Question

With an initial deposit of $100, and and interest rate of 12%, the balance in a bank account after t years is f(t) = 100(1.12)^t dollars.

(a) Find the average rate of change over [0, 0.5] and [0, 1]. (Round your answers to two decimal places.)

(c) Estimate the instantaneous rate of change at t = 0.5 by computing the average rate of change over intervals to the left and right of t = 0.5. (Round your answers to four decimal places.)

Interval	[0.5,0.51]	[0.5,0.501]	[0.5,0.5001]
Average Rate of Change	?	?	?
Interval	[0.49,0.5]	[0.499,0.5]	[0.4999,0.5]
Average Rate of Change	?	?	?

Answer 1

(a) The average rate of change (ROC) of f(t) = 100(1.12)^t over the time interval [t₁, t₂] is given by

   ∆f/∆t = f(t₂) - f(t₁) / t₂ - t₁

Time interval [0, 0.5] ; Avg ROC = f(0.5) - f(0) / (0.5 - 0) = (103.923 - 100)/0.5 =11.66 = 11.66 $/yr

Time interval [0, 1] ; Avg ROC = f(1) - f(0) / (1 - 0) = (112 - 100)/1 = 12$/yr

(b)

Time interval [0.5, 0.51] ;

Avg ROC = f(0.51) - f(0.5) / (0.51 - 0.5) = (105.95 - 105.83)/0.01 = 0.12/0.01 = 12.00038 $/yr

Time interval [0.5, 0.501] ;

Avg ROC = f(0.501) - f(0.5) / (0.501 - 0.5) = (105.842 - 105.83)/0.001 = 0.011994/0.001 = 11.99426 $/yr

Time interval [0.5, 0.5001] ;

Avg ROC = f(0.5001) - f(0.5) / (0.5001 - 0.5) =11.99365 $/yr

Time interval [0.49, 0.5] ;

Avg ROC = f(0.5) - f(0.49) / (0.5 - 0.49) = 0.11986787/0.01 =11.9867 $/yr

Time interval [0.499, 0.5] ;

Avg ROC = f(0.5) - f(0.499) / (0.5 - 0.499) = 0.0119929/0.001 = 11.9929 $/yr

Time interval [0.4999, 0.5] ;

Avg ROC = f(0.5) - f(0.4999) / (0.5 - 0.4999) =11.99351275 $/yr

The rate of change at t = 0.5 is approximately 12$/yr.