Chapter 1, Problem 1.23P

To determine

(a)

The given data and determine the best function.

Answer to Problem 1.23P

Plot for the given data is:

The function is linear and given by:

$y=50x-1250$.

Explanation of Solution

Given Information:

Data is given by:

$\begin{array}{cccccc}x:& 25& 30& 35& 40& 45\\ y:& 0& 250& 500& 750& 1000\end{array}$.

Calculation:

From the given data plot is:

From this plot, it is known that the function is linear with slope:

$\begin{array}{l}a=\frac{1000-0}{45-25}\\ \Rightarrow a=50\end{array}$

So, the output function:

$\begin{array}{l}y=ax+b\\ \Rightarrow y=50x+b(1)\end{array}$

Substitute the first pair of values from given data:

$\begin{array}{l}0=50\times 250+b\\ \Rightarrow b=-1250\end{array}$

Substitute this value in first equation:

$y=50x-1250$.

Conclusion:

Plot for the given data is:

The function is linear and given by:

$y=50x-1250$.

To determine

(b)

The given data and determine the best function.

Answer to Problem 1.23P

Plot for given data is:

The output is power function, and is given by:

$y=3639x^{-0.97}$.

Explanation of Solution

Given Information:

Data is by:

$\begin{array}{ccccccccccccc}x:& 2.5& 3& 3.5& 4& 4.5& 5& 5.5& 6& 7& 8& 9& 10\\ y:& 1500& 1220& 1050& 915& 810& 745& 690& 620& 520& 480& 410& 390\end{array}$.

Calculation:

Plot for the given data is:

The plot is not linear.

For getting linear relationship, we can take log to the input out output. So, the data becomes:

$\begin{array}{ccccccccccccc}\log x:& 0.3979& 0.4771& 0.5441& 0.6021& 0.6532& 0.699& 0.7782& 0.7404& 0.8451& 0.9031& 0.9542& 1\\ \log y:& 3.176& 3.086& 3.021& 2.961& 2.908& 2.872& 2.792& 2.839& 2.716& 2.681& 2.613& 2.591\end{array}$

Plot for this data:

This plot is linear with slop:

$\begin{array}{l}a=\frac{2.591-3.176}{1-0.3979}\\ \Rightarrow a=-0.97\end{array}$

So, the equation:

$\begin{array}{l}\log y=a\log x+b\\ \Rightarrow \log y=-0.97\log x+b\end{array}$

Substitute a pair of values:

$\begin{array}{l}2.591=-0.97\times 1+b\\ \Rightarrow b=3.561\end{array}$

So, the equation becomes:

$\begin{array}{l}\log y=-0.97\log x+3.561\\ \Rightarrow \log y=-0.97\log x+\log 3639\\ \Rightarrow y=3639x^{-0.97}\end{array}$.

Conclusion:

Plot for given data is:

The output is power function, and is given by:

$y=3639x^{-0.97}$.

To determine

(c)

The given data and determine the best function.

Answer to Problem 1.23P

The plot for given data is:

Function is exponential, and is given by:

$y=89.567\times {10}^{-0.000337x}$.

Explanation of Solution

Given Information:

Data is by:

$\begin{array}{cccccc}x:& 1000& 2000& 3000& 4000& 5000\\ y:& 41.2& 18.62& 8.62& 3.92& 1.86\end{array}$.

Calculation:

Plots for the given data,

This plot is not linear.

Take log of output, the data become,

$\begin{array}{cccccc}x:& 1000& 2000& 3000& 4000& 5000\\ \log y:& 1.6149& 1.27& 0.9355& 0.5933& 0.2659\end{array}$

Plots for these values:

This plot is linear, and slope is:

$\begin{array}{l}a=\frac{0.2659-1.6149}{5000-1000}\\ \Rightarrow a=-0.00033725\end{array}$

So,

$\begin{array}{l}\log y=ax+b\\ \Rightarrow \log y=-0.00033725x+b\end{array}$

Substitute one pair of values:

$\begin{array}{l}1.6149=-0.00033725\times 1000+b\\ \Rightarrow b=1.95215\end{array}$

So, equation becomes:

$\begin{array}{l}\log y=-0.00033725x+1.95215\\ \Rightarrow \log y=-0.00033725x\times \log 10+\log 89.567\\ \Rightarrow y=89.567\times {10}^{-0.000337x}\end{array}$.

Conclusion:

The plot for given data is:

Function is exponential, and is given by:

$y=89.567\times {10}^{-0.000337x}$.