C4 QUESTIONS FROM PAST PAPERS DIFFERENTIATION

Transcription

1 C4 QUESTIONS FROM PAST PAPERS DIFFERENTIATION. A curve C has equation + y = y Find the eact value of at the point on C with coordinates (, ). (Total 7 marks). The diagram above shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m. Water is flowing into the tank at a constant rate of 0.48π m min. At time t minutes, the depth of the water in the tank is h metres. There is a tap at a point T at the bottom of the tank. When the tap is open, water leaves the tank at a rate of 0.6 h m min. (a) Show that t minutes after the tap has been opened dh h (5) When t = 0, h = 0. (b) Find the value of t when h = 0.5 (6) (Total marks). The curve C has the equation cos + cosy =,, y 6 (a) Find in terms of and y. () The point P lies on C where =. 6 City of London Academy

2 (b) Find the value of y at P. () (c) Find the equation of the tangent to C at P, giving your answer in the form a + by + cπ = 0, where a, b and c are integers. () (Total 9 marks) 4. The area A of a circle is increasing at a constant rate of.5 cm s. Find, to significant figures, the rate at which the radius r of the circle is increasing when the area of the circle is cm. (Total 5 marks) 5. The curve C has the equation ye = + y. (a) Find in terms of and y. (5) The point P on C has coordinates (0, ). (b) Find the equation of the normal to C at P, giving your answer in the form a + by + c = 0, where a, b and c are integers. (4) (Total 9 marks) 6. A curve C has the equation y y = + 8. (a) Find in terms of and y. (b) Hence find the gradient of C at the point where y =. (4) () (Total 7 marks) City of London Academy

3 7. A container is made in the shape of a hollow inverted right circular cone. The height of the container is 4 cm and the radius is 6 cm, as shown in the diagram above. Water is flowing into the container. When the height of water is h cm, the surface of the water has radius r cm and the volume of water is V cm. (a) Show that 4h V 7. () [The volume V of a right circular cone with vertical height h and base radius r is given by the formula V r h. ] Water flows into the container at a rate of 8 cm s. (b) Find, in terms of π, the rate of change of h when h =. (5) (Total 7 marks) 8. 5 The diagram above shows a right circular cylindrical metal rod which is epanding as it is heated. After t seconds the radius of the rod is cm and the length of the rod is 5 cm. The cross-sectional area of the rod is increasing at the constant rate of 0.0 cm s. City of London Academy

4 (a) Find when the radius of the rod is cm, giving your answer to significant figures. (4) (b) Find the rate of increase of the volume of the rod when =. (4) (Total 8 marks) 9. A curve has equation y + y = 4. The points P and Q lie on the curve. The gradient of the 8 tangent to the curve is at P and at Q. (a) Use implicit differentiation to show that y = 0 at P and at Q. (b) Find the coordinates of P and Q. (6) () (Total 9 marks) 0. A curve is described by the equation 4y = y. (a) Find the coordinates of the two points on the curve where = 8. () (b) Find the gradient of the curve at each of these points. (6) (Total 9 marks). A curve has parametric equations = 7cos t cos7t, y = 7 sin t sin 7t, t. 8 (a) Find an epression for in terms of t. You need not simplify your answer. () (b) Find an equation of the normal to the curve at the point where t. 6 Give your answer in its simplest eact form. (6) (Total 9 marks). A set of curves is given by the equation sin + cos y = 0.5. City of London Academy 4

5 (a) Use implicit differentiation to find an epression for. () For π < < π and π < y < π, (b) find the coordinates of the points where = 0. (5) (Total 7 marks). (a) Given that y =, and using the result = e ln, or otherwise, show that = ln. () ( (b) Find the gradient of the curve with equation ) y at the point with coordinates (,6). (4) (Total 6 marks) 4. A curve C is described by the equation y + y + 5 = 0. Find an equation of the normal to C at the point (0, ), giving your answer in the form a + by + c = 0, where a, b and c are integers. (Total 7 marks) 5. f() = ( + ) ln, > 0. (a) Use differentiation to find the value of f'() at = e, leaving your answer in terms of e. (4) (b) Find the eact value of e f ( ) (5) (Total 9 marks) 6. A curve C is described by the equation + 4y + 6y 5 = 0. Find an equation of the tangent to C at the point (, ), giving your answer in the form a + by + c = 0, where a, b and c are integers. (Total 7 marks) 7. The volume of a spherical balloon of radius r cm is V cm, where V = 4 r. (a) Find dr City of London Academy 5

6 () The volume of the balloon increases with time t seconds according to the formula 000 (t ), t 0. (b) Using the chain rule, or otherwise, find an epression in terms of r and t for dr. () (c) 000 Given that V = 0 when t = 0, solve the differential equation (t ), to obtain V in terms of t. (4) (d) Hence, at time t = 5, (i) find the radius of the balloon, giving your answer to significant figures, () (ii) show that the rate of increase of the radius of the balloon is approimately.90 0 cm s. () (Total marks) 8. The value V of a car t years after the st January 00 is given by the formula V = (.5) t. (a) Find the value of the car on st January 005. () (b) Find the value of when t = 4. () (c) Eplain what the answer to part (b) represents. () (Total 6 marks) 9. A curve has equation + y y + 6 = 0. Find the coordinates of the points on the curve where = 0. (Total 7 marks) City of London Academy 6

7 0. The curve C with equation y = k + ln, where k is a constant, crosses the -ais at the point A,0 e. (a) Show that k =. (b) Show that an equation of the tangent to C at A is y = e. () (4) (c) Complete the table below, giving your answers to significant figures ln () (d) Use the trapezium rule, with four equal intervals, to estimate the value of ( ln ). (4) (Total marks). f() = +,. 5 e (a) Find f (). () The curve C, with equation y = f(), crosses the y-ais at the point A. (b) Find an equation for the tangent to C at A. () e (c) Complete the table, giving the values of to decimal places e () (d) Use the trapezium rule, with all the values from your table, to find an approimation for the value of City of London Academy 7

8 0 e 5. (4) (Total marks). A drop of oil is modelled as a circle of radius r. At time t r = 4( e t ), t > 0, where is a positive constant. (a) Show that the area A of the circle satisfies da = (e t e t ). (5) In an alternative model of the drop of oil its area A at time t satisfies da A t, t > 0. Given that the area of the drop is at t =, (b) find an epression for A in terms of t for this alternative model. (7) (c) Show that, in the alternative model, the value of A cannot eceed 4. () (Total marks). The curve C has equation 5 + y y + = 0. The point P on the curve C has coordinates (, ). (a) Find the gradient of the curve at P. (5) (b) Find the equation of the normal to the curve C at P, in the form y = a + b, where a and b are constants. () (Total 8 marks) City of London Academy 8

9 4. y P M R O N The curve C with equation y = e + 5 meets the y-ais at the point M, as shown in the diagram above. (a) Find the equation of the normal to C at M in the form a + by = c, where a, b and c are integers. (4) This normal to C at M crosses the -ais at the point N(n, 0). (b) Show that n = 4. () The point P(ln 4, ) lies on C. The finite region R is bounded by C, the aes and the line PN, as shown in the diagram above. (c) Find the area of R, giving your answer in the form p + q ln, where p and q are integers to be found. (7) (Total marks) 5. y A O City of London Academy 9

10 The diagram above shows a graph of y = sin, 0 < <. The maimum point on the curve is A. (a) Show that the -coordinate of the point A satisfies the equation tan + = 0. (4) The finite region enclosed by the curve and the -ais is shaded as shown in the diagram above. A solid bo S is generated by rotating this region through radians about the -ais. (b) Find the eact value of the volume of S. (7) (Total marks) 6. The function f is given by ( ) f() =,,,. ( )( ) (a) (b) Epress f() in partial fractions. Hence, or otherwise, prove that f() < 0 for all values of in the domain. () () (Total 6 marks) 7. A curve has equation y 4 + y 5 = 0. Find an equation of the normal to the curve at the point (4, ), giving your answer in the form a + by + c = 0, where a, b and c are integers. (Total 8 marks) MARK SCHEME. d ln. B ln. y y M A = A Substituting, 8ln M 4ln Accept eact equivalents M A7 [7] City of London Academy 0

11 . (a) h M A V dh 9h 9 B dh h M dh Leading to h cso A5 75 (b) d h d t separating variables M 45h When t 0, h 0. 5ln 4 5h t C 5ln 4 5h t C M A 5ln C M t 5ln 5ln 4 5h When h 0.5 t 5ln 5ln.5 5ln 5ln.5 awrt 0.4 M A Alternative for last marks t 5ln (4 5h) = 5ln.5 + 5ln M M = 5ln = 5n awrt 0.4 A6.5 []. (a) sin sin y =0 M A d y sin sin Accept, sin y sin y City of London Academy

12 sin sin y A (b) At, cos cosy M 6 6 cos y A y y awrt 0.49 A 9 sin sin 6 (c) At,, M 6 9 sin sin 9 y 9 6 M Leading to 6 + 9y π = 0 A [9] da 4..5 B da A r r dr B When A = r r da da dr dr dr.5 r d t M M dr awrt 0.99 A [5] 5. (a) e d y ye y A correct RHS *M A d d ( ye ) e d d ye B City of London Academy

13 (e d y) y ye *M ye e y A5 eº (b) At P, 4 M eº Using mm = m' 4 y ( 0) 4 M M 4y + 4 = 0 or any integer multiple A4 Alternative for (a) differentiating implicitly with respect to y. e ye y A correct RHS *M A d ( ye ) e ye B ( + ye ) e y *M e y ye ye e y A5 [9] 6. (a) C: y y = + 8 Differentiates implicitly to include either y ±ky or.(ignore.) M Correct equation. A A correct (condoning sign error) attempt to d (y ) y combine or factorise their 'y,. M City of London Academy

14 Can be implied. y y A oe4 (b) y = 9 () = + 8 Substitutes y = into C. M = 8 = A 4 fromcorrect working. (, ) (4) 4 6 Also can be ft using their value and y = in the correct part (a) of y Aft (b) final A. Note if the candidate inserts their value and y = into, then an answer of their, may indicate y a correct follow through. [7] r 7. (a) Similar triangles 6 r h Uses similar triangles, h 4 ratios or trigonometry to find either one of these two epressions oe. M V r h h h 4 h 7 AG r h Substitutes into the formula for the volume of water V. A (b) From the question, 8 8 B dh h 7 4h 9 dh h 7 4h 9 B dh Candidate s M; ; dh 4h h dh h or 8 or oe A 4h h City of London Academy 4

15 When h =, dh or 44 8 A oe isw5 Note the answer must be a one term eact value. Note, also you can ignore subsequent working 8 after 44 [7] da 8. (a) From question, = 0.0 da A da da 0.06 (0.0) ; 0.06 When = cm, Hence, = (cm s ) da = 0.0 seen or implied from working. B by itself seen or implied from working B da 0.0 Candidate s ; M; awrt A cso 4 (b) V = (5) = 5 = 5 d V ;{ 0.4} When = cm, = 0.4() = 0.48 (cm s ) V = (5) = 5 B = 5 or ft from candidate s V in one variable Bft Candidate s ; Mft 0.48 or awrt 0.48 A cso 4 City of London Academy 5

16 [8] 9. (a) y + y = 4 (eqn *) 6 y 0 d y d 6 y 6 y or y y 8 6 y 8 y giving 8 y = 8 6y giving y = 6 Hence, y = y = 0 not necessarily required. Differentiates implicitly to include either ky or. (Ignore ) M Correct application ( ) of product rule B ( y ) 6 y and (4 0) A 8 Substituting into their equation. M* Attempt to combine either terms in or terms in y together to give either a or by. dm* simplifying to give y = 0 AG A cso 6 (b) At P & Q, y =. Substituting into eqn * gives () + () = 4 Simplifying gives, = 4 = ± y = y = 4 Hence coordinates are (, 4) and (, 4) Attempt replacing y by in at least one of the y terms in eqn * Either = or = M A Both (, 4) and (, 4) A [9] 0. (a) 4y = y (eqn *) = 8 5 4y = ( 8)y 5 4y = 96y City of London Academy 6

17 Substitutes = 8 (at least once) into * to obtain a three term quadratic in y. Condone the loss of = 0. M 4y 96y + 5 = 0 y 4y + 8 = 0 (y 6)(y 8) = (8) y y = 6 or y = 8. An attempt to solve the quadratic in y by either factorising or by the formula or by completing the square. dm Both y = 6 and y = 8. or ( 8, 8) and ( 8, 6). A (b) 8y ; y y 8y (64) (8) ( 8, 8),, ( 8) 8(8) (64) (6) ( 8, 6), 0. ( 8) 8(6) Differentiates implicitly to include either ky or. Ignore... M Correct LHS equation; A; Correct application of product rule (B) not necessarily required. Substitutes = 8 and at least one of their y-values to attempt to find any one of. dm One gradient found. A Both gradients of and 0 correctly found. A cso 6 Aliter Way City of London Academy 7

18 8y; y y 8y (64) (8) ( 8, 8),, ( 8) 8(8) (64) (6) ( 8, 6), 0. ( 8) 8(6) Differentiates implicitly to include either k or y. Ignore... M Correct LHS equation; A; Correct application of product rule (B) not necessarily required. Substitutes = 8 and at least one of their y-values to attempt to find any one of or. dm One gradient found. A Both gradients of and 0 correctly found. A cso 6 Aliter Way 4y = y (eqn *) 4y + y = 0 y y 4 9 y 8 y (9 8 4(4)( 6 ) ) City of London Academy 8

19 @ = 8 ( (64) ; (8 4(9 ) 8( 8) (64) 4(9(64) ( 5)), ) ) A credible attempt to make y the subject and an attempt to differentiate either or (9 ). M d y k(9 y (9 ) (g( )) d ) ; (8 ) A A Substitutes = 8 find any one of. dm One gradient correctly found. A Both gradients of and 0 correctly found. A 6 [9]. (a) = 7cos t cos 7t, y = 7 sin t sin 7t, 7 sin t 7 sin 7t, 7 cost 7 cos7t 7 cost 7 cos7t 7 sin t 7 sin 7t Attempt to differentiate and y with respect to t to give in the form ± A sin t ± B sin 7t in the form ± C cos t ± D cos 7t Correct and Candidate s M A Bft City of London Academy 9

20 Aliter Way = 7cos t cos 7t, y = 7 sin t sin 7t, 7 sin t 7 sin 7t, 7 cost 7 cos7t 7 cost 7 cos7t 7( sin 4t sin t) tan4t 7 sin t 7 sin 7t 7( cos4t sin t) Attempt to differentiate and y with respect to t to give in the form ± A sin t ± B sin 7t in the form ± C cos t ± D cos 7t Correct and Candidate s M A Bft 7 7 cos 7 cos 6 6 (b) When t =, m( T) ; 7 6 d 7 sin 7 sin Hence m(n) = or = awrt awrt.7 6 When t =, cos cos y 7 sin sin N: y 4 = ( 4 ) N: y or y or y or 4 = (4 ) c c Hence N: y or y or y City of London Academy 0

21 Substitutes t = or 0 into their epression; M 6 to give any of the four underlined epressions oe (must be correct solution only) Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A cso Aftoe. The point (4,4) or (awrt 6.9,4) B Finding an equation of a normal with their point and their normal gradient or finds c by using y = (their gradient) + c. M Correct simplified EXACT equation of normal. This is dependent on candidate using correct (4 Aliter Way 4 When t =, m( T) tan ; 6 d 6 Hence m(n) = or = awrt 0.58 When t =, 6 N: y 4 = N: or 4 = Hence N: () () 7 7 cos cos y 7 sin sin 6 6 ( 4 awrt ) y or y or y (4 ) c c y or y or y 4,4) A oe6 Substitutes t = or 0 into their epression; M 6 City of London Academy

22 to give any of the four underlined epressions oe (must be correct solution only) Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. A cso Aftoe. The point (4,4) or (awrt 6.9,4) B Finding an equation of a normal with their point and their normal gradient or finds c by using y = (their gradient) + c. M Correct simplified EXACT equation of normal. This is dependent on candidate using correct (4,4) A oe6 Beware: A candidate finding an m(t) = 0 can obtain Aft for m(n), but obtains M0 if they write y 4 = ( 4 ). If they write, however, N: = 4, then they can score M. Beware: A candidate finding an m(t) = can obtain Aft for m(n) = 0, and also obtains M if they write y 4 = 0( 4 ) or y = 4. [9]. (a) sin + cos y = 0.5 (eqn *) cos sin y = 0 (eqn #) cos sin y Differentiates implicitly to include ± sin y. (Ignore ( =).) M cos sin y A cso (b) cos 0 0 cos 0 sin y giving or When =, sin cos y 0.5 When =, sin cos y 0.5 cos y =.5 y has no solutions City of London Academy

23 cos y = 0.5 y = or In specified range (, y) =, and, Candidate realises that they need to solve their numerator = 0 or candidate sets = 0 in their (eqn #) and attempts to solve the resulting equation. Mft both, or = 90 or awrt = ±.57 required here A Substitutes either their or into eqn * M Only one of y or - or 0 or 0 or awrt.09 or awrt.09 A Only eact coordinates of, and, A Do not award this mark if candidate states other coordinates inside the required range. A5 [7]. (a) Way y = = e ln ln.e ln M Hence ln.( ) ln AG A cso Aliter Way ln y = ln( ) leads to ln y = ln ln y Hence y ln ln AG Takes logs of both sides, then uses the power law of logarithms City of London Academy

24 ... and differentiates implicitly to give ln y M ln AG A cso (b) d ( ) y ( ) y..ln When =, d y () 4 ln = 64 ln = ( ) A. ( ).ln or.y.ln if y is defined M A d y ( ) ( ) Substitutes = into their which is of the form k or A M 64 ln or awrt 44.4 A4 Aliter Way ln y = ln( ) leads to ln y = ln y.ln When =, y () d 4 = 64 ln = ln A.ln y.ln y M A d y ( ) ( ) Substitutes = into their which is of the form k or A M 64 ln or awrt 44.4 A4 [6] 4. = 6 4y 0 City of London Academy 4

25 Differentiates implicitly to include either ky or. (ignore.) M Correct equation. A 6 4y not necessarily required. At (0, ), Substituting = 0 & y = into an equation involving ; dm to give or A cso Hence m(n) or Aftoe. 7 Either N: y = ( 0) 7 or: N: y 7 Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. y m( 0) + with their tangent or normal gradient ; or uses y = m + with their tangent or normal gradient ; M; N: 7 + y = 0 Correct equation in the form a + by + c = 0, where a, b and c are integers. A oe cso [7] Beware: 7 does not necessarily imply the award of all the first four marks in this question. So please ensure that you check candidates initial differentiation before awarding the first A mark. Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes the final line then they must be awarded A0. Beware: A candidate finding an m(t) = 0 can obtain Aft for m(n) =, but obtains M0 if they write y = ( 0). If they write, however, N: = 0, then can score M. Beware: A candidate finding an m(t) = can obtain Aft for m(n) = 0, and also obtains M if they write y = 0( 0)or y =. Beware: The final cso refers to the whole question. City of London Academy 5

26 Aliter Way = 6 4y 0 Differentiates implicitly to include either k or (ignore.) M Correct equation. A 4y 6 not necessarily required. At (0, ), Substituting = 0 & y = into an equation involving to give 7 ; dm A cso 7 Hence m(n) = or Aftoe. 7 Uses m(t) or to correctly find m(n). Can be ft using.. Either N: y = or N: y = 7 7 ( 0) y = m( 0) with their tangent, or uses y = m + with their tangent, or normal gradient ; or normal gradient ; M N: 7 + y = 0 Correct equation in the form a + by + c = 0, where a, b and c are integers. Aoe cso [7] Aliter Way y + y 5 = y 4 6 y 49 ) 6 4 City of London Academy 6

27 Differentiates using the chain rule; M y 49 ( ) d 6 Correct epression for ; A oe At (0, ), d y 49 6 Hence m(n) = Substituting = 0 into an equation involving ; dm to give or A cso Uses m(t) to correctly find m(n). Can be ft from their tangent gradient. Aft Either N: y or N: y 7 7 ( 0) y = m( 0) with their tangent or normal gradient ; or uses y = m + with their tangent or normal gradient M N: 7 + y = 0 A oe Correct equation in the form a + by + c = 0, where a, b and c are integers. [7] 5. (a) f() = ( + ) + ln M A f(e) = (e + ) + e = e + M A 4 e e (b) ( )n ( ) M A = ( )n ( ) e = ( )n ( ) A 9 = e 0 M A [9] 6. Differentiates City of London Academy 7

28 to obtain: 6 + 8y,... + (6 + 6y) = 0 +(B) 6 6y 6 8y 8 Substitutes =, y = into epression involving, to give = M, A 0 Uses line equation with numerical gradient y ( ) = (their gradient) M ( ) or finds c and uses y = (their gradient) + c To give 5y = 0 (or equivalent = 0) Aft [7] 7. (a) 4r B dr dr (b) Uses. dr in any form, = 000 M, A 4r (t ) (c) V = 000 (t + l) and integrate to p(t + ), = 500(t + ) (+c) M, A Using V = 0 when t = 0 to find c, (c = 500, or equivalent) M V = 500( ) t (any form) A 4 (d) (i) Substitute t = 5 to give V, then use r = V to give r, = 4.77 M, A 4 (ii) Substitutes t = 5 and r = their value into their part (b) M dr (.90 0 ) (cm/s) AG A [] 8. (a) Substitutes t = 4 to give V, = 975. or or 975 or 980 ( s.f) M, A (b) = ln.5 V; = or or 80 M A; A M needs ln.5 term (c) rate of decrease in value on st January 005 B [6] City of London Academy 8

29 9. + y 6y = 0 M (A) A = 0 + y = 0 (or equivalent) M Eliminating either variable and solving for at least one value of or y. M y y y + 6 = 0 or the same equation in y = or = A ( ), (, ) A 7 Note: y d y Alternative: y y ( + 6) = 0 (6 9) y = 6 8. (6 9) 8 = 0 = M (6 9) M A A 64 = M A = A 7 (, ), (, ) [7] 0. (a) 0 = k + ln 0 = k k = (*) M A e (Allow also substituting k = and = into equation and e showing y = 0 and substituting k = and y = 0 and showing =.) e (b) B At A gradient of tangent is = e e M Equations of tangent: y = e e M Simplifying to y = e (*) cso A 4 (c) y =.69, y =.9 B, B City of London Academy 9

30 (d) ( ln ) (...) ( ( )) ft their (c) M Aft 4.7 A 4 accept 4.67 [] B e. (a) Differentiating; f() = + M;A 5 (b) A: 0, 5 B Attempt at y f(0) = f(0); M 6 y or equivalent one line termed equation 5 5 A ft (c).4,.55,.86 B(,0) (d) 0.5 Estimate = ; ( ) [( ) + ( )] B M A ft = ,.4.49 A 4 []. (a) A = r dr, = 4e t B, B da dr da = r, = 4( e t ) 4e t M, M da = ( e t e t ) Acso 5 t (b) A da = t d M A Separation t = (+ c) M, A = + c Use of (, ) M c = A So A t = + A = Attempt A = or A = M t t 4t i.e. A = (or equivalent) A 7 ( t) City of London Academy 0

31 t (c) Because < or t < ( + t) ( A < 4) B ( t) []. (a) 0, +(y + ), 6y = 0 M, (B), A At (, ) 0 + (4 + ) = 0 M 4 7 =.4 or or A 5 (b) 5 The gradient of the normal is 7 M 5 Its equation is y = ( ) 7 M (allow tangent) y = + or y = Acao [8] 4. (a) M is (0, 7) B = e M Attempt gradient of normal is M ft their y(0) or = (Must be a number) equation of normal is y 7 = ( 0) or +y 4 = 0 + y = 4 o.e. A 4 (b) y = 0, = 4 N is (4, 0) (*) B cso (c) R T (e + 5) = [e + 5] some correct M City of London Academy

32 ln 4 0 R = (e + 5) = ( ln 4) ( + 0) M limits used = ln 4 A T = (4 ln 4) B Area of T T = (7 ln) ; R = ln Use of ln 4 = ln B R = T + R, R = 97 ln M, A 7 [] 5. (a) d y sin (sin ) cos M, A At A sin (sin ) cos = 0 dm sin + cos = 0 (essential to see intermediate line before given answer) tan + = 0 (*) A 4 (b) V = y = sin M 0 0 = cos cos M A = cos sin sin M 0 = cos sin cos A = [ ] M = [ 4] A 7 [] ( ) A B 6. (a), and correct method for finding A or B M ( )( ) A =, B = A, A (b) f() = M A ( ) ( ) Argument for negative, including statement that square terms are positive for all values of. (f.t. on wrong values of A and B) A ft [6] 7. Differentiates w.r.t. to give City of London Academy

33 , + y, 4 + y = 0 M, B, A At (4, ) 48 (8y + 6) 4 + 7y = 0 M 8 y = = 9 A Gradient of normal is M y = ( 4) M i.e. y 6 = 4 y + = 0 A8 [8] ============================================================================= EXAMINERS REPORTS. This question was also well answered and the general principles of implicit differentiation were well understood. By far the commonest source of error was in differentiating ; eamples such as, ln and were all regularly seen. Those who knew how to differentiate difficulty in finding d (y) nearly always completed the question correctly, although a few had correctly. A minority of candidates attempted the question by taking the logs of both sides of the printed equation or a rearrangement of the equation in the form = y y. Correctly done, this leads to quite a neat solution, but, more frequently, errors, such as ln( + y ) = ln + ln y, were seen. It was noteworthy that a number of correct solutions were seen using partial differentiation, a topic which is not in the A level Mathematics or Further Mathematics specifications. These were, of course, awarded full marks.. Many found part (a) difficult and it was quite common to see candidates leave a blank space here and proceed to solve part (b), often correctly. A satisfactory proof requires summarising the information given in the question in an equation, dt such as = 0.48π 0.6πh, but many could not do this or began with the incorrect = 0.48π 0.6πh. Some also found difficulty in obtaining a correct epression for the volume of water in the tank and there was some confusion as to which was the variable in epressions for the volume. Sometimes epressions of the form were differentiated with respect to r, which in this question is a constant. If they started appropriately, nearly all candidates could use the chain rule correctly to complete the proof. Part (b) was often well done and many fully correct solutions were seen. As noted in the introduction above, some poor algebra was seen in rearranging the equation but, if that was done correctly, candidates were nearly always able to demonstrate a complete method of solution although, as epected, slips were made in the sign and the constants when integrating. Very few candidates completed the question using definite integration. Most used a constant of integration (arbitrary constant) and showed that they knew how to evaluate it and use it to complete the question.. As has been noted in earlier reports, the quality of work in the topic of implicit differentiation has improved in recent years and many candidates successfully differentiated the equation and rearranged it to find dh V r h. Some, however, City of London Academy

34 forgot to differentiate the constant. A not infrequent, error was candidates writing = sin siny and then incorporating the superfluous (cosy) =. on the left hand side of the equation into their answer. Errors like were also seen. Part (b) was very well done. A few candidates gave the answer 0, not recognising that the question required radians. Nearly all knew how to tackle part (c) although a few, as in Q, spoilt otherwise completely correct solutions by not giving the answer in the form specified by the question. 4. Connected rates of change is a topic which many find difficult. The eaminers reported that the responses to this question were of a somewhat higher standard than had been seen in some recent eaminations and the majority of candidates attempted to apply the chain rule to the data of the question. Among those who obtained a correct relation,.5 = πr or an equivalent, a common error was to use r =, instead of using the given A = to obtain r =. sin y dr Unepectedly the use of the incorrect formula for the area of the circle, A = πr, was a relatively common error. 5. As noted above work on this topic has shown a marked improvement and the median mark scored by candidates on this question was 8 out of 9. The only errors frequently seen were in differentiating implicitly with respect to. A few candidates failed to read the question correctly and found the equation of the tangent instead of the normal or failed to give their answer to part (b) in the form requested. 6. A significant majority of candidates were able to score full marks on this question. In part (a), many candidates were able to differentiate implicitly and eaminers noticed fewer candidates differentiating 8 incorrectly with respect to to give 8. In part (b), many candidates were able to substitute y = into C leading to the correct -coordinate of. Several candidates either rearranged their C equation incorrectly to give = or had difficulty finding the cube root of 8. Some weaker candidates did not substitute y = into C, but substituted y = into the epression to give a gradient of. 7. A considerable number of candidates did not attempt part (a), but of those who did, the most common method was to h use similar triangles to obtain r and substitute r into V = r h to give V 4 h Some candidates 7 used trigonometry to find the semi-vertical angle of the cone and obtained correctly used similar shapes to compare volumes by writing down the equation from this. A few candidates Part (b) discriminated well between many candidates who were able to gain full marks with ease and some candidates who were able to gain just the first one or two marks. Some incorrectly differentiated V = r h to give V dh d r. 8 Most of the successful candidates used the chain rule to find by applying The final answer was sometimes carelessly written as. Occasionally, some candidates solved the differential 4 h equation 8 and equated their solution to and then found or differentiated implicitly to find 7 dh dh. 8. At the outset, a significant minority of candidates struggled to etract some or all of the information from the question. 8 These candidates were unable to write down the rate at which this cross-sectional area was increasing, = 0.0; or the cross-sectional area of the cylinder A = and its derivative = π; or the volume of the cylinder V = 5π City of London Academy 4 da y e h r dh V 6 h dh da

35 and its derivative = 5π. In part (a), some candidates wrote down the volume V of the cylinder as their cross-sectional area A. Another popular error at this stage was for candidates to find the curved surface area or the total surface area of a cylinder and write down either A = 0π or A = π respectively. At this stage many of these candidates were able to set up a correct equation to find and usually divided 0.0 by their and substituted = into their epression to gain out of the 4 mark available. Another error frequently seen in part (a) was for candidates to incorrectly calculate Finally, rounding the answer to significant figures proved to be a problem for a surprising number of candidates, with a value of 0.00 being seen quite often; resulting in loss of the final accuracy mark in part (a) and this sometimes as a consequence led to an inaccurate final answer in part (b). Part (b) was tackled more successfully by candidates than part (a) maybe because the chain rule equation is rather more straight-forward to use than the one in part (a). Some candidates struggled by introducing an etra variable r in addition to and obtained a volume epression such as V = πr (5). Many of these candidates did not realise that r and were then unable to correctly differentiate their epression for V. Other candidates incorrectly wrote down the volume as V = π (5). Another common error was for candidates to state a correct V, correctly find, then substitute = to arrive at a final answer of approimately About 0% of candidates were able to produce a fully correct solution to this question. 9. This question was generally well done with a majority of candidates scoring at least 6 of the 9 marks available. In part (a), implicit differentiation was well handled with most candidates appreciating the need to apply the product rule to the y term. A few candidates failed to differentiate the constant term and some wrote =... before starting to differentiate the equation. After differentiating implicitly, the majority of candidates rearranged the resulting equation to make the subject before substituting as rather than substituting for in their differentiated equation. Many candidates were able to prove the result of y = 0. A surprising number of candidates when faced with manipulating the equation + y = 8 & y = and then proceeded to solve these equations simultaneously., separated the fraction to incorrectly form two equations 6 Some candidates, who were unsuccessful in completing part (a), gave up on the whole question even though it was still possible for them to obtain full marks in part (b). Other candidates, however, did not realise that they were epected to substitute y = into the equation of the curve and made no creditable progress with this part. Those candidates who used the substitution y = made fewer errors than those who used the substitution = 6 y 8 y as. The most common errors in this part were for candidates to rewrite y as either 4 or ; or to solve the equation = 4 to give only = or even = ±4. On finding = ±, some candidates went onto substitute these values back into the equation of the curve, forming a quadratic equation and usually finding etra unwanted points rather than simply doubling their two values of to find the corresponding two values for y. Most candidates who progressed this far were able to link their values of and y together, usually as coordinates. 0. This question was generally well done with many candidates scoring at least seven or eight of the nine marks available. da In part (a), the majority of candidates were able to use algebra to gain all three marks available with ease. It was disappointing, however, to see a significant minority of candidates at A level who were unable to correctly substitute y = 8 into the given equation or solve the resulting quadratic to find the correct values for y. In part (b), implicit differentiation was well handled, with most candidates appreciating the need to apply the product rule to the y term although errors in sign occurred particularly with those candidates who had initially rearranged the given equation so that all terms were on the LHS. A few candidates made errors in rearranging their correctly 8 y 8 City of London Academy 5

36 differentiated equation to make manipulating their correctly substituted the subject. Also some candidates lost either one or two marks when epressions to find the gradients.. In part (a), many candidates were able to apply the correct formula for finding in terms of t. Some candidates erroneously believed that differentiation of a sine function produced a negative cosine function and the differentiation of a cosine function produced a positive sine function. Other candidates incorrectly differentiated cos 7t to give either sin 7t to give either cos 7t or cos 7t. sin 7 t or sin 7t and also incorrectly differentiated In part (b), many candidates were able to substitute into their gradient epression to give, but it was not uncommon to see some candidates who made errors when simplifying their substituted epression. The majority of candidates were able to find the point (, 4). Some candidates, however, incorrectly evaluated cos ( ) and sin 7 ( ) as and respectively and found the incorrect point (, ). Some candidates failed to use the 6 gradient of the tangent to find the gradient of the normal and instead found the equation of the tangent, and so lost valuable marks as a result. It was pleasing to see that a significant number of candidates were able to epress the equation of the normal in its simplest eact form.. In part (a), the majority of candidates were able to successfully differentiate the given equation to obtain a correct epression for, although there were a small proportion of candidates who appeared to forget to differentiate the constant term of 0.5. Some candidates, as was similar with Q, produced a sign error when differentiating sin and cos y with respect to. These candidates then went on to produce the correct answer for, but lost the final accuracy mark. A few candidates incorrectly believed that the epression could be simplified to give cot. In part (b), the majority of candidates realised that they needed to set their numerator equal to zero in order to solve = 0. Most candidates were then able to obtain at least one value for, usually =, although = was not always found. A surprising number of candidates did not realise that they then needed to substitute their value(s) back into the original equation in order for them to find y. Of those who did, little consideration was given to find all the solutions in the specified range, with a majority of these candidates finding y = also finding y = 7, but only a minority of candidates. Therefore it was uncommon for candidates to score full marks in this part. Some candidates also incorrectly set their denominator equal to zero to find etra coordinates inside the range. Also another small minority of candidates stated other incorrect coordinates such as or in addition to the 7 two sets of coordinates required. These candidates were penalised by losing the final accuracy mark. 4 t 6 cos sin y,, 7 6. In part (a), candidates either replaced with e ln and applied the chain rule; or took logs of both sides of the given equation and then differentiated implicitly. A majority of the candidates were equally likely to correctly apply either one of these two methods. Weaker candidates, however, seemed oblivious to the fact that ln and ln are, in fact, different, and wrote them almost interchangeably. City of London Academy 6

37 Part (b) proved challenging for a significant number of candidates. Those candidates who used implicit differentiation in parts (a) and (b) were more likely to achieve the correct gradient. Such ( an ) approach avoided the ( errors seen when candidates ( ) were trying to handle indices. Such errors included either =. =. or =. = 4. ( ) or. Another common error was for some candidates to argue that since the derivative of is.ln, then the derivative of must be ln. 4. This question was successfully completed by the majority of candidates. Whilst many demonstrated a good grasp of the idea of implicit differentiation there were a few who did not appear to know how to differentiate implicitly. Candidates who found an epression for ( ) ( ) in terms of and y, before substituting in values of = and y =, were prone to errors in manipulation. Some candidates found the equation of the tangent and a number of candidates did not give the equation of the normal in the requested form. 5. The product rule was well understood and many candidates correctly differentiated f() in part (a). However, a significant number lost marks by failing to use ln e = and fully simplify their answer. Although candidates knew that integration by parts was required for part (b), the method was not well understood with common wrong answers involving candidates mistakenly suggesting that ln and attempting to use u = dv ln + and in the formula. dv u uv du v ) e e Candidates who correctly gave the intermediate result ln often failed to use l l a bracket for the second part of the epression when they integrated and went on to make a sign error by giving rather than `This question was generally well answered with most candidates showing good skills in differentiating eplicitly. Candidates who found an epression for in terms of and y, before substituting in values, were more prone to errors in manipulation. Some candidates found the equation of the normal and a number of candidates did not give the equation of the tangent in the requested form. It was quite common to see such statements as presentation. 6 8y 6 6y 0, but often subsequent correct working indicated that this was just poor 7. The fact that this question had so many parts, with a good degree of independence, did enable the majority of candidates to do quite well. All but the weakest candidates scored the first mark and the first were gained by most. The integration in part (c ) did cause problems: eamples of the more usual mistakes were to write (t ) k or or or t t t (t ) equal to zero; two of the mistakes which came more into the howler category were ln000v or V ln and, or to omit the constant of integration or assume it d ( ) t t 4t 4t 4t... 4t. Many candidates were able to gain the method marks in parts (d) and (e). 8. (a) Most understood the contet of this problem and realised that they needed to use t = 4, although t = 0, or 5 were often seen. City of London Academy 7

38 (b) (c) Very few had any idea at all about how to differentiate V (many gave their answer as t(.5) -t-, or had a term (.5) -t ). The comments made in answer to the request to interpret their answer to part (b) were usually too generalised and vague. The eaminers required a statement that the value of rate of change of value on st January 005 which had been found represented the 9. Almost all candidates could start this question and the majority could differentiate implicitly correctly. This is an area of work which has definitely improved in recent years. Many, having found to find a substantial number of students in this relatively advanced A module proceeding from, could not use it and it was disturbing to + y = y. Those who did obtain y = often went no further but those who did could usually obtain both correct points, although etra incorrect points were often seen. 0. In part (a), the log working was often unclear and part (b) also gave many difficulty. The differentiation was often incorrect. was not unepected but epressions like were also seen. Many then failed to substitute e into their and produced a non-linear tangent. Parts (c) and (d) were well done. A few did, however, give their answers to an inappropriate accuracy. As the table is given to decimal places, the answer should not be given to a greater accuracy.. For many candidates this was a good source of marks. Even weaker candidates often scored well in parts (c) and (d). In part (a) there were still some candidates who were confused by the notation, often interpreted as, and common wrong answers to the differentiation were and +. The most serious error, which occurred far too 5 e frequently, in part (b) was to have a variable gradient, so that equations such as y = were common. 5 5 The normal, rather than the tangent, was also a common offering.. There were two common approaches used in part (a); substituting for r to obtain a formula for A in terms of t or using the chain rule. The inevitable errors involving signs and were seen with both methods and the eaminers were disappointed that some candidates did not seem to know the formula for the area of a circle: were common mistakes. Part (b) proved more testing. Most could separate the variables but the integration of negative powers caused problems for some who tried to use the ln function. Many did solve the differential equation successfully though sometimes they ran into difficulties by trying to make A the subject before finding the value of their arbitrary constant. The final two marks were only scored by the algebraically deterous. There was some poor work here and seeing followed by was not uncommon. The final part eluded most. Those who had a correct answer to part (b) sometimes looked at the effect on A of t but only a small minority argued that since t >0 then, and therefore A <4.. This was usually well done, but differentiation of a product caused problems for a number of candidates. Many still insisted on making r, r and 4 r unnecessary algebraic errors. e the subject of their formula before substituting values for and y. This often led to 4. Whilst the majority of answers to part (a) were fully correct, some candidates found difficulties here. A small number failed to find the coordinates of M correctly with (0, 5) being a common mistake. Others knew the rule for perpendicular gradients but did not appreciate that the gradient of a normal must be numerical. A few students did not show clearly that the gradient of the curve at = 0 was found from the derivative, they seemed to treat y = e + 5 and assumed the gradient was always. Some candidates failed to obtain the final mark in this section because they did not observe the City of London Academy 8 e y 0 y - f f A t 4 A or t A t t ( t)

39 instruction that a, b and c must be integers. For most candidates part (b) followed directly from their normal equation. It was disappointing that those who had made errors in part (a) did not use the absence of n = 4 here as a pointer to check their working in the previous part. Most preferred to invent all sorts of spurious reasons to justify the statement. Many candidates set out a correct strategy for finding the area in part (c). The integration of the curve was usually correct but some simply ignored the lower limit of 0. Those who used the simple half base times height formula for the area of the triangle, and resisted the lure of their calculator, were usually able to complete the question. Some tried to find the equation of PN and integrate this but they usually made no further progress. The demand for eact answers proved more of a challenge here than in 6(c) but many candidates saw clearly how to simplify e ln4 and convert ln 4 into ln on their way to presenting a fully correct solution. 5. Most candidates made some attempt to differentiate sin, with varying degrees of success. sin + cos was the most common wrong answer. Having struggled with the differentiation, several went no further with this part. It was surprising to see many candidates with a correct equation who were not able to ti up the terms to reach the required result. Most candidates went on to make an attempt at πy. The integration by parts was generally well done, but there were many of the predictable sign errors, and several candidates were clearly not epecting to have to apply the method twice in order to reach the answer. A lot of quite good candidates did not get to the correct final answer, as there were a number of errors when substituting the limits. 6. Most answered part (a) correctly, and errors were usually because candidates had miscopied a sign, or written + instead of ( +). The differentiation was usually correct, though a minority integrated and some misunderstood the notation and found the inverse function instead of the derived function. Those who returned to the original epression and differentiated gave themselves more work. Many found it difficult to answer the final part of this question. The eaminers were looking for a statement that square terms are always positive 7. No Report available for this question. City of London Academy 9