dh  1 M1  = × =  2 3 × their  dt dh dt  5 3 2 A1 5

This question in 0606/21 May/June 2018

2 The variables x and y are such that y = ln 3x - 1 for x 2 . 3 dy (i) Find . [2] dx (ii) Hence find the approximate change in x when y increases from ln 1.2 to ln 1.2 + 0. 125 . [3] ^ h ^ h

5 marks

Mark scheme: 2(i) 1 M1 k × 3 x − 1 1 A1 3 × 3 x − 1 2(ii) 11 B1 x = soi 15 d y M1 0.125 ≈ their × δx oe d x x = their 11 15 0.05 nfww A1

This question in 0606/23 May/June 2018

12 In this question all lengths are in metres. x A B 30° h 5 C A water container is in the shape of a triangular prism. The diagrams show the container and its cross-section. The cross-section of the water in the container is an isosceles triangle ABC, with angle ABC = angle BAC = 30°. The length of AB is x and the depth of water is h. The length of the container is 5. (i) Show that x = 2 3 h and hence find the volume of water in the container in terms of h. [3] (ii) The container is filled at a rate of 0.5 m3 per minute. At the instant when h is 0.25 m, find (a) the rate at which h is increasing, [4] (b) the rate at which x is increasing. [2]

9 marks

Mark scheme: 12(i) h M1 tan30 = oe x 2 Correct completion to given answer A1 V = 5 3 h 2 isw B1 12(ii)(a) dV 5 3 B1 FT theirV = k h 2 = their10 3 h or dh 2 dh dh dV M1 = × soi dt dV dt dh 1 M1 = × 0.5 dt  d V  their    dh  0.115 or 0.11547 to 0.1155 oe A1 12(ii)(b)  dx dx dh  1 M1  = × =  2 3 × their  dt dh dt  5 3 2 A1 5

This question in 0606/23 May/June 2018

2 Two variables x and y are such that y = for x 2 0. x3 dy 1 - 3 ln x (i) Show that = 4 . [3] dx x (ii) Hence find the approximate change in y as x increases from e to e + h, where h is small. [2]

5 marks

Mark scheme: 2(i) d 1 B1 (ln x ) = soi dx x 3  1 2 M1 x  − 3 x ln x d y  x  = d x 3 2 x ( ) −3  1 −4 or x + −3 x ln x   ( )  x  Completion to given answer: A1 dy 1 − 3ln x = dx x 4 2(ii)  1 − 3lne  M1  4  h  e  −h2 oe or −0.0366h awrt A1 e 4

This question in 0606/21 May/June 2019

7 The variables x, y and u are such that y = tan u and x = u 3 + 1. (i) State the rate of change of y with respect to u. [1] (ii) Hence find the rate of change of y with respect to x, giving your answer in terms of x. [4]

5 marks

Mark scheme: 7(i) sec 2 u B1 7(ii) dy dy du M1 Attempts = × dx du dx dy dy dx or = ÷ dx du du dy their sec 2 u A1 FT their (i) = dx 3u 2 u = 3 x − 1 soi B1 sec 2 ( 3 x − 1) A1 final answer cao 3( 3 x − 1) 2 If B1 only then SC1 for 2 1 − k ( x − 1) 3 sec 2 ( x − 1) 3

This question in 0606/21 May/June 2019

9 Variables x and y are such that y = . Use differentiation to find the approximate change in y x2 as x increases from 0.5 to 0.5+ ,h where h is small. [6]

6 marks

Mark scheme: 9 d(e 3 x ) 3 x B1 = 3e soi d x Applies product rule to e.g. numerator: M1 or to x −2 sin x : x −2 cos x + ( −2 x −3 )sin x their(3e3x)sinx + e3x cosx 3 x 2 or to e × x− : e 3 x × ( −2 x −3 ) + their (3e 3 x ) × x −2 Correct quotient rule: M1 − 2 x ( e 3 x sin x ) x 2 ( their ( 3e 3 x sin x + e 3 x cos x ) ) 4 or applies product rule for a second x time e.g. : x −2 (their ( 3e 3 x ) sin x + e 3 x cos x ) + ( −2 x −3 )( e 3 x sin x ) Fully correct derivative; isw A1  d y  M1 δ y = their   × h  d x x =0.5  7.14h A1 or 7.137[66...]h with coefficient rot to 4 or Answer only, without working, scores more figs SC1 isw

This question in 0606/22 Feb/March 2020

1 Variables x and y are such that y = sin x + e -x . Use differentiation to find the approximate change in y as x increases from r to r + ,h where h is small. [4] 4 4

4 marks

Mark scheme: Question Answer Marks Partial Marks 1 dy − x B2 B1 for cos x or –e–x = cos x − e dx dy M1 δy = their × h dx x = π 4 0.251h A1

This question in 0606/22 May/June 2020

8 (a) Differentiate y = tan( x + 4) - 3 sin x with respect to x. [2] ln( 2x + 5) (b) Variables x and y are such that y = 3 x . Use differentiation to find the approximate 2e change in y as x increases from 1 to 1 + h, where h is small. [6]

8 marks

Mark scheme: 8(a) sec 2 ( x + 4) − 3cos x B2 B1 for each 8(b) d(ln(2x +5)) 2 B1 = dx 2 x + 5 d(2e 3 x ) 3 x B1 = 6e d x d y M1 FT their derivatives of ln(2x + 5) and 2e3x = d x  2e 3 x 3 x ln(2 x + 5)  their 2 − their 6e  2 x + 5  4e 6 x d y A1 = d x  2e 3 x 3 x ln(2 x + 5)  2 − 6e  2 x + 5  4e 6 x d y M1 δy = their × h d x x =1 −0.138h A1

This question in 0606/23 May/June 2020

11 In this question all lengths are in centimetres. 1 2 The volume, V, of a cone of height h and base radius r is given by V = r r h. 3 R w 90 180 The diagram shows a large hollow cone from which a smaller cone of height 180 and base radius 90 has been removed. The remainder has been fitted with a circular base of radius 90 to form a container for water. The depth of water in the container is w and the surface of the water is a circle of radius R. (a) Find an expression for R in terms of w and show that the volume V of the water in the container is r 3 given by V = w + 180 - 486000r . [3] 12 ` j (b) Water is poured into the container at a rate of 10 000 cm3s−1. Find the rate at which the depth of the water is increasing when w = 10. [4]

7 marks

Mark scheme: 11(a) 1 B1 R = ( w + 180) 2 1 2 M1 V = π ( their R ) ( w + 180 ) 3 1 2 − π ( 90 ) (180 ) 3 Correct completion to given answer: A1 π V = ( w + 180 )3 − 486000π 12 11(b) dV π B1 = 3 ( w + 180 )2 oe dw 12 dw dw dV M1 = × soi dt dV dt d w 1 M1 = × 10000 d t  d V  their    dw  w =10 0.353 [cms−1] A1 or 0.3526[97...] [cms−1] rot to four or more figs

This question in 0606/23 May/June 2020

5 A cube of side x cm has surface area S cm2. The volume, V cm3, of the cube is increasing at a rate of 480cm 3 s -1 . Find, at the instant when V = 512, (a) the rate of increase of x, [4] (b) the rate of increase of S. [2]

6 marks

Mark scheme: 5(a)  d(x 3 )  2 3 B1  =  3 x and x = 512 soi  dx  OR  d( 3 V )  1 − 23  =  V  dV  3 dx dV dx B1 = × oe, soi dt dt dV 480 M1 dV 2 oe FT their = k (8) 2 3(8) dx d x − 23 or = k (512) k ≠ 0 d V 2.5oe A1 5(b) 12(8) ×their 2.5 soi M1 FT their 8 provided it is not 512 240 A1 FT provided at least M1 earned in (a)

This question in 0606/22 Feb/March 2021

7 Variables x and y are such that y = . Use differentiation to find the approximate change x in y when x increases from 1.9 to 1.9+ ,h where h is small. [6]

6 marks

Mark scheme: 7 d B1 (sin3 x )  3cos3 x soi d x u  (1  sin3 x ) 4 M1 FT their 3cos3x d u 3  4(1  sin3 x ) (3cos3 x ) dx soi 3 4 1  12 M1 FT their d u ortheir d v but x  4(1  sin3 x  (3cos3 x )   (1  sin3 x )  x d y 2 d x d x  2 not both dx x   Correct derivative A1 Evaluates their derivative at x = 1.9 and M1 multiplies by h 0.651h A1

This question in 0606/21 May/June 2022

9 In this question all lengths are in metres. x h 5 The diagram shows a water container in the shape of a triangular prism. The depth of water in the container is h. The container has length 5. The water in the container forms a prism with a uniform cross-section that is an equilateral triangle of side x. 5 3 h 2 (a) Show that the volume, V, of the water is given by V = . [4] 3 (b) Water is pumped into the container at a rate of 0.5 m3 per minute. Find the rate at which the depth of the water is increasing when the depth of the water is 0.1 m. [4]

8 marks

Mark scheme: 9(a) 2 M1 Correct expression 2  x  h  x   connecting x and h  2 oe h or cos30  oe x 1 1 2 or xh = x sin60 2 2 2 h A1 Must be x = x  oe 3 2 M1 1  2 h  V =  their  sin60 5   2  3  1 2 h or h their  5 2 3 5 3 h 2 A1 Correct completion to given answer V  3 9(b) 10 3 B1 Correct derivative of V e.g. h 3 d h d V d h B1 V   soi Not d t d t d V h  10 3  M1 0.5    0.1   3 oe 0.866 (metres per minute) A1 3 or 0.8660[25…] rot to 4 or more sf Allow isw 2

This question in 0606/21 May/June 2022

4 Variables x and y are related by the equation y = 1 + + 2 where x 2 0 . Use differentiation to find x x the approximate change in x when y increases from 4 by the small amount 0.01. [5] x 3- 1 2

5 marks

Mark scheme: 4 Substitutes y = 4 and rearranges to correct 3- B1 term quadratic 3 x 2  2 x 1 0 oe Solves their 3-term quadratic in x as far as M1 x = … d y  2  3 B1  2 x  2 x oe, isw d x 0.01  d y  M1 FT their derivative and their x, providing  their   or better their x > 0 and their x ≠ 4 unless 4 is a δx  d x x = their 1  genuine solution of their 3-term quadratic; must see a power decrease for attempted differentiation in two out of the three terms 1 A1 dep on all previous marks being awarded  oe as the only solution 400 4 Alternative method 1 (B2) (  1) 2 x  B1 for y x or better y  1 x 2 1  y 2  4  4( y  1)( 1) or x  or for x  y  1 2( y  1) 1 y 2 4  4(1  y ) or x  oe or for x  1  y 2(1  y ) 1   2 dx 2  1  12  (M1)  1  ( 1 y )( 1)  y  1 oe isw (1  y )  y  y        2 dy 2 d x     or  1  d y (1  y )  1  2 2  (1  y ) ( y  1)  y  oe   2 dx   or  oe d y ( y  1) 2   (M1) FT their derivative ; δx dx  their   or better 0.0 1  d y   y  4  must have attempted derivative 1 (A1) dep on all previous marks being awarded  oe as the only solution 400

This question in 0606/22 May/June 2022

7 (a) Variables x and y are such that y = . Use differentiation to find the approximate tan x r r change in y as x increases from to + h , where h is small. [5] 4 4 2 1 d y 1 d y ( x + 1)( x - 4)(b) Given that y = 3 show that y - - 2 can be written as 5 . [4] ( x - 3) dx 3 f d x p ( x - 3)

9 marks

Mark scheme: 7(a) d 2 B1 (cos x ) = −2cos x sin x soi dx Attempts the quotient rule M1 d 2 2 2 FT their (cos x ) dy −2cos x sin x tan x − (1 + cos x )sec x dx = dx tan 2 x Fully correct isw A1 d 2 FT their (cos x ) only dx δy dy M1  their h dx x = π 4 δy − 4 h cao A1 7(a) Alternative method 1 d 3 2 (B1) (cos x ) = −3cos x sin x soi dx Attempts the quotient rule: (M1) d 3 2 3 FT their (cos x ) dy (sin x )( − sin x − 3cos x sin x ) − (cos x + cos x )cos x dx = dx sin 2 x Fully correct isw (A1) d 3 FT their (cos x ) only dx δy dy (M1)  their h dx x = π 4 δy − 4 h cao (A1) Alternative method 2 (B1) d  −2 sec 2 x  2 = −2(tan x ) dx  tan x  Attempts the product rule (M1) d  2  FT their   dy d x  tan x  = −2(tan x ) −2 sec 2 x − (sin x ( − sin x ) + cos x (cos x )) dx Fully correct isw (A1) d  2  FT their   only d x  tan x  δy dy (M1)  their h dx x = π 4 δy − 4 h cao (A1) 7(b) dy −4 B1 = −3( x − 3) oe, soi dx d 2 y −5 B1 = −−4 3( x − 3) oe, soi dx 2 ( x − 3) 2 + 3( x − 3) − 4 M1 dy −4 FT = k ( x − 3) ( x − 3) 5 dx d 2 y −5  x −+3 3 4  x ( x − 3) − 4 and = m ( x − 3) where k 2 or  4 − 5 =  5 dx  ( x − 3) ( x − 3)  ( x − 3) and m are constants Correct completion to given answer: A1 x 2 − 3 x − 4 ( x + 1)( x − 4) = ( x − 3)5 ( x − 3)5

This question in 0606/22 Feb/March 2023

4 Variables x and y are related by the equation y = 2 + tan ( 1 - x) where 0 G x G r . Given that x is 2 increasing at a constant rate of 0.04 radians per second, find the corresponding rate of change of y when y = 3 . [6]

6 marks

Mark scheme: 4 dy 2 B2 must be seen  sec (1  x ) oe dx dy 2 B1 for  k sec (1  x ) oe, k  1 dx or dy 2 SC1 for  sec 1  x or dx dy 2  sec (1  x )  c dx Solves 3  2  tan(1  x ) as far as M1 1 – x = tan–11 π A1 1  x  isw or 0.7853[98…] 4 π x 1 isw or 0.2146[01…] 4 Correct use of chain rule and correctly writes M1 dep on at least B1 and an attempt to in terms of cosine or tangent: solve 3  2  tan(1  x ) their   1  0.04 oe, soi or 2  π  cos  their   4   2  π   their  1 1  tan  their    0.04 oe soi   4   0.08 oe, nfww A1 dep on all previous marks awarded

This question in 0606/22 May/June 2023

7 (a) f ( x) = 3 + ( 4 x - 2) 5 where x 2 1. Find an expression for f l ( x) , giving your answer as a simplified algebraic fraction. [3] 5x (b) Variables x and y are related by the equation y = . Using differentiation, find the 3x + 2 approximate change in x when y increases from 10 by the small amount 0.01. [4] (c) (i) Differentiate y = x 3 ln x with respect to x. [2] x 2 (ii) Hence find ( 2 + 3 ln x) dx . [3] y e 6 o

12 marks

Mark scheme: 7(a) 10(4 x  2) 4 10(4 x  2) 4 3 B2 for correct unsimplified form e.g. or isw 1 5 5 3  (4 x  2) 5 2 2  5  4 x – 2  4  4  3  (4 x  2)   1  3  (4 x  2)  1 2 or 1  5 2 10  3  (4 x  2)   4 x – 2  4 or B1 for 5(4x – 2)4 4 soi  1 5 1 or  3  (4 x  2)  2  g( x ) 2 7(b) dy 5(3 x  2)  3(5 x ) B1  2 oe isw or d x (3 x  2) dy 2 1  5 x   3 x  2   3  5  3 x  2  oe isw   dx [ y = 10] x = 0.8 B1 0.01  dy  M1 FT their x providing their x ≠ 10 or  their  oe 0.01and their genuine attempt at a δx  dx x 0.8  derivative using product or quotient rule 1 A1 dep on all previous marks awarded or 0.00016 or 1.6  104 isw 6250 Alternative method 2 y (B1) x  oe 3 y  5 dx 2(3 y  5)  3( 2 y ) (B1)  2 oe isw or dy (3 y  5) dx 2 1 2 y   3 y  5   3 ( 2)  3 y  5    dy   (M1) FT their genuine attempt at a derivative δ x d x  their oe using product or quotient rule  0.01  d y y 10   1 (A1) dep on all previous marks awarded or 0.00016 or 1.6  104 isw 6250 7(c)(i) 1 B2 1 3x2lnx + x3  or better, isw B1 for (their 3x2)lnx + x3  (their ) x x 7(c)(ii) x 3 ln x x 3 B3 must have arbitrary constant   c oe, isw 6 18 x 3 ln x x 3 B2 for  or 6 18 x 3 ln x kx 3  +c, k > 0 nfww 6 3 or B1 for 1  3 x 2 ln x  x 2  dx  1 x 2 dx soi 6  6  x 3 ln x x 2 or dx soi 6  6 or x 2 d x soi 3 ln x  3 x 2 ln x  d x  x    2 3 x 3 or soi 3 x ln x  dx  x ln x    3

This question in 0606/22 May/June 2023

7 Variables x and y are such that y = . Use differentiation to find the approximate change 1– x in y as x increases from 0.1 to 0.1+ h , where h is small. [6]

6 marks

Mark scheme: 7 d(4 x 3  2sin8 x ) 2 B2 B1 for 12x2+ kcos8x, where k > 0  12 x  16cos8 x soi d x Correct quotient rule: M1 or applies correct product rule to t hei r (1  x ) 12 x 2  16cos8 x  (4 x 3 + 2sin8 x )( 1) (4 x 3  2sin8 x )(1  x )1 :     (1  x ) 2 (4 x 3  2sin8 x )  (1  x ) 2 1    their (12 x 2  16cos8 x ) (1  x ) 1   Fully correct derivative; isw A1 FT their 12x2 + 16cos8x δ y  d y  M1  their   h  d x x =0.1  14.3h A1 or 14.29[54...]h with coefficient rot to 4 or more figs isw

This question in 0606/23 May/June 2023

11 A cylinder, open at both ends, has base radius r cm and height 4r cm. Its curved surface area is S cm2. dS dr Given that r varies with time t, find S at the instant when = 6 . [5] dt dt

5 marks

Mark scheme: 11 dS dS dr dS B1 =  or = 6 soi dt dr dt dr 2 B1 S = 2πr (4r ) or 8πr 16πr = 6 M1 FT their S = kr2 with k a positive integer to give 2kπr = 6 6 A1 r = oe, isw 16π 9 A1 S = oe, isw 8π

This question in 0606/22 Feb/March 2024

6 Variables x and y are such that y = cos x sin 2 x . Use differentiation to find the approximate change in y as x increases from 3 to 3+ h , where h is small. [5]

5 marks

Mark scheme: 6 d 2 B1 (sin x )  2sin x cos x soi d x cos x  their (2sin x cos x )  M1 d 2 FT their (sin x ) (their  sin x )  sin 2 x d x cos x  their (2sin x cos x ) ( sin x )  sin 2 x A1 d 2 FT their (sin x ) isw d x δy 2 3 M1 FT their derivative  their (2sin3cos 3  sin 3) or better h 0.274h or A1 dep on correct derivative seen 0.2738[08...]h where the coefficient of h is rot to 4 or more sf

This question in 0606/21 May/June 2024