Chapter 4.3, Problem 51HP

To determine

To find: A real life case that have a constant rate of change.

Answer to Problem 51HP

The required equation of a real life case that have a constant rate of change in point slope form is $y-150=50\left(x-3\right)$ and in slope intercept form is $y=50x+0$

Explanation of Solution

Given information:

The equation undergoes a constant rate of change.

Consider the formula of simple interest

  $I=\mathrm{Pr}t$

Substitute $1000$ for P , $0.05$ for r and $3$ for t ,.

  $\begin{array}{c}P=1000\times 0.05\times 3\\ =150\end{array}$

The result of the expression is $\left(x_1,y_1\right)=\left(3,150\right)$ .

Substitute $1000$ for P , $0.05$ for r and $5$ for t ,.

  $\begin{array}{c}P=1000\times 0.05\times 5\\ =250\end{array}$

The result of the expression is $\left(x_2,y_2\right)=\left(5,250\right)$ .

Now, consider the point slope form of equation

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Substitute $250$ for $y_2$

  $150$ for $y_1$ , $5$ for $x_2$ and $3$ for $x_1$ in the above equation.

  $\begin{array}{l}y-150=\frac{250-150}{5-3}\left(x-3\right)\\ y-150=50\left(x-3\right)\end{array}$

Now, consider the slope intercept form of equation

  $y=mx+b$

Consider the above equation.

  $\begin{array}{c}y-150=50\left(x-3\right)\\ y=50x+0\end{array}$

Hence, the required equation of a real life case that have a constant rate of change in point slope form is $y-150=50\left(x-3\right)$ and in slope intercept form is $y=50x+0$