Chapter 7. Exercise 7A. dy dx = 30x(x2 3) 2 = 15(2x(x 2 3) 2 ) ( (x 2 3) 3 ) y = 15

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1 Chapter 7 Exercise 7A. I will use the intelligent guess method for this question, but my preference is for the rearranging method, so I will use that for most of the questions where one of these approaches is suitable.. Guess y x+ d x+ x + Adjust by a factor of : x + x + y x + x + x +. This can not be rearranged to the form f xfx n so we expand:.. xx + x + x y x + x x + x + x + x + x y + x x x x x x y x. This can not be rearranged to the form f xfx n so we expand: x + x x + x + x x + x + x y x + x + x x + x + x 7. xx xx x y x. This can not be rearranged to the form f xfx n so we expand: xx xx x + 9 x x + 9x y x x + 9x x x + 9x x + x + x + y x + x + x + y x + x + x x y x x x x x x y x x x Note that the last step is valid because

2 Exercise 7A Solutions to A.J. Sadler s Such a simplification would not be possible if we were raising to an even power. + x + x y x + x x + x y + x + x + x + x y + x + x x + x x + x + x y + x This question is probably just as easy to do by first expanding: x + x x + x y x + x x + x Although these two solutions may not look quite the same, you should satisfy yourself that they are, in fact, both correct. Remember, c is an arbitrary constant. x x x y x x x x y x x + x + x + y x + x + x + x + y x + x + x x + xx + xx + y x + x + x + x + x + x + y x + x x + xx + x + y x + x x x y x x

3 Unit C Specialist Mathematics Exercise 7B.. 7. x + x + x + y x + y x + x + x + x + x + x + x x x y x x x. 9. dp dt + t + t P + t + c P + t + dp dt tt t P t c 7 P t + 7 Exercise 7B. cos x sin x. sin x sin x sin x cos x sin x cos x cos x sin x cos x cos x sin x cos x cos x sin x sin x sin x cos x cos x cos x sin x sin x sin x cos x cos x cos x sin x sin x sin x cos x 9 cos x

4 Exercise 7B Solutions to A.J. Sadler s. For this question and the next, you should note that sin x + π cos x and cos x π sin x. You may make the substitution at the beginning and then antidifferentiate, or first antidifferentiate and then substitute.. See previous question.. cos x + π sin cos x + π sin x sin x cos x cos tan x x cos x cos x tanx x + π cos tanx x cosx + sin x cos x sin x sin x cos x cosx sin x cos x + sin x sin x + cos x x + cos x + cos x x + cos x + cos x x + sin x + sin x. Although this looks long, you should by now be able to antidifferentiate it in a single step, simply working term by term. + x x x + x x cos x cos x sin x sin x sin x cos x so the overall antiderivative is x + x x + sin x + cos x sin x cos x cos x cos x sin x cos x cos x sin x cos x sin x sin x cos x sin x sin x cos x sin x cos x sin 7 x cos x sin x sin x sin x cos x sin x cos x sin x sin x sin x cosx sin x cos x sin x sin x sin x cos x sin x cos x sin x sin x cos x sinx cos x sin x cos x cos x

5 Unit C Specialist Mathematics Exercise 7B sin x sin xsin x sin x cos x sin x sin x cos x cos x os x cos x cos xcos x cos x sin x cos x cos x sin x sin x sin x cos x cos xcos x cos xcos x cos x sin x cos x sin x + sin x cos x cos x sin x os x sin x sin x sin x + sin x cos x cos x cos x + cos x + sin x + x sin x + x sin x sin x sin x cos x sin x x sin x + x sin x sin x sin x sin x sin x sin x cos x cos x cos x + cos x cos x + cos x cos x + cos x cos x + sin x sin x + x sin x sin x + x cos x + sin x x Compare this with your answers for questions 7 and. cos x sin x cos x sin x Compare this with your answers for questions 7 and. sin x cos x os x sin x sinx + x sin 7x cos 7x 7 You need to know the angle sum trig identities well enough to recognise them. sin x cos x cos x sin x sinx x sin x cos x

6 Exercise 7B.. cos x cos x sin x sin x cosx + x cos 7x sin 7x 7 cos x cos x + sin x sin x cosx x cos x sin x. Refer to your answers to questions and There are at least three different ways of approaching this question. You can treat it as. 9. fxf x [fx] where fx sin x i.e. sin x ; fxf x [fx] where fx cos x i.e. cos x ; or Recognise the double angle formula for sine: sin x cos x. You should satisfy yourself that these answers are equivalent differing only in the value of the constant of integration. sin x cos x sin x sin x cos x sin x cos x cos x sin x cos x sin x cos x cos x os x cos x sin x cos x cos x sin x cos x sin x sin x cos x sin x cos x sin x sin x. C sin p dp cos p + k cos + k k C cos p sin x Solutions to A.J. Sadler s Note that I use k for the constant of integration here rather than the more commonly used c so as to avoid confusion with C.. P cos x. sin x sin π c P sin x + sec x tan x sec x. By the chain rule, so. By the chain rule,.. 7. so d sec x sec x tan x sec x tan x sec x d cot x cosec x cosec x cot x sin x cosec x cot x sin x cos x sin x cos x cosx sec x tan x sec x cos x sin x cosec x cot x cosec x. This is of the form fx n f x where fx cosec x: cosec x cot xcosec x cosec x cosec x 9. cosec x cot x cosec x cot xcosec x cosec x

7 Unit C Specialist Mathematics Exercise 7C Exercise 7C In the following solutions I have used notation like du x. This is convenient in helping us make the variable substitutions, but it is important to remember that du and are meaningless outside the context of integration. We must not fall into the trap of treating them like real quantities.. u x du x xx u u x u du x xx + u u du u u du u u du u 7 7 u u7 u 9 u u 7 x + x + 7 x + x + 7 x + x. u x du x u du. u x du x xx u x u du x x u u du uu du u u du u u u u u u x x x x x x +. u x + du x u u x xx + u x u du u u x +. u x + du x u du. u x du x u + du 7

8 Exercise 7C Solutions to A.J. Sadler s xx u + u du u + u du u7 7 + u u7 7 + u 7 u u x x x x x x + 9. u x + du x u du xx + u u du u u u u x + x + x + x 7. u x du x u du x x uu du u u du u u. u x du x u du u u u u x x x x x x x x + x x u u du u u du. u x + du x u du x x + u u du u u du u u u u x + x + x + 9x. u x + du x u7 7 u u u x x x x + du u u x + x x x x +. u x du x u du

9 Unit C Specialist Mathematics Exercise 7D x u x du u u du u u du u u u u u uu x x 7 cos 7 x sin x 9u 7 du 9 u 9 cos x. u x + du x x sinx + sin u du cos u cosx + xx + xx +. u sin x du cos x sin x cos x u du. u cos x du sin x u sin x. u x + du x u du x + x + u + u du u + u du u u u u + 7 x + x x + x + 9 Exercise 7D.. x + sin x x x cos x. cos x + sin xcos x sin x cos x sin x cos x sin x. cos x sin x. x + x x x + x x + x 9

10 Exercise 7D. 7.. sin x sin x cos x cos x + cos x cos x cos x os x + cos x 7 + cos x 7 os x x sin x 7 x + sin x x sin x cos x cos x 9. u x du x x sinx. u + x du sin u du cos u cosx u + x du u. u + x du x u du + x x + x u u du u u du u u u 9 u 9 u u 9 + x + x 9 + x + 9x 9 + x 9x. sin x cos x. u x + 7 du x u 7 Solutions to A.J. Sadler s sin x du xx+7 u 7u du u 7u du 7 u7 7 u u u 9 sin x cos x x + 7 x x + 7 x 7 x + 7 x + 7 x + 7 x x x xx 7 xx 7 x cos x + sin x sin x cos x. u x du 9. x u + du xx 7 u + u 7 du u + u 7 du x x 9 u9 + u u u + 9 x x + 9 x x +

11 Unit C Specialist Mathematics Exercise 7D. u + x du + x. u + x du... x u du u du u + x u + x x u du u du u u u du u u u u uu + x + x + xx sin x cos x sin x cos x cos x sin x cos x cos x cos x sin x cos x sin x cos x sin x sin x cos x os x cos x + cos x x + x + x + 9 x + x + 9. u x + du x x sinx + x sin u sin u du cos u cosx +. u x du x u + du x + x u + + u u + u u + u u u u u u x + du x. x x + 77 x x + 7 x + x + x x u du x 9. u x du x u + u x + x x x xx x u du u + u du u + u du u7 7 + u x x x x + 7 x x +

12 Exercise 7E... cos x sin x x x cos x cos x sin x x x x sin x cos x sin x. u x du x u + cos x x x x u du u + u du u + u + u du u 7 + u + u du u + u7 7 + u u Solutions to A.J. Sadler s u + u + x x + x + x x x + + 9x + x x x + + 9x x x + x + Exercise 7E. x.. y x x x y x y x y x x. sin y x sin y x cos y x cos y x 7. y x + y x + y y x + x. y x y x. y x x y x x y y x x. y x y x y x 9. cos y x cos y x sin y x

13 Unit C Specialist Mathematics Exercise 7E. yy + x yy + x. y + x y + x x y x c y x +. y x y x. y x. x y x os y x + y + sin y x + x π + + c π y + sin y x + x + π. y + x + x y + x + x. y + y x + x + + c y + y x + x + v dv s ds v s v s c v s + v + v ± 7 Ignore the negative solution, because v is a speed so can not be negative.. y sin x y sin x y cos x cosπ/ c y cos x + a a cos π + a + Point A is π, b b cosπ/ + b + Point B is π/, + At B, sinπ/ + / V dv dt V dv dt V t c V t + a V + 9 V cm b t + t + t t

14 Exercise 7F Solutions to A.J. Sadler s Exercise 7F.... x [ x ] [x] x [ ] x x [ x x ]. It is not necessary to do any work for this. Any integral with equal upper and lower bounds is equal to zero x [ x + x ] x [ ] x x 9. u x + du x u x [ x ]. du..... π π π π π π π π x + 9 u [ u ] 9 sin x [ cos x] π cos x [sin x] π sin x π π. u du. sin x π π [ sin x. sin π... π cos x ] π x π π sin π + π + π π π cos x [ sin x ] π π sin π sin π [ cos x sin x cos x. x x so a b c ] π π cos π cos π x [ x ] x [ x ] x [ x ] 7

15 Unit C Specialist Mathematics Exercise 7F. x + x x + x so a x + x [x + x b c ] x + x [x + x ] x + x [x + x ] sin x cos x so a b. a b c π π sin x [ cos x] π cos π os sin x [cos x] π cos os π [ ] x x x x 9. u x + du x u du x x x + u du u [ ] u. Using the same substitutions as the previous question, xx +. u x + du x u du x x + u u u u u du u du [ ] u u + 9. u u u du u u [ u u du ] [ u u ]. u sin x du cos x π sin x cos x u [ u ] u du

16 Exercise 7G Solutions to A.J. Sadler s. u x + du x u du x x + u [ u u u du u u du u u du u [ u u ] [ uu ] ] u x du x u + du x + x u u + u du u + u du [ u + u [ u + u [ ] ] u + u [ ] uu + ] + + Exercise 7G. Area is always positive, but b a fx is signed, so they will only be equal where i.e. a, b, e, and f fx x [a, b]. Since x > for x [, ], the area is given by A x [ x ] 7 units. y x crosses the x axis at x, so a A x ] [x x + units b A x + x ] + [x x units. a Area [x ] units b Area [x ] + [x ] units. y sin x is positive between x and x π so a Area[ cos x] π + units b Area[ cos x] π + units. x intercepts are at x and x so we must find the area in two pieces: x [, ] and

17 Unit C Specialist Mathematics Exercise 7G x [, ]. x x x x x. x A y + y [. x ] + [. x ] unit 7. x intercepts are at x.. a b A x ] [x x + + units x [ x ] x x c Roots are given by x x x x. x Thus the area is to be found in two parts:. A x [ ] x. x A x. [ ] x x. + + A A + A. units 9. We know that y sin x is non-negative for all x. Referring to earlier work e.g. 7B question to integrate sin x we get A π sin x ] π [ x sin x π + π units. We want c fx gx, so this immediately a includes both e and g. Between a and b fx gx and between b and c gx fx so both parts of b are positive so we include that. Both d and f will give us the difference between the area from a to b and that from b to c and c gives us the absolute value of this difference, so these are wrong. Part a deals with areas between one or other curve and the x axis so this is entirely wrong. 7

18 Exercise 7G Solutions to A.J. Sadler s. A π π π sin x sin x sin x sin x [ cos x] π + unit. First find where the curves intersect: x + x + x x x x + x or x Decide which curve is greater between x and x by choosing any convenient point. At x, x + > x +. Thus the area is given by A x + x + [ x x + x ] units. First determine the points of intersection at x {,, }. A x x + x x + x x x + x + [ ] x x + x + x A x x + x x + x [ ] x x + x + x.. A A + A. units. First determine the points of intersection: x, i.e. x A x [ ] x x frac +. a A units π sin x x π ] π [ cos x x π cos π π π cos π 7π + π π + units b It should be clear from your graph that the line and curve enclose only two regions. From the symmetry of the curve and line it should also be clear that these two regions have equal area. Thus the total area enclosed is A π + π + units. a By the null factor theorem points A, B and C must be where either cos x or sin x, i.e. A π,, B π, and C π,. b cos x sin x sin x

19 Unit C Specialist Mathematics Exercise 7G and A π [ sin x ] π A π π cos x sin x [ sin x ] π π cos x sin x total A A + A units 7. a + +..m b A.m. x x + x +. [.x x +.x +.x ] m. a First consider the point, from the persepctive of the curved edge: y ax a a. The gradient at, is m ax.. The equation of the line is b The area can be found by considering the curved section and straight section separately: A A [.x.x ].x [.x x ] A A + A m 9. First find the points of intersection between the curve and the line. x x x x x x + x x x or x Now find the x values where the curve meets the water level. Now d: y.x. c d. m x x x x x or x which values could have been deduced from the symmetry of the problem. 9

20 Exercise 7G Solutions to A.J. Sadler s Now find the area in three parts: x x A ] [x x + x x x A [ ] x x 7 x A A by symmetry 9. A total A + A + A m to the nearest square metre. The coordinates of point B in the cross-section as shown are,. This gives the coordinate of the curve as y ax a a y x From this the area of region ABC is A x ] [x x cm Hence the total area of the cross section is cm and the capacity is V cm L or.m. A [ x m + x x + x x + x x + x x x x x ]. The first two x intercepts can be easily determined to be where x and x. For the third, Thus the area is π sin x π x π x x.7 A sinπx [ ] cosπx π cos π cos π π.7 A π sin x [ ].7 cos π x π cos π cos π π π A total A + A π units

21 Unit C Specialist Mathematics Miscellaneous Exercise 7 Miscellaneous Exercise x + x + sin x + cos x sin x cos xx sin x x sin xx cos x sin x x cos x os x + sin x sin x os x cos x + cos x + sin x + sin x os x cos x + sin x + cos x + sin x os x cos x + sin x + os x cos x + sin x sin x cos x + sin x cos x + sin x cos x sin x cos x + sin x cos x + sin x. y + x + y x x + y x y x + y x y x y x + y 7. dt t dt t dt dt t t. cos y + x sin y sin x + y cos x cos y y cos x sin x + x sin y cos y ycosx sin x + x sin y 9. Given sin A, we can first deduce cos A ± depending on whether A is in the first or second quadrant. a sin A sin A cos A ± ± b cos A sin A ± 7 sin A c tan A cos A ± 7. a x + y + x y x + y y x x + y y x +. a 7 y y mx x y 7 x y 7x 7x y b x + y b sin x x y 9 y y mx x y x 9 9y x 9y 7 x + x + 9y + x c Let u x +. sin x x u du du

22 Miscellaneous Exercise 7 d e f x x + u u du u + 9u du u u du u u 7 7 u 9 7 u 7 u x + 77 u 77 x + x + x sin x sin x sin x. sin x. sin x. cos x.sin x x x sin x cos x cos x x cos cos x sin x cos x cos x x sin sin x x sin sin x sin x. a Change the sign of the argument: z cis π Solutions to A.J. Sadler s e When dividing complex numbers in polar form, subtract the arguments and divide the moduli: π cis π cis π. L.H.S.: sin θ cos θ + cos θ sin θ sin θ + cos θ sin θ cos θ sin θ + cos θ os θ sin θ cos θ cos θ sin θ cos θ cos θ sin θ cos θ sin θ R.H.S.. L.H.S. d x x + x + h x + h + x x + lim h h x + xh + h x h + x + x lim h h xh + h h lim h h lim x + h h x R.H.S.. To prove: for z a non-zero complex number, z z z Proof: Let z a + bi for a, b R. Then z a bi and z a + b. b A complex number plus its conjugate gives double the real component: cos π cis c A complex number minus its conjugate gives double the imaginary part: i sin π i cis π d When multiplying complex numbers in polar form, add the arguments and multiply the moduli: π cis + π cis L.H.S. z a + bi a + bi a bi a bi a bi a b i a bi a + b z z R.H.S.

23 Unit C Specialist Mathematics Miscellaneous Exercise 7. y cos x is non-negative between x and x π so the area is the simple integral A π π π π cos x cos xcos x cos x sin x cos x cos x sin x [ sin x sin x ] π units 7. O E AB AO + OB D A b a AC. AB F.b a OC OA + AC a +.b a.a + b a DF DO + OF ha + m OC ha +.ma + b.m ha +.mb b FE FO + OE m OC + kb c.ma + b + kb.ma + k.mb DF FE.m ha +.mb.ma + k.mb m ha k mb m h m h and k m k m h k m C B d. a DE DO + OE ha + kb hb a h AB DE AB dt t dt t dt dt t t t t d y b t t t t t t t t t t t t t t t t t t t 9. a For the first statement: L.H.S. cosa B osa + B cos A cos B + sin A sin B + cos A cos B sin A sin B cos A cos B R.H.S. Similarly the second statement: L.H.S. cosa B cosa + B cos A cos B + sin A sin B cos A cos B sin A sin B sin A sin B R.H.S. b From the preceding we can draw the following simplifications: and sin x sin x cosx x cosx + x cos x cos x cos 7x cos x cos7x x From this, os7x + x cos x os 9x

24 Miscellaneous Exercise 7. a b i. ii. π π sin x sin x cos x cos x sin x sin x cos 7x cos x cos x os 9x sin x + sin 9x. sin x [.x os x] π π os π os π π. sin x π + π If you had to do this without a calculator, you should be able to estimate the value of π about or and more than and less than sufficiently to be able to be confident that π <. c. sin x crosses the x axis where sin x., i.e. at x π. We need to find the area in two parts: A π. sin x [.x os x] π π os π os π + π + A π π. sin x [.x os x] π π π os π π + π π π A total A + A π + π π units Solutions to A.J. Sadler s + π. John s conjecture is easily disproven by finding a single counter-example e.g. x or x. However even these give a result that is a multiple of six. Conjecture: x x is a multiple of for x. Proof: Consider the factorisation x x xx x + At least one of these factors must be even. If x is not even, then both x and x + are even. Similarly, and one factor must be a multiple of. If x has a remainder of when divided by, then x is a multiple of. If x has a remainder of when divided by, then x + is a multiple of. If x has no remainder when divided by, it is itself a multiple of. There are no other possibilities. Since x x is has both a factor that is even and a factor that is a multiple of, it must be a multiple of.. Let A be the given position of the shuttle, O be the position of the satellite and P be the position of the shuttle at closest approach t seconds later. OP AP OA + AP AP 9 + t 7 OP r + v r + tv v + t 9m to the nearest m t s Let B be the position of the shuttle at t. Let w be the new velocity. OB r + v m 7 w. 7 ms 9

25 Unit C Specialist Mathematics Miscellaneous Exercise 7. Starting with the first expression stripped of the +c: sin x sin x x + + x x x x x + sin x cos x + sin x cos x + sin x cos x sin x cos x + sin x cos x + sin x cos x + sin x cos x + sin x cos x + + sin x cos x cos x + sin x cos x sin x cos x + sin x cos x sin x cos x + x sin x cos x