(Got a little tricky Calculus question on my hands)

Historical data shows that the amount of money sent out of Canada for interest and dividend payments during the period from 1967 to 1979 can be approximated by the model P = (5 x 10^8)e^0.20015t, where t is measured in years (t = 0 in 1967) and P is the total payment in Canadian dollars.

a) Determine and compare the rates of increase for the years 1968 and 1978.

b) Assuming this trend continues, compare the rate of increase for 1988 with the rate of increase for 1998.

(Calculus - Transcendental Functions)

Question

(Got a little tricky Calculus question on my hands)

Historical data shows that the amount of money sent out of Canada for interest and dividend payments during the period from 1967 to 1979 can be approximated by the model P = (5 x 10^8)e^0.20015t, where t is measured in years (t = 0 in 1967) and P is the total payment in Canadian dollars.

a) Determine and compare the rates of increase for the years 1968 and 1978.

b) Assuming this trend continues, compare the rate of increase for 1988 with the rate of increase for 1998.

(Calculus - Transcendental Functions)

anonymous · Answer

$$P=5\times10^8\times e^{0.20015t}$$
rate of increase is the first derivative 
so, rate of increase of payment = $${dP\over dt}=?$$