Question

The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan−1(t2), where t is measured in days. What is the instantaneous rate of change of the mass of the colony, in grams per day, at the moment the colony reaches a mass of 6 grams?

Answer 1

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

$\displaystyle \frac{dy}{dx}$

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

$P(t)=2+5atan(t^2)$

Let's find its first derivative, recalling that

$[atan(u)]'=\displaystyle \frac{u'}{1+u^2}$

$P'(t)=\displaystyle 5\frac{(t^2)'}{1+(t^2)^2}$

$P'(t)=\displaystyle 5\frac{2t}{1+t^4}$

$P'(t)=\displaystyle \frac{10t}{1+t^4}$

We need to know the value of t, so we use the provided condition P=6 gr

$P(t)=2+5atan(t^2)=6$

$\displaystyle atan(t^2)=\frac{4}{5}$

$\displaystyle (t^2)=tan\left ( \frac{4}{5} \right )$

$\displaystyle t=\sqrt{tan\left ( \frac{4}{5} \right )}$

$\displaystyle t=1,015\ days$

We use this value in the derivative:

$P'(1.015)=\displaystyle \frac{10(1.015)}{1+(1.015)^4}$

$P'(1.015)=4.93\ grams$