Rates of change

1. Rates of change Examples a) A circular stain gets larger as time goes on. Its radius r cm at time t s is given by r = 0.1t2. 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.

2. 1. Answer • r = 0.1t2 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

3. b) Area = r2 =  (0.1 t2)2 A = 0.01 t4 dA/dt= 0.04 t3 when t = 0.5, dA/dt = 0.005  cm2/s = 0.0156 cm2/s when t = 1.5, dA/dt = 0.135  cm2/s = 0.424 cm2/s This tells us how quickly the area of the stain is growing at the particular instants

4. 2. A water tank is being filled so that the volume of water in the tank at time t secs is V m3 where V = 0.15t2 + 0.16t. Calculate the rate at which the tank is filling when t = 6.

5. 2. answer V = 0.15t2 + 0.16t We want : dv/dt = 0.03t + 0.16 When t = 6, dv/dt = 1.96 m3/s

6. 3. The distance s metres moved by an object in a time t seconds is given by s = 2t3 + 3t2 – 6t + 2 Find the velocity and acceleration when t = 4

7. 3. answer s = 2t3 + 3t2 – 6t + 2 v= ds/dt = 6t2 +6t - 6 a = dv/dt = 12t + 6 when t = 4, v = 6  42 + 6  4 – 6 = 114 m/s a = 12  4 + 6 = 54 m/s

8. 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?

9. 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

10. b)The car stops when v = 0, so v = ds/dt = 15 - (10/3)t =0 10t = 45 t = 4.5 secs Distance travelled in 4.5 secs is s = 15  4.5 – 5/3  4.52 = 33.75 m Notice that acceleration a = dv/dt = -10/3 < 0 ie the car is slowing down