Increase or Decrease at a Constant Rate

Increase at a Constant Rate A fish is now 8 cm long. The length of the fish increases at a rate of 10% per month. How long will it be after n months? Increases by 10% After the 1st month Increases by 10% After the 2nd month … 8 cm This question involves successive percentage changes at a constant rate.

Increase at a Constant Rate Increases by 10% After the 1st month Increases by 10% After the 2nd month … 8 cm Length after the 1st month 8 cm 10%) (1 + × = Length after the 2nd month = 8 10%) (1 + × cm 10%) (1 + × 8 10%)2 cm (1 + × = Length after the 3rd month = + 8 10%)2 (1 × 10%) cm (1 + × 8 10%)3 cm (1 + × = … ∴ Length after the n th month 8 10%)n cm (1 + × =

If a value P increases at a constant rate r % in each period, the new value A after n periods can be calculated by the following formula: A = P  (1 + r %)n (1 + r %) is called the growth factor, which must be greater than 1.

The number of cells after 3 hours = 1000  (1 + 20%)3 For example, there are 1000 cells at the beginning of an experiment and the number of cells grows at a rate of 20% every hour. The number of cells after 3 hours = 1000  (1 + 20%)3 Substitute P = 1000, r = 20 and n = 3 Into the formula A = P  (1 + r %)n. = 1000  1.23 = 1728

Follow-up question 1. Amy is now 100 cm tall. If her height increases at a rate of 5% per year, find Amy’s height after 2 years. Solution Amy’s height after 2 years = 100  (1 + 5%)2 cm = 100  1.052 cm = 110.25 cm

Follow-up question (cont’d) 2. The volume of a certain gas in a room is 40 mL. If the volume increases at a rate of 6% per second, find the volume of the gas in the room after 5 seconds. (Give your answer correct to 2 decimal places.) Solution The volume of the gas in the room after 5 seconds = 40  (1 + 6%)5 mL = 40  1.065 mL = 53.53 mL (cor. to 2 d.p.)

Decrease at a Constant Rate Similarly, if a value P decreases at a constant rate r % in each period, the new value A after n periods can be calculated by the following formula: A = P  (1 – r %)n (1 – r %) is called the decay factor, which must be less than 1.

the length of a pencil was 10 cm. The length of For example, the length of a pencil was 10 cm. The length of the pencil was decreased by 3% after each use. The length of the pencil after being used 2 times = 10  (1 – 3%)2 cm Substitute P = 10, r = 3 and n = 2 Into the formula A = P  (1 – r %)n. = 10  0.972 cm = 9.409 cm

and the percentage decrease is called the depreciation rate. In our daily life, the values of machines and electrical appliances decrease after they have been used. Such a drop in value is called depreciation, and the percentage decrease is called the depreciation rate. For example, the value of a washing machine is $6000 now and its depreciation rate is 30% per year. $6000 The value of the washing machine 3 years later 30%) (1 $6000 -  = 3 Depreciation means decrease. 0.7 $6000  = 3 $2058 =

Follow-up question 1. The population of a city is 200 000. If the population is decreased at a rate of 15% per year, find the population of the city after 3 years. Solution The population of the city after 3 years = 200 000  (1 – 15%)3 = 200 000  0.853 = 122 825

Follow-up question (cont’d) 2. The value of a car is $600 000 now and its depreciation rate is 8% per year. Find its value after 5 years. (Give your answer correct to the nearest dollar.) Solution The value of the car after 5 years = $600 000  (1 – 8%)5 = $600 000  0.925 = $395 449 (cor. to the nearest dollar)