The number of times a new pop song has been downloaded $t$ weeks after its initial release is given by $f(t)=\frac{3,900,000}{1+400 e^{-0.75 t}}$.
a) After how many weeks is the rate of change of the number of downloads maximized?
b) What is the rate of change of the number of downloads at the time found in part (a)?
c) How many times has the song been downloaded at the time found in part (a)?
a) The rate of change of the number of downloads is maximized after 7.99 weeks.
(Do not round until the final answer. Then round to two decimal places as needed.)
b) The rate of change of the number of downloads is songs per week. (Round to the nearest whole number as needed.)
Answer
Therefore, the rate of change of the number of downloads is maximized after approximately 7.99 weeks, at which point the rate of change is approximately 1,000,000 songs per week and the song has been downloaded approximately 3,900,000 times.
Step 1 ：The function given is \(f(t)=\frac{3,900,000}{1+400 e^{-0.75 t}}\).
Step 2 ：To find the maximum rate of change, we need to find the derivative of this function, which represents the rate of change.
Step 3 ：The derivative of \(f(t)\) is \(f'(t)\).
Step 4 ：Using the quotient rule, we get \(f'(t) = \frac{(1+400 e^{-0.75 t}) * 0 - 3,900,000 * (0 - 400 * -0.75 * e^{-0.75 t})}{(1+400 e^{-0.75 t})^2}\).
Step 5 ：Simplifying, we get \(f'(t) = \frac{3,900,000 * 300 * e^{-0.75 t}}{(1+400 e^{-0.75 t})^2}\).
Step 6 ：To find the maximum rate of change, we need to find the maximum of \(f'(t)\). This occurs when the derivative of \(f'(t)\) is 0.
Step 7 ：The derivative of \(f'(t)\) is \(f''(t)\).
Step 8 ：Using the quotient rule again, we get \(f''(t) = \frac{(1+400 e^{-0.75 t})^2 * 0 - 2 * (1+400 e^{-0.75 t}) * (3,900,000 * 300 * e^{-0.75 t})}{(1+400 e^{-0.75 t})^4}\).
Step 9 ：Simplifying, we get \(f''(t) = -\frac{2 * 3,900,000 * 300 * e^{-0.75 t} * (1+400 e^{-0.75 t})}{(1+400 e^{-0.75 t})^4}\).
Step 10 ：Setting \(f''(t)\) equal to 0 and solving for \(t\), we get \(t = \ln(\frac{1}{400}) / -0.75\).
Step 11 ：Evaluating this expression, we get \(t \approx 7.99\) weeks.
Step 12 ：Substituting \(t = 7.99\) into \(f'(t)\), we get \(f'(7.99) \approx 1,000,000\) songs per week.
Step 13 ：Substituting \(t = 7.99\) into \(f(t)\), we get \(f(7.99) \approx 3,900,000\) songs.
Step 14 ：Therefore, the rate of change of the number of downloads is maximized after approximately 7.99 weeks, at which point the rate of change is approximately 1,000,000 songs per week and the song has been downloaded approximately 3,900,000 times.
