Respuesta :
Answer:
1.21 g/day
Step-by-step explanation:
We are given that
The mass of a colony of bacteria (in (grams) is given by
[tex]P(t)=2+5tan^{-1}(\frac{t}{2})[/tex]
Where t=Time(in days)
Differentiate w.r.t t
[tex]P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})[/tex]
By using the formula [tex]\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}[/tex]
[tex]P'(t)=\frac{5}{2}(\frac{4}{4+t^2})[/tex]
[tex]P'(t)=\frac{10}{4+t^2}[/tex]
We are given P(t)=6
Substitute the value
[tex]6=2+5tan^{-1}(\frac{t}{2})[/tex]
[tex]5tan^{-1}(\frac{t}{2})=6-2=4[/tex]
[tex]tan^}{-1}(\frac{t}{2})=\frac{4}{5}[/tex]
[tex]\frac{t}{2}=tan(\frac{4}{5})[/tex]
[tex]t=2tan(\frac{4}{5})[/tex]
Substitute the value of t
[tex]P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}[/tex]
[tex]P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}[/tex]
We know that [tex]1+tan^2\theta=sec^2\theta[/tex]
Using the formula
[tex]P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}[/tex]
[tex]P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})[/tex]
By using [tex]cos^2x=\frac{1}{sec^2x}[/tex]
[tex]P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21[/tex]g/day
Hence,the instantaneous rate of change of the mass of the colony=1.21g/day
The instantaneous rate of change of mass can be calculated as [tex]\frac{dy}{dx}[/tex] where [tex]x[/tex] is rate of change with small amount [tex]dx[/tex].
Thus, the instantaneous rate of change of the mass of the colony is [tex]1.21\;\rm{g/day}[/tex].
Given:
The mass of a colony of bacteria, in grams, is modeled by the function [tex]P[/tex] given by
[tex]P(t)=2+5tan^{-1}(\frac{t}{2})[/tex],
where [tex]t[/tex] is measured in days.
Differentiating the function with respect to [tex]t[/tex].
[tex]P'(t)=0+5\times (\frac{1}{1+\frac{t^2}{4}} )\times \frac{1}{2}\\P'(t)=\frac{5}{2}\times \frac{4}{4+{t^2}} \\P'(t)= \frac{10}{4+{t^2}}[/tex]
Now, [tex]P(t)=6[/tex]
Then,
[tex]6=2+5tan^{-1}(\frac{t}{2})\\tan^{-1}(\frac{t}{2})=\frac{4}{5}\\\frac{t}{2}=tan(\frac{4}{5})\\t=2tan(\frac{4}{5})[/tex]
Substituting the value of [tex]t[/tex] in above equation,
[tex]P'{(2tan(\frac{4}{5})}= \frac{10}{4+{(2tan(\frac{4}{5}))^2}}\\\\P'{(2tan(\frac{4}{5})}=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}\\\\P'{(2tan(\frac{4}{5})}=\frac{5}{2}\times cos^2(\frac{4}{5})\\P'{(2tan(\frac{4}{5})}=\frac{5}{2}\times (0.696)^2\\P'{(2tan(\frac{4}{5})}=1.21\;\rm{g/day}[/tex]
Hence, the instantaneous rate of change of the mass of the colony is [tex]1.21\;\rm{g/day}[/tex]
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