Chapter 11.7, Problem 44PPE

a.

To determine

To graph the function and check the values for what distances is $s\le 1$ .

Answer to Problem 44PPE

  $s\le 1$ for $d\ge 40$

Explanation of Solution

Given information:

As radio signals move away from a transmitter, they become weaker, the function $s=\frac{1600}{d^2}$ gives the strength s of a signal at a distance of d miles from a transmitter.

Calculation:

As radio signals move away from a transmitter, they become weaker, the function $s=\frac{1600}{d^2}$ gives the strength s of a signal at a distance of d miles from a transmitter.

The graph of the functions is shown below

  

From the graph it is clear that $s\le 1$ for $d\ge 40$ .

b.

To determine

The signal strength at 10 mi, 1mi and 0.1 mi.

Answer to Problem 44PPE

  $\begin{array}{l}\text{At 10 mi, }\text{16}\\ \text{At 1 mi, }\text{1600}\\ \text{At 0}\text{.1 mi, }\text{160000}\end{array}$

Explanation of Solution

Given information:

As radio signals move away from a transmitter, they become weaker, the function $s=\frac{1600}{d^2}$ gives the strength s of a signal at a distance of d miles from a transmitter.

Calculation:

In order to find the signal strength at 10 mi, substitute $d=10$ in the given function and solve for s as shown below,

  $\begin{array}{c}s=\frac{1600}{d^2}\\ =\frac{1600}{{\left(10\right)}^2}\\ =\frac{1600}{100}\\ =16\end{array}$

So, at 10 mi the signal strength is 16.

Similarly, to find the signal strength at 1 mi, substitute $d=1$ in the given function and solve for s as shown below,

  $\begin{array}{c}s=\frac{1600}{d^2}\\ =\frac{1600}{{\left(1\right)}^2}\\ =\frac{1600}{1}\\ =1600\end{array}$

So, at 1 mi the signal strength is 1600.

And to find the signal strength at 0.1 mi, substitute $d=0.1$ in the given function and solve for s as shown below,

  $\begin{array}{c}s=\frac{1600}{d^2}\\ =\frac{1600}{{\left(0.1\right)}^2}\\ =\frac{1600}{0.01}\\ =160000\end{array}$

So, at 0.1 mi the signal strength is 160000.