HSC Mathematics - Extension 1. Workshop E4

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1 HSC Mathematics - Extension 1 Workshop E4 Presented by Richard D. Kenderdine BSc, GradDipAppSc(IndMaths), SurvCert, MAppStat, GStat School of Mathematics and Applied Statistics University of Wollongong Moss Vale Centre October 2009

2 HSC Mathematics - Extension 1 Workshop E4 Richard D Kenderdine University of Wollongong October Applications of Calculus In these topics we are frequently required to find derivatives with respect to time. Related rates This topic applies the chain rule to practical situations. Suppose we want to find dv and we have V linked to another variable, say x, then we can use dv = dv dx dx (1) 1. Exercises: (a) A conical container with base angle π 3 is being filled with water at the rate of 5 litres/sec. If the base is r metres and the height h metres, find the height at which the rate of change in depth of water in the container is 1.19 metres/sec. (b) A policeman is 2 km due south of a road intersection. He observes a car travelling due west towards the intersection. Find the speed of the car, in kph, if the angle between the car and the intersection is reducing at 0.01 radians/sec when the angle is π 6. (c) A fixed point P lies on the circumference of a circle of radius 20 cm and centre O. Point Q moves around the circumference of the circle such that the angle P ÔQ increases at 0.2 radians/sec. Find the rate of increase in the perimeter of the triangle P OQ when P ÔQ is 50. Extended exponential growth and decay In simple exponential growth/decay the rate of change in a quantity is proportional to the quantity. With extended exponential growth/decay the rate of change is proportional to the difference between the quantity and a constant (eg the temperature of the surrounding air). The constant is the asymptote of the function. 1 rdk301@uow.edu.au 1

3 The basic form of the relationship is for which the function is dt = k(t A) (2) T = A + Ce kt (3) 2. Exercises: (a) A drink is removed from a cooler at 8 and warms to 16 after 30 mins. If the drink is assumed to warm at a rate proportional to the difference between its temperature and the temperature of the surrounding air, which is at 25, find (i) its temperaure after 1 hour (ii) how long it took to reach 20. (b) The population growth curve which often arises in biology is satisfies dn = k(c N(t)) C where N(t) is the population at time t, k is the rate of growth and C is the maximum sustainable population (carrying capacity). (i) Derive the equation for N(t) (4) (ii) If the initial population is 1000, the carrying capacity is 2000 and it took 10 years to reach a population of 1200, find (α) the population after 50 years, and (β) how long it will take to have a population of (c) Consider the function N(t) = be kt (5) (i) Find values for b and k given that N(0) = 20 and N(20) = 50 (ii) Find t such that N(t) = 60 (iii) Show that dn = AN(C N) where A and C are constant and specify the values of the constants. 2

4 Acceleration in terms of displacement We use the relationship ẍ = d dx (1 2 v2 ) (6) when acceleration is defined in terms of displacement rather than time. 3. Exercises: (a) Starting with ẍ = dv, prove Eqn. 6. (b) A particle moves in a straight line according to ẍ = 2 x. Initially it is at x = 1 cm and moving with velocity 2 2 cm/s. (i) Find the equation for velocity in terms of displacement. (ii) Find the equation for displacement in terms of time. (iii) Describe the motion of the particle. (c) A particle moves in a straight line according to ẍ = kẋ 2 where k is a positive constant and ẋ > 0. If the initial speed is V cm/s at x = 0 cm show that (i) ẋ = V 1+kV t (ii) x = 1 k ln(1 + kv t) (iii) ẋ = V e kx (d) A particle moves in a straight line from its initial position x = 1 cm with velocity v = 9 x2 x (i) Find an expression for the particle s acceleration in terms of displacement. (ii) Find an expression for time in terms of displacement and hence an expression for displacement in terms of time. (iii) Sketch a graph of displacement against time. (iv) Find the time to travel from x = 1 cm to x = 2 cm. (e) A particle is moving in a straight line from its initial position x = π 2 cm with velocity given by v = sin2 x. (i) Find an expression for acceleration in terms of displacement. 3

5 (ii) Find an expression for displacement in terms of time. (iii) Sketch a graph of displacement against time. Simple Harmonic Motion SHM is an example of motion where the acceleration can be defined in terms of displacement. We model SHM with the general cos function where, as usual, amplitude = a period = 2π n phase shift parameter = α centering parameter = c x = a cos(nt + α) + c (7) The motion is centered at x = c and α allows the body to have its initial position at any point over the range of motion. 4. Exercise: If α = 0 what is the initial position of the body? We then find velocity ẋ = an sin(nt + α) (8) and acceleration ẍ = an 2 cos(nt + α) = n 2 (x c) (9) If c = 0 we have the usual ẍ = n 2 x If we start with ẍ = n 2 (x c) and use ẍ = d dx ( 1 2 v2 ) we obtain 5. Exercise: v 2 = n 2 [a 2 (x c) 2 ] (10) Show the working to obtain Equation 10. 4

6 SHM has the following properties: (a) speed (magnitude of velocity) is zero at the endpoints and maximum at the centre (b) acceleration is maximum at the endpoints and zero at the centre 6. Exercise: Check these properties of velocity and acceleration when x = c (centre) and x = c + a (endpoint: maximum +ve displacement). Students often have difficulty with SHM. It is useful to remember that you are just working with the general cos function. 7. Exercise: (a) The velocity (v ms 1 ) of a particle moving in SHM along the x-axis is given by v 2 = 5 + 6x x 2 where x is the displacement in metres. (1990 HSC) (i) What are the endpoints of the motion? (ii) What is the centre of motion? (iii) What is the maximum speed? (iv) Find the acceleration in terms of x (v) Find the period of the motion. (b) A particle moves in a straight line with position at time t given by x = 1 + sin 4t + 3 cos 4t (11) (i) Prove that the particle is undergoing SHM with centre x = 1 (ii) Find the amplitude (iii) When does the particle first reach maximum speed after t = 0? (1998 HSC) (c) A particle moves in a straight line with position at time t given by (i) When is the particle first at x = 1 2? x = sin 2 2t (12) (ii) In which direction is the particle travelling at x = 1 2? (iii) Find the acceleration in terms of displacement. 5

7 (iv) State whether the particle is undergoing SHM. (v) Determine the period of motion. (d) A particle moves in a straight line under SHM. Initially it is 3 cm to the right of the origin and travelling with a velocity of 2 cm/sec. Find the maximum velocity and acceleration if the period is 8 sec. Projectile Motion All the results have to be derived from the model (zero horizontal acceleration, vertical acceleration due to gravity) and the initial conditions (velocity, angle of projection and position): ẍ = 0 ÿ = g Initial velocity = V Initial angle of projection = α Initial angle of projection = (x 0, y 0 ) if (x 0, y 0 ) = (0, 0) then the equations for position are 8. Exercise: x = V t cos α y = V t sin α 1 2 gt2 (a) A projectile is thrown from a 20m high vertical cliff with a velocity of 15m/s at an angle of elevation of 20. (i) Derive the equations for velocity and position in terms of time. (ii) Calculate the distance of the point of impact from the base of the cliff and the time of flight. (iii) Calculate the maximum height of the projectile. (iv) Calculate the angle between the path of the projectile and the ground at the point of impact. (v) Determine the cartesian form of the path of the projectile. (b) A particle is projected from the origin with a velocity of 40 m/s and angle of elevation θ. Assuming g = 10 m/s 2 : (i) Determine the cartesian equation of the path of the projectile. (ii) Determine the condition for there to be two values of θ in the range (0, π 2 ) for the projectile to pass through a given point P with coordinates (X, Y ). (iii) If P also lies on the line y = x and α and β are the two values of θ, find the value of α + β. 6