Methods of Integration

U96-b)! Use the substitution u = - to evaluate U95-b)! 4 Methods of Integration d. Evaluate 9 d using the substitution u = + 9. UNIT MATHEMATICS (HSC) METHODS OF INTEGRATION CSSA «8» U94-b)! Use the substitution u = log e to evaluate e (log e ) d. U9-b)! 5 Evaluate 9 d using the substitution u = - 9. U9-b)! U9-b)! U89-a)! Use the substitution u = -, to evaluate d. Evaluate the definite integral 6 «5» d b means of one of the substitutions u = 4 or = sin. 4 Find the value of d, b means of the substitution u = +. «-» «5» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) METHODS OF INTEGRATION CSSA U87-a)! tan Using the substitution u = tan, show that tan.sec d C. Hence evaluate 4 tan sec d. U86-iii)! U86-iv)! Use the substitution u = - 4 to find an epression for Evaluate U85-i)! Evaluate a. sin cos d. 4 d using the substitution u = 4 -. b. d using the substitution = sin. d. 4 «4 C» «a) b) 4» U84-i)! Evaluate 9 d using the substitution = u. «Ln» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

U95-5a)! Find sin d. Primitive of sin and cos UNIT MATHEMATICS (HSC) PRIMITIVE OF SIN X AND COS X CSSA «sin 4 C» 8 CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD CSSA Equation dn dt k (N - P) Velocit and Acceleration as a Function of Projectile Motion Simple Harmonic Motion U96-4a)! N is the number of animals in a certain population at time t ears. The population size satisfies the equation dn k( N ), for some constant k. dt i. Verif b differentiation that N = + Ae -kt, A constant, is a solution of the equation. ii. Initiall there are 5 animals but after ears there are onl left. Find the values of A and k. iii. Find when the number of animals has fallen to. iv. Sketch the graph of the population size against time. «i) Proof ii) A = 5, k = 5 Ln t iii) 4.4 ears (to d.p.) iv) 4 U96-6b)! A particle moving in a straight line is performing Simple Harmonic Motion about a fied point O on the line. At time t seconds the displacement metres of the particle from O is given b: = a cos nt, where a > and < n <. After second the particle is metre to the right of O, and after seconds the particle is metre to the left of O. i. Find the values of n and a. ii. Find the amplitude and period of the motion. N 5 «i) n =, a = ii) amplitude = metres, period = 6 seconds»» U96-7b)! A B C h V O In the diagram an aircraft is fling with constant velocit U at a constant height h above horizontal ground. When the plane is at A it is directl over a gun at O. When the plane is at B a shell is fired from the gun at the aircraft along OB. The shell is fired with initial velocit V at an angle of elevation. i. If and are the horizontal and vertical displacements of the shell from O at time t seconds, show that if g is the acceleration due to gravit, = Vt cos and = Vt sin - gt. CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD CSSA h ii. Show that if the shell hits the aircraft at time T at point C, then VTcos = UT tan. iii. Show that if the shell hits the aircraft then U(Vcos - U)tan = gh. «Proof» U95-c)! A particle moves in a straight line so that its displacement metres from an origin O at time t seconds is given b = sin t. i. Show that d. dt 4 ii. State the amplitude and the period of the motion. iii. Find the maimum speed of the particle. «i) Proof ii) amplitude = m, period = 4s iii) 5 ms -» U95-5b)! At time t the temperature T of a bod in a room of constant temperature is decreasing according to the equation dt k( T ) for some constant k >. dt i. Verif that T = + Ae -kt, A constant, is a solution of the equation. ii. The initial temperature of the bod is 9 and it falls to 7 after minutes. Find the temperature of the bod after a further 5 minutes. «i) Proof ii) 6 (to nearest degree)» U95-7)! V R O A stone is projected from O with velocit V at an angle above the horizontal. A straight road goes through O at an angle above the horizontal, where <. The stone strikes the road at R. Air resistance is to be ignored, and the acceleration due to gravit is g. i. If the stone is at the point (, ) at time t, find epressions for and in terms of t. Hence show that the equation of the path of the stone is = tan - g sec. V ii. If R is the point (X,Y), epress X and Y in terms of OR and. Hence show that the range OR of the stone up the road is given b OR = V cos sin( ). gcos iii. Show that OR is a maimum when =, and interpret this result geometricall. CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

iv. Hence show that the maimum value of OR is «i) = Vt cos, = Vt sin - gt UNIT MATHEMATICS (HSC) APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD CSSA V g( sin ). ii) X = R cos, Y = R sin iii) Proof. For maimum range, the angle of projection of the stone bisects the angle + iv) Proof» U94-5a)! A bod is moving in a straight line. At time t seconds its displacement is metres from a fied point O on the line and its velocit is v ms -. If v = find its acceleration when =.5. «-8 ms -» U9-c)! A particle is moving in a straight line with Simple Harmonic Motion. If the amplitude of the motion is 4cm and the period of the motion is seconds, calculate: i. the maimum velocit of the particle; ii. the maimum acceleration of the particle; iii. the speed of the particle when it is cm from the centre of the motion. «i) 8 cms ii) 6 cms iii) 4 cms» 9 U9-c)! i. At an time t the rate of cooling of the temperature T of a bod, when the surrounding temperature is P, is given b the equation dt k( T P), for some constant k. Show that dt the solution T = P + Ae -kt, for some constant A, satisfies this equation. ii. A metal bar has a temperature of 4 C and cools to C in minutes, when the surrounding temperature is 5 C. Find how much longer it will take the bar to cool to 6 C, giving our answer correct to the nearest minute. «i) Proof ii) 9 minutes» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD CSSA U9-7b)! A particle moves in a straight line. At time t its displacement from a fied point O on the line is, its velocit is v and its acceleration is a. i. Show that a = d v. d ii. If a = 4-4 and when t =, = 6 and v = 8 show that v = 4-8 -. iii. Use the epression for v to find the set of possible values of. iv. Describe the motion of the particle in each of the cases. when t =, = 6 and v = 8.. when t =, = 6 and v = -8. «i) ii) Proof iii) 4 iv) ) The particle starts 6 units to the right of O. It accelerates to the right. ) The particle starts 6 units to the right of O. It moves to the left, slows to a stop 4 units to the right of O, the accelerates to the right.» U9-5a)! i. A ball is thrown from a point O on the edge of a cliff which is metres above a beach. The ball is thrown with speed 5 ms - at an angle of 45 above the horizontal. Taking ii. g = ms - show that the ball hits the beach at a point 6 metres along the beach. A second ball is thrown horizontall from and hits the beach at the same point as the first ball. Taking g = ms - find the speed of projection of the second ball. (Standard results about projectile motion can be quoted without proof.) «i) Proof ii) 6 ms -» U9-4a)! O is a fied point on a given straight line. A particle moves along this line and its displacement cms, from O at a given time, t secs, after its start of motion is given b: = + cos t. i. Show that the acceleration is given b: = - 4. ii. iii. iv. State the centre of motion. State the first two occasions when the particle is at rest and the displacements on these occasions. State the amplitude and period of motion. «i) Proof ii) = 5 iii) t =, = and t =, = iv) Amplitude = cm, Period = secs» U9-5a)! A stone is thrown from a point O which is at the top of a cliff metres above a horizontal beach. The stone is thrown at an angle of elevation above the horizontal and with a speed of 5ms -. The stone hits the beach at a point which is distant 4 metres horizontall from the point O. i. Taking g, the gravitational constant, as ms -, show that tan = 4 or tan =. ii. Hence find the two possible times for which the stone is in the air, giving our answers in eact form. «i) Proof ii) 4 seconds and 5 seconds» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD CSSA U9-5d)! A certain particle moves along the -ais in accordance with the law t = - 5 + where is measured in cm and t in seconds. Initiall, the particle is.5 cm to the right of O and moving awa from O. i. Prove that the velocit, v cm/sec, is given b v = 4 5. ii. Find an epression for the acceleration, a cm/sec, in terms of. iii. Find the velocit and acceleration of the particle when:. = cm.. t = 6 sec. iv. Describe carefull in words the motion of the particle. 4 «i) Proof ii) iii) ) v = ( 4 5) cms-, a = 4 7 cms- ) v = 7 cms-, a = 4 4 cms- iv) The particle moves in the postive direction with a negative acceleration retarding its motion.» U89-b)! A particle moves in a straight line and at time t seconds, its distance cm from a fied origin point O, on the line is given b: = + cost. i. Sketch a graph of as a function of t in the domain t ii. Show that the motion of the particle is Simple Harmonic Motion. iii. State the centre of motion of the particle. iv. Find the displacements of the particle when it is at rest and thus determine the length of its path. v. State the period of motion for the particle. CSSA OF NSW 984-996 EDUDATA: DATAVER. 996 «i) ii) Proof iii) = iv) =,,,,... Length of path = cm v) secs» U88-5)! A particle is projected from a point, O, on ground level with the velocit of metres per second at an angle of 6 to the horizontal. After a time T seconds, it reaches a point P, on its upward path, where the direction of the flight is at to the horizontal. Taking the acceleration due to gravit, g, to be m/s, i. show that T =. ii. find the height of P above ground level. iii. find the greatest height reached b the particle. «i) Proof ii) 4 m iii) 5m» U87-7b)! A particle moves in a straight line and its displacement, cm, from a fied origin point after t seconds is determined b the function: = sin t - sin t cos t - t. i. Find the initial displacement and velocit of the particle. t

UNIT MATHEMATICS (HSC) APPLICATIONS OF CALCULUS TO THE PHYSICAL WORLD CSSA ii. Show that the particle never comes to rest and alwas moves in one particular direction, stating what this direction is. iii. Show that the particle initiall has zero acceleration and find the first occasion after this when zero acceleration occurs again. «i) = cm, v = - cms - ii) Proof, negative direction iii). seconds» U86-5iii)! The speed v centimetres/second of a particle moving with simple harmonic motion in a straight line is given b v = 6 + 4 -, where cm is the magnitude of the displacement from a fied point O. a. Show that = -( - ). b. Find the centre of the motion. c. Find the period of the motion. d. Find the amplitude of the motion. «a) Proof b) = c) secs d) cm» U85-5ii)! A bod is moving with simple harmonic motion in a straight line. It has an amplitude of metres and a period of seconds. How long would it take for the bod to travel from one of the etremities of its path of motion to a point 4 metres awa? «.5 seconds» U85-6ii)! C A 8m m B m D AB and CD are two buildings situated metres apart on level ground. Their heights are 8m and m respectivel. An object is projected from point A at an angle of 45 to the horizontal, and this object strikes point C. Take the acceleration due to gravit, g, as m/sec. Show that the time taken for the object to get from A to C is 4 seconds, and find the value of the initial velocit of projection. «Proof, 5 ms -» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS CSSA Inverse Functions and Inverse Trigonometric Functions U96-c)! i. Find the value of such that sin - = cos -. ii. On the same aes sketch the graphs of = sin - and = cos -. iii. On the same diagram as the graphs in (ii), draw the graph of = sin - + cos -. (-, ) = cos - U96-a)! «i) = ii) iii),, = sin - + cos - = sin - O,,, 4 (, ) 8 f( ) 4 i. Show that f is an even function, and the ais is a horizontal asmptote to the curve = f(). ii. Find the co-ordinates and nature of the stationar point on the curve = f(). iii. Sketch the graph of the curve showing the above features. iv. Find the eact area of the region in the first quadrant bounded b the curve = f() and the line =. (, )» «i) Proof ii) (, ) is a maimum turning point iii) iv) units» U95-b)! i. If f() = e +, find the inverse function f - (). ii. State the domain and range of f - (). iii. On one diagram sketch the graphs of f() and f - (). = e + = Ln - «i) f - () = Ln - ii) Domain: >, Range: All real iii)» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

U95-4b)! UNIT MATHEMATICS (HSC) INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS CSSA L P L O The diagram shows the graph of the function f() = 4. i. Find the equations of the asmptotes L and L. ii. B comparing the values of f(-) and f() show that f is an even function. What is the geometrical significance of this result? iii. Find the eact equation of the tangent to the curve at the point P where =. iv. Find the eact area of the shaded region. «i) = -, = ii) The function is smmetrical about the -ais iii) = - iv) units» U94-c)! The diagram below shows the graph of = + sin -. A B C i. Write down the co-ordinates of the endpoints A and C. ii. Write down the co-ordinates of the point B. iii. Find the equation of the tangent to the curve = + sin - at the point B. U9-a)! Find U9-a)! 9 d. «i) A(, ), C(, ) ii) B(, ) iii) 6 - + =» «sin - + C» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

Given that UNIT MATHEMATICS (HSC) INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS CSSA d k, find the value of the constant k. «8» U9-c)! i. Sketch the graph of the function = sin -. ii. State the domain and the range of the function. iii. Find the eact equation of the tangent to the curve = sin - at the point where =., «i) U9-a)!, ii) Domain: -, Range: iii) - 6 + - =» Find the eact value of sin(tan - ). «4 5» U9-a)! 6 d d Show that 6. 4 4 «Proof» U9-a)! d Evaluate, leaving the answer in eact form. 9 U9-e)! Find the value of the epression cos sin in terms of. «5» 6 CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS CSSA U9-d)! i. Show that the function f() = 4 ( ) is an increasing function for all values of in its domain. ii. Sketch the graph of the function, showing clearl the co-ordinates of an points of intersection with the -ais and the -ais, and also the equations of an asmptotes. iii. Find the inverse function, f - (), and state its range. = 4 «i) Proof ii) U9-5c)! Use the substitution u = = to evaluate ii) f - () = 4, Range: All real, ecept =» d, giving the answer in eact form. ( ) «6» U89-a)! Evaluate 5. d, leaving our answer in eact form. 9 U89-b)! State the domain and range for the function: = cos - (). 8 «Domain:, Range:» U89-a)! If f() = cos - - sin - ( - ), find f (). Hence, or otherwise, show that: cos - - sin - ( - ) =. «f () =» U88-b)! State the range and domain of the function = sin - ( ) and draw a sketch of the function, carefull labelling the etremities of both the range and the domain. CSSA OF NSW 984-996 EDUDATA: DATAVER. 996

UNIT MATHEMATICS (HSC) INVERSE FUNCTIONS AND INVERSE TRIGONOMETRIC FUNCTIONS CSSA U88-c)! Evaluate «Range: -, Domain: - 6 cos 4sin d using the substitution u = sin.» «4» U87-c)! Show that the two curves = cos - and = tan - ( - ) cut the -ais at the same point and have a common tangent at this point. «Proof» U86- ii)! Given that = sin -, show that d. d sin «Proof» U85-4iii)! sin if A function is defined b the rules f() =. cos if a. Sketch the function for - b. Evaluate f(- ) + f() - f( ). - «a) U84-ii)! a. Prove that sin - + cos - =. b. Find the eact values of and which satisf the simultaneous equations sin - - cos - = ; cos- + sin 5. «a) Proof b) = b)», =» CSSA OF NSW 984-996 EDUDATA: DATAVER. 996