Review exercise

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1 Review eercise y cos sin When : 8 y and 8 gradient of normal is 8 y When : 9 y and 8 Equation of normal is y 8 8 y y y y y e ln( ) e ln e When : y e ln and e Equation of tangent is y e ln e y e ln e e y e ln 5e gradient of normal is 9 Equation of normal is y 9 y y 9 y e a du dv and v e e dv du u v Let u e e e e or or When, y When, y e coordinates of stationary points are, and,e. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

2 5 a y sin du dv and v sin cos Let u du dv v u v sin cos sin sin cos cos sin When : y and Equation of tangent is y = y a ycosec sin Let u sin du cos and y u du u Using the chain rule: du du cos sin sin tan cosec cot cosec y cosec y cot y cosec ycot cot y 7 y arcsin sin y cos y and cos y y sin ycos y cos y sin y d Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

3 8 a cot t, y sin t cosec t, sin t cos t sin tcost cosec t sin tcost When t : and y equation of tangent is y ( ) y c cot t cot t y y sin t sin t and cosec t y cosec t cot y y 8 y t As t,cot t cot t so the domain of the function is. 9 a, y t t Using the chain rule:, d t ( t) d t ( t) ( t) ( t) When t : and y 9 () 9 d ( ) equation of tangent is y 9 y 9 8 t t Sustitute into y y t : Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

4 y y 5 Differentiating with respect to : y d d Sustituting, y : 7 7 gradient of normal at (, ) is Equation of normal is 7 y ( ) 7 y 7 y a sin cos y.5 7 Differentiating with respect to : cos sin y d cos sin y cos When : cos y.5 cos y.5 y or y When : cos y.5 cos y.5 (no solutions) the only stationary points in the given range are at, and,. y e e e e d y e e e d y For C to e conve,. e, and for all, a a d y for all. Hence C is conve for all. dv V r r dr Using the chain rule: d V d d d V r r r dr dv dr r d t (t ) dr (t ) r 5 (t ) r g( ) g(.)... g(.5) The change of sign implies that the root is in.,.5. g(.55).5... g(.5)... The change of sign implies that the root satisfies.55.5, and so. correct to decimal places. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

5 5 a p( ) cos e a.7 p(.7) cos.7 e p(.8) cos.8 e... The change of sign implies that the root is in.7,.8. p(.755) cos.755 e p(.75) cos.75 e The change of sign implies that the root satisfies , and so.7 correct to decimal places. f( ) e 5 e 5 ln 5 ln 5, for 5 5 f( ) c Using : All correct to decimal places. Using : ln ln ln All correct to decimal places. d f(.55) f(.55) a f( ) f(.)..... f(.) a The change of sign implies that the root satisfies.55.55, and so.5 correct to decimal places. f( ) e As A is a stationary point, the gradient at A is zero. f ( a). The change of sign implies that the root is in.,.. The Newton Raphson process uses f ( ) as a denominator. Division y zero is undefined so a cannot e used to find an approimation for. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 8 f ( ) e.. 9 a Using.9 : f( ) f ( ) f(.9).9 f (.9) ( d.p.) f( ) i f(.) f(.) ii The change of sign implies that there is in.,.. a root f(.) f(.7) The change of sign implies that there is in.,.7. a root f( ) c Using.5: d a All correct to decimal places. 9 f ( ) Using.: f( ) f ( ) f(.). f (.) ( d.p.) v( ).cos.5sin 5 5 Rcos 5 Rcos 5 Rcos cos sin sin 5 5. cos.5sin 5 5 Rcos. and Rsin.5.5 tan.5 ( d.p.). R R cos sin..5.9 so R.7 Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

7 c d v( ).7 cos.5 5 v ( ).7sin sin v (.7).8sin v (.8).8sin The change of sign implies that there is a stationary point in the interval.7,.8. v ( ).8cos Using.: v (.). v''(.) a a a 78 9 a 9 8 a 79 a 9 a a a cos 5 cos5 cos sin 5 sin cos 5 cos5 cos sin 5 sin Adding: cos 7 cos cos5 cos cos5 cos cos 7 cos sin 7 sin c sin cos ( d.p.) e v (.5) 5..8sin Consider y e e m m m e e m m m e m m m 8 e e m m 8 e e m m v (.75) 5.5.8sin The change of sign implies that there is a stationary point at.7 correct to decimal places. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 Let I 5 5 Let I d Let u u u d replace with u du. u u and u 5 u I u du u u d u u u du Let u d u replace with. u u du u u d u u u u du u I u u 8 8 Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 e Let I ( )ln du Let u n dv and v Using the integration y parts formula: e e I n e e e e e 9 e e e e 9 9 e 9 9 (e ) 9 5 A B 7 a ( )( ) e A( ) B( ) ()( ) 5 A( ) B( ) 5 ()( ) n( ) n( ) n 9 n 8 lnn ln 9 ln 8 ln 8 n 9 n n 7 n n 5 8 Area R Consider 8sin cos d y cos sin cos 8sin cos cos Let : 7 B( 7) so B Let : 7 A so A 5 ( )( ) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 9 a When : y sin.57 ( d.p.) : a Let When : y sin.88 ( d.p.) I sin.5( ( ( s.f.) I e du Let u dv and e v e.).8) Using the integration y parts formula: I e e e e e e e When =., y =.89 When =.8, y =.9 c Area of R.( ( ) 7.89) (s.f.) d Percentage error in answer from part c e.8 %.7% e a c A B ( )( ) A( ) B( ) Let : B B Let : A A ( )( ) ( )( ) ( ) y Separating the variales: y ( )( ) n y d d n n c n n( ) n A ( ) n A the general solution is A( ) y y A( ) When, y so A( ) A the particular solution is ( ) y ( ) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

11 a dv V r r dr Using the chain rule: dv dv dr dr dr r k dr r V dr k r r k r 5 k B Separating the variales: 5 k r dr r k t A 9k r t A 8 9k r t A 8 Separating the variales: dv kv n kv t c k When t, V so ln c k Comining the ln terms: kv n t k kv n kt kv kt e kv e kt V e k k A and B k k kt a Rate of change of volume is dv cm s Increase is cm s Decrease is kv cm s, where k is a constant of proportionality. the overall rate of change is dv kv Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

12 c kt dv V e e k k kt Sustitute d V when t 5 : e e 5k 5k Taking natural logarithms: 5k n or 5k ln k n.8 ( d.p.) 5 V n n When t : V n n 75 n 8. ( d.p.) the volume is 8 cm ( s.f.). t 5 5 c When t, C C so A = C C C e kt When t, C C so k C Ce k e k n k n PQ (8 ( ) ( ) ( k ) 9 ( 8) ( k ) 7 5 k 8 5 k k k or k AB 5 AC a d C is the rate of change of concentration. BC AC AB i 8j 7k BC 9 9 The concentration is decreasing so the rate of change is negative. dc C or d C kc, where k is a positive constant of proportionality. Separating the variales: d C k d t C so ln C kt ln A, where n A is a constant. ln C kt A C A e kt the general solution is C Ae kt cos BAC BAC. ( d.p.) 7 a Let O e the fied origin. PQ OQ OP i 5j k PQ 5 9 c Unit vector in direction of PQ 5 i j k cos z.7 9. ( d.p.) z Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

13 7 d AB i 5j k There is no scalar, say m, for which AB mpq, so AB and PQ are not parallel. 8 MN i 5j k MN 5 MP k i jk MP k k 5 NP k 8 i j 7k NP k 8 7 If MN k 8 5 MP then k 5 k k k or k k (since k is positive) If MN NP then. i jk pi 8 qr j prk 9 Comparing coefficients of i: p p Comparing coefficients of k: pr pr 8 r Comparing coefficients of j: 8 qr qr q 8 p, q 8, r a Particle is in equilirium so F F F Comparing coefficients of i: a9 a Comparing coefficients of j: 5 7 Comparing coefficients of k: c5 c R F F ( 9 ) i (5 7) j ( 5) k ( i j k) N c F ma i j k a a i j k ms Acceleration k 8 5 k 8 85 there are no integer solutions for k if MN NP d a 9 ms If MP NP then k k k k k k 9 k 7 there are no positive solutions for k if MP NP k Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

14 Challenge a ay y y Differentiating with respect to : d d d a y y y y y a y y y a y y Sustituting for in the original equation: ay y 8y y ay 5y a ya 5y y or y 5 When y, y a a When y, y 5 5 a a at, and at,. 5 5 a y y a a y y Sustituting for y in the original equation: a a a a a a a 8 a a a 5 a a ac a a a a (as a ) so 5 a has no solutions. Hence d for all. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

15 y sin and y cos Curves intersect when sin cos sin cos sin sin sin sin sin sin or sin the intersections are at 5, and Shaded area up to is cos sin cos sin sin cos Shaded area etween 5 cos sin sin cos 5 ( ) the total shaded area is 5 and is Shaded area etween 5 sin cos cos sin 5 and is 5 Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 5