Chapter 9.6, Problem 59SR

a.

To determine

Verify the formula for exponential decay or growth.

Answer to Problem 59SR

Negative.

Explanation of Solution

Given information:

The power output $P$ (in watts) is given by $P=50e^{-\frac{t}{250}}$ , where $t$ is the time in days.

Calculation:

The power output from a radioisotope is given by,

  $P=50e^{-\frac{t}{250}}\mathrm{......}\left(1\right)$

Where $P$ is the power point in watts and $t$ is the time in days.

The formula $\left(1\right)P=50e^{-\frac{t}{250}}$ for power output is an example of “Exponential Decay”.

“Because the exponent in the formula is negative.”

b.

To determine

Find the power available after $100$ days.

Answer to Problem 59SR

  $33.5watt$

Explanation of Solution

Given information:

The power output $P$ (in watts) is given by $P=50e^{-\frac{t}{250}}$ , where $t$ is the time in days.

Calculation:

Substituting $t=100$ in formula $\left(1\right)P=50e^{-\frac{t}{250}}$ and solve for $P$ as follows,

  $\begin{array}{l}P\left(100\right)=50e^{-\frac{100}{250}}\\ =50\times 0.67032\\ =33.5160023\\ \approx 33.5watt\end{array}$

Hence, about $33.5watt$ power is available after $100$ days.

c.

To determine

Find the days that the satellites continue to operate.

Answer to Problem 59SR

  $402days$

Explanation of Solution

Given information:

The power output $P$ (in watts) is given by $P=50e^{-\frac{t}{250}}$ , where $t$ is the time in days.

Calculation:

Substituting $P=10$ watts in formula $\left(1\right)P=50e^{-\frac{t}{250}}$ and solve for $t$ as follows,

  $\begin{array}{l}10=50e^{-\frac{t}{250}}\\ e^{-\frac{t}{250}}=\frac{10}{50}\\ e^{-\frac{t}{250}}=\frac{1}{5}\\ e^{\frac{t}{250}}=5\end{array}$

Apply the property of equality for logarithmic function $x=y\Rightarrow \log x=\log y,$

  $\begin{array}{l}\log \left(e^{\frac{t}{250}}\right)=\log \left(5\right)\\ \frac{t}{250}=\log (5)\left[\log (e^x)=x\right]\\ \frac{t}{250}=\mathrm{1.6094379......}\end{array}$

  $t=250\times \mathrm{1.6094379....}$

Simplify,

  $\begin{array}{l}t=\mathrm{402.35947....}\\ \approx 402\end{array}$

Hence, the satellite can continue to operate for about $402days$ .