Rates of change Examples 1 a A circular

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Rates of change Examples 1 a) A circular stain gets larger as time goes

Rates of change Examples 1 a) A circular stain gets larger as time goes on. Its radius r cm at time t s is given by r = 0. 1 t 2. Find the rate of change of r with respect to t when t = 0. 5 s and when t = 1. 5 s 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. 5 s and when t =1. 5 s.

1. Answer a) r = 0. 1 t 2 dr/dt = 0. 2 t

1. Answer a) r = 0. 1 t 2 dr/dt = 0. 2 t 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

b) Area = r 2 = (0. 1 t 2)2 A = 0. 01 t 4 d. A/dt= 0. 04 t 3 when t = 0. 5, d. A/dt = 0. 005 cm 2/s = 0. 0156 cm 2/s when t = 1. 5, d. A/dt = 0. 135 cm 2/s = 0. 424 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

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. 15 t 2 + 0. 16 t. Calculate the rate at which the tank is filling when t = 6.

2. answer V = 0. 15 t 2 + 0. 16 t We want

2. answer V = 0. 15 t 2 + 0. 16 t We want : dv/dt = 0. 03 t + 0. 16 When t = 6, dv/dt = 1. 96 m 3/s

3. The distance s metres moved by an object in a time t seconds

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

3. answer s = 2 t 3 + 3 t 2 – 6 t

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

4. If the distance s metres travelled by a new car in time t

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

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 km/hr = 54 km/hr

b)The car stops when v = 0, so v = ds/dt = 15 -

b)The car stops when v = 0, so v = ds/dt = 15 - (10/3)t =0 10 t = 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