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) = 30t − 3t^2; t = 3
a. 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.)

Answer 1

-0.061 dollars per day.

Explanation:

(R(t + h) - R(t)) / h

We can use this formula to find the average rate of change of R(t) over various time intervals h, starting with t = 3:

For h = 1 day:

R(3 + 1) = 30(3 + 1) - 3(3 + 1)^2 = 27 - 6 = 21

Average rate of change = (21 - 30(3) + 3(3)^2) / 1 = (21 - 27) / 1 = -6

For h = 0.1 day:

R(3 + 0.1) = 30(3 + 0.1) - 3(3 + 0.1)^2 = 29.7 - 8.91 = 20.79

Average rate of change = (20.79 - 30(3) + 3(3)^2) / 0.1 = (20.79 - 27) / 0.1 = -0.61

For h = 0.01 day:

R(3 + 0.01) = 30(3 + 0.01) - 3(3 + 0.01)^2 = 29.93 - 8.9699 = 20.9601

Average rate of change = (20.9601 - 30(3) + 3(3)^2) / 0.01 = (20.9601 - 27) / 0.01 = -0.061

As we can see, as the value of h decreases, the average rate of change approaches a more accurate value. The average rate of change of R(t) at t = 3 days is approximately -0.061 dollars per day.