Answer:
P'(1.015)=4.93 grams
Explanation:
Instantaneous Rate Of Change
If one variable y is a function of another variable x, the rate of change can be measured by changing x by a small amount (dx) and computing the change in y (dy). The rate of change is
[tex]\displaystyle \frac{dy}{dx}[/tex]
When dx tends to zero, we call it the instantaneous rate of change and is easily computed as the first derivative of y
Note: We'll be using the standart notation atan for the inverse tangent
We are given the relation between the mass of a colony of bacteria in grams P(t) and the time t in days
[tex]P(t)=2+5atan(t^2)[/tex]
Let's find its first derivative, recalling that
[tex][atan(u)]'=\displaystyle \frac{u'}{1+u^2}[/tex]
[tex]P'(t)=\displaystyle 5\frac{(t^2)'}{1+(t^2)^2}[/tex]
[tex]P'(t)=\displaystyle 5\frac{2t}{1+t^4}[/tex]
[tex]P'(t)=\displaystyle \frac{10t}{1+t^4}[/tex]
We need to know the value of t, so we use the provided condition P=6 gr
[tex]P(t)=2+5atan(t^2)=6[/tex]
[tex]\displaystyle atan(t^2)=\frac{4}{5}[/tex]
[tex]\displaystyle (t^2)=tan\left ( \frac{4}{5} \right )[/tex]
[tex]\displaystyle t=\sqrt{tan\left ( \frac{4}{5} \right )}[/tex]
[tex]\displaystyle t=1,015\ days[/tex]
We use this value in the derivative:
[tex]P'(1.015)=\displaystyle \frac{10(1.015)}{1+(1.015)^4}[/tex]
[tex]P'(1.015)=4.93\ grams[/tex]
