Question

The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), 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:

1.21 g/day

Step-by-step explanation:

We are given that

The mass of a colony of bacteria (in (grams) is given by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Differentiate w.r.t t

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

By using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

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

$P'(t)=\frac{10}{4+t^2}$

We are given P(t)=6

Substitute the value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Substitute the value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We know that $1+tan^2\theta=sec^2\theta$

Using the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By using $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$g/day

Hence,the instantaneous rate of change of the mass of the colony=1.21g/day

Answer 2

Answer:

$P'(1.015)=2.5\ grams/day$

Step-by-step explanation:

Rate Of Change

If we have y as a function of x, the rate of change can be obtained by changing x by a small amount ($\Delta x$) and computing the change in y ($\Delta y$). The rate of change is

$\displaystyle \frac{\Delta y}{\Delta x}$

In calculus, when $\Delta x$ tends to zero it becomes dx, and the function has infinitesimal changes called dy. The instantaneous rate of change is computed as the first derivative of y

$\displaystyle y'=\frac{dy}{dx}$

The first derivative of the inverse tangent is

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

Note: The standard notation atan will be used for the inverse tangent

We have the relation between the mass of a colony of bacteria in grams P(t) and the time t expressed in days

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

Computing the first derivative:

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

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

$P'(t)=\displaystyle \frac{10}{4+t^2}$

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

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

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

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

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

$\displaystyle t=2.06\ days$

Replacing this value in the first derivative:

$P'(2.06)=\displaystyle \frac{10(2.06)}{4+(2.06)^2}$

$P'(2.06)=2.5\ grams/day$