Question

Consider the function as representing the value of an ounce of palladium in U.S. dollars as a function of the time t in days.

R(t) = 220 + 40t³; t = 1

Find the average rate of change of R(t) over the time intervals [t, t + h], where t is as indicated and h = 1, 0.1, and 0.01 days. (Use smaller values of h to check your estimates.)

h	1	0.1	0.01
average rate

Estimate the instantaneous rate of change of R at time t, specifying the units of measurement.

R'(1)= $ per ___

Answer 1

R(t) = 220 + 40t³

R(1) = 220 + 40(1)³ = 260

R(t + h) = 220 + 40(t + h)³ = 220 + 40t³ + 40h³ + 120t²h + 120th²

R(1 + h) = 220 + 40(1)³ + 40h³ + 120(1)²h + 120(1)h² = 260 + 40h³ + 120h + 120h²

For h = 1

Substituting h = 1 in R(1 + h) we get

R(1 + 1) =  260 + 40(1)³ + 120(1) + 120(1)² = 540

=> R(2) = 540

R(1) = 260

Therefore,

Average Rate = (R(2) - R(1)) / (2 - 1) = 540 - 260 = 280

For h = 0.1

Substituting h = 0.1 in R(1 + h) we get

R(1 + 0.1) =  260 + 40(0.1)³ + 120(0.1) + 120(0.1)² = 260 + 0.04 + 12 + 1.2 = 273.24

=> R(1.1) = 273.24

R(1) = 260

Therefore,

Average Rate = (R(1.1) - R(1)) / (1.1 - 1) = (273.24 - 260) / 0.1 = 132.4

For h = 0.01

Substituting h = 0.01 in R(1 + h) we get

R(1 + 0.01) =  260 + 40(0.01)³ + 120(0.01) + 120(0.01)² = 260 + 0.004 + 1.2 + 0.12 = 261.324

=> R(1.01) = 261.324

R(1) = 260

Therefore,

Average Rate = (R(1.01) - R(1)) / (1.01 - 1) = (261.324 - 260) / 0.01 = 132.4

h	1	0.1	0.01
Average Rate	280	132.4	132.4

R(1) = 260

R(1 + h) = 260 + 40h³ + 120h + 120h²

R(1 + h) - R(1) = 40h³ + 120h + 120h²

(R(1 + h) - R(1)) / h = (40h³ + 120h + 120h²) / h = 40h² + 120h + 120

R ' (1) = lim_{h -> 0} (R(1 + h) - R(1)) / h = lim_{h -> 0} (40h² + 120h + 120) = 120

Therefore,

R ' (1) = $120 per day