Rates of change Examples 1a) A circular stain gets larger as time goes on. Its radius r cm at time t s is given by r = 0.1t 2. Find the rate of change of r with respect to t when t = 0.5s and when t = 1.5s b) If the area of the stain at time t is A sq cm, express A in terms of t. Find the rate of change of A w.r.t when t = 0.5s and when t =1.5s.

1. Answer a) r = 0.1t 2 dr/dt = 0.2t when t = 0.5, dr/dt = 0.2 x 0.5 = 0.1 cm/s when t = 1.5, dr/dt= 0.2 x 1.5 = 0.3 cm/s This tells us how quickly the radius of the stain is growing at these particular instants

b) Area =  r 2 =   (0.1 t 2 ) 2 A = 0.01  t 4 dA/dt= 0.04  t 3 when t = 0.5, dA/dt =  cm 2 /s = cm 2 /s when t = 1.5, dA/dt =  cm 2 /s = cm 2 /s This tells us how quickly the area of the stain is growing at the particular instants

2. A water tank is being filled so that the volume of water in the tank at time t secs is V m 3 where V = 0.15t t. Calculate the rate at which the tank is filling when t = 6.

2. answer V = 0.15t t We want : dv/dt = 0.03t When t = 6, dv/dt = 1.96 m 3 /s

3. The distance s metres moved by an object in a time t seconds is given by s = 2t 3 + 3t 2 – 6t + 2 Find the velocity and acceleration when t = 4

3. answer s = 2t 3 + 3t 2 – 6t + 2 v= ds/dt = 6t 2 +6t - 6 a = dv/dt = 12t + 6 when t = 4, v = 6   4 – 6 = 114 m/s a = 12  = 54 m/s

4.If the distance s metres travelled by a new car in time t seconds after the brakes are applied is given by a) What is the speed (in km/hr) at the instant the brakes are applied? b) How far does the car travel before it stops?

a) v = ds/dt = 15-(10/3)t When brakes are applied, t = 0 therefore v = 15 m/s To change to km/hr, v =15 m/s = 15/1000  60  60 km/hr = 54 km/hr

b)The car stops when v = 0, so v = ds/dt = 15 - (10/3)t =0 10t = 45t = 4.5 secs Distance travelled in 4.5 secs is s = 15  4.5 – 5/3  = m Notice that acceleration a = dv/dt = -10/3 < 0 ie the car is slowing down