Chapter 3.3, Problem 67E

TELEVISION VIEWERSHIP The number of viewers of a television series introduced several years ago is approximated by the function

N(t) = (60 + 2t)^2/3 (1 ≤ t ≤ 26)

where N(t) (measured in millions) denotes the number of weekly viewers of the series in the till week. Find the rate of increase of the weekly audience at the end of week 2 and at the end of week 12. How many viewers were there in week 2? In week 24?

To determine

To find The rate of increase of the weekly audience and the number of viewers of a television series.

Answer to Problem 67E

The rate of increase of the weekly audience at the end of week 2 is $0.333\text{million}$ and at the end of week 12 is $0.304452\text{million}$. The number of viewers in week 2 is $16\text{million}$ and in week 24 is $22.67\text{million}$

Explanation of Solution

Given information:

The given function is $N\left(t\right)={\left(60+2t\right)}^{\frac{2}{3}}$.

Formula used:

The power rule for differentiation is,

$\frac{d}{dx}{\left[f\left(x\right)\right]}^n=n{\left[f\left(x\right)\right]}^{n-1}{f}^{\prime }\left(x\right)$

Calculation:

Section1:

Differentiate the given function to find the rate of increase of the weekly audience.

$\begin{array}{c}{N}^{\prime }\left(t\right)=\frac{d}{dt}{\left(60+2t\right)}^{\frac{2}{3}}\\ =\frac{2}{3}{\left(60+2t\right)}^{\frac{2}{3}-1}\frac{d}{dt}\left(60+2t\right)\\ =\frac{2}{3}{\left(60+2t\right)}^{-\frac{1}{3}}\left(2\right)\\ =\frac{4}{3}{\left(60+2t\right)}^{-\frac{1}{3}}\end{array}$

The simplified form of the derivative function is,

$N^{\prime }\left(t\right)=\frac{4}{3}{\left(60+2t\right)}^{-\frac{1}{3}}$

Substitute 2 for t in the above function,

$\begin{array}{c}{N}^{\prime }\left(2\right)=\frac{4}{3}{\left(60+4\right)}^{-\frac{1}{3}}\\ =\frac{4}{3}\cdot \frac{1}{4}\\ =\frac{1}{3}\\ =0.333\end{array}$

The rate of increase of the weekly audience at the end of week 2 is $0.333\text{million}$

Substitute 12 for t in the above function,

$\begin{array}{c}{N}^{\prime }\left(12\right)=\frac{4}{3}{\left(60+24\right)}^{-\frac{1}{3}}\\ =\frac{4}{3\sqrt[3]{84}}\\ =\frac{4}{13.13836}\\ =0.304452\end{array}$

The rate of increase of the weekly audience at the end of week 2 is $0.304452\text{million}$

Thus, the rate of increase of the weekly audience at the end of week 2 is $0.333\text{million}$ and at the end of week 12 is $0.304452\text{million}$.

Section2:

Substitute 2 for t in equation (1) to find the number of viewers.

$\begin{array}{c}N\left(2\right)={\left(60+4\right)}^{\frac{2}{3}}\\ ={\left(64\right)}^{\frac{2}{3}}\\ =16\end{array}$

The number of viewers of a television series in week 2 is $16\text{million}$.

Substitute 24 for t in equation (1) to find the number of viewers.

$\begin{array}{c}N\left(24\right)={\left(60+48\right)}^{\frac{2}{3}}\\ ={\left(108\right)}^{\frac{2}{3}}\\ =22.67\end{array}$

The number of viewers of a television series in week 24 is $22.67\text{million}$

Thus, the number of viewers of a television series in week 2 is $16\text{million}$ and in week 24 is $22.67\text{million}$