Question

1. while playing his latest video game, mr. henderson notices that he is feeding the enemy team in such a way that the number of feeds can be calculated with the following formula: $f = \frac{\pi m^3}{m + 1}$, where m is the number of minions on the screen and f is the total number of feeds. mr. henderson knows that both the number of feeds and the number of minions are changing with time. furthermore, he observes that there are currently 9 minions on the screen and the number of minions is increasing by 2 minions per second. at this very moment in the game, what rate is mr. henderson feeding the enemy team?

Explanation:

Step1: Differentiate \(F\) with respect to \(t\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = \pi m^{3}\), \(u^\prime=3\pi m^{2}\frac{dm}{dt}\), and \(v=m + 1\), \(v^\prime=\frac{dm}{dt}\).
So \(\frac{dF}{dt}=\frac{3\pi m^{2}(m + 1)\frac{dm}{dt}-\pi m^{3}\frac{dm}{dt}}{(m + 1)^{2}}\)
Simplify the numerator: \(\frac{dF}{dt}=\frac{\pi m^{2}(3(m + 1)-m)\frac{dm}{dt}}{(m + 1)^{2}}=\frac{\pi m^{2}(2m + 3)\frac{dm}{dt}}{(m + 1)^{2}}\)

Step2: Substitute \(m = 9\) and \(\frac{dm}{dt}=2\)

\(\frac{dF}{dt}=\frac{\pi\times9^{2}\times(2\times9 + 3)\times2}{(9 + 1)^{2}}\)
First calculate the values in the formula:
\(9^{2}=81\), \(2\times9+3=18 + 3=21\), \((9 + 1)^{2}=100\)
Then \(\frac{dF}{dt}=\frac{\pi\times81\times21\times2}{100}=\frac{\pi\times81\times42}{100}=\frac{3402\pi}{100}=\frac{1701\pi}{50}\approx\frac{1701\times3.14}{50}=\frac{5341.14}{50}=106.8228\)

Answer:

\(\frac{1701\pi}{50}\approx106.82\) feeds per second