Chapter 3.1, Problem 35E

To determine

To find: the reason that the parabolas “everywhere equidistant”.

Answer to Problem 35E

The required solution is shown below.

Explanation of Solution

Consider the general form of a parabola.

  $y=a{\left(x-x_0\right)}^2+k$

Such that $a\ne 0$ and $\left(x_0,k\right)$ is its vertex. Suppose two parabolas with equations given as,

  $\begin{array}{c}{y}_1=a{\left(x-x_0\right)}^2+k_1\\ y_2=a{\left(x-x_0\right)}^2+k_2\end{array}$

For $k_1\ne k_2$ and $a,h$ are equal. Then one parabola is the vertical translation of the other, such that they are horizontal aligned. For example, the parabolas,

  $\begin{array}{l}f\left(x\right)={\left(x-1\right)}^2\\ g\left(x\right)={\left(x-1\right)}^2+2\end{array}$

  

Find the derivative of the function $g\left(x\right)$ .

  $\begin{array}{c}g\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{g\left(x+h\right)-g\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[{\left(x+h-1\right)}^2+2-\left[{\left(x-1\right)}^2+2\right]\right]}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[{\left(x+h-1\right)}^2+2-{\left(x-1\right)}^2-2\right]}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[{\left(x+h-1\right)}^2-{\left(x-1\right)}^2\right]}{h}\end{array}$

Notice at this point, the $k$ coordinate of the vertex is eliminated from the equation. If calculated the derivative of $f\left(x\right)$ .

  $f\text{'}\left(x\right)=g\text{'}\left(x\right)=2x-2$

Graph the equation.

  

Finally, the graph are vertically equidistant.

  $\begin{array}{c}g\left(x\right)-f\left(x\right)=\left[{\left(x-1\right)}^2+2\right]-{\left(x-1\right)}^2\\ =2\end{array}$

There are always at a distance of $2$ for any point $x$ .

More generally, two parabolas with equation.

  $\begin{array}{l}{y}_1=a{\left(x-x_0\right)}^2+k_1\\ y_2=a{\left(x-x_0\right)}^2+k_2\end{array}$

Will have a distance of

  $\left|y_1-y_2\right|=\left|k_1-k_2\right|$

For any $x$ and their derivative will be,

  $\begin{array}{c}\frac{d{y}_1}{dx}=\frac{d{y}_2}{dx}\\ =2a\left(x-x_0\right)\end{array}$

Therefore, the required solution is shown above.