Question

What is the maximum vertical distance between the line \(y = x + 2\;\) and the parabola \(y = {x^2}\;\;\)for \( - 1 \le x \le 2\;\;\) ?

Short answer

\(d = \frac{9}{4}\)

Step-by-step solution

Given Data

The line \(y = x + 2\) and the parabola\(y = {x^2}\) are given.

Determination of the vertical distance

It can be seen from the graph that for \( - 1 \le x \le 2\) , it follows

\(x + 2 > {x^2}\)

The straight line \(y = x + 2\) lies above the curve \(y = {x^2}\)

Therefore, \(x + 2 - {x^2} > 0\)

Now, define the vertical distance as

\(d = x + 2 - {x^2} = d\left( x \right)\)

It is needed to find the maximum of d, since it is always positive.

To find the critical points, solve

\(d'\left( x \right) = 0\)for x.

\(d = x + 2 - {x^2}\)

\(d' = 1 - 2x\)

The critical point is:

\(\begin{aligned}{c}1 - 2x &= 0\\2x &= 1\\x &= \frac{1}{2}\end{aligned}\)

And \(1 - 2x > 0\)

\(x > \frac{1}{2}\)

Therefore, \(d'\left( x \right) > 0\) for \( - 1 < x < \frac{1}{2}\)

And \(d'\left( x \right) < 0\)for \(\frac{1}{2} < x < 2\)

It means that d increases on the interval \(\left( { - 1,\frac{1}{2}} \right)\) and decreases on \(\left( {\frac{1}{2},1} \right)\).

Therefore, It shows that d has an absolute maximum at \(x = \frac{1}{2}\) on the interval \(\left( { - 1,2} \right)\) .

Now, find \(d\left( {\frac{1}{2}} \right)\)

\(\begin{aligned}{c}d\left( {\frac{1}{2}} \right) &= \frac{1}{2} - 1 - {\left( {\frac{1}{2}} \right)^2}\\ &= \frac{9}{4}\end{aligned}\)

Find \(d\left( { - 1} \right)\) and \(d\left( 2 \right)\), to determine the absolute maximum .

\(d\left( x \right) = x + 2 - {x^2}\)

Therefore,

\(d\left( { - 1} \right) = - 1 + 2 - {\left( { - 1} \right)^2} = 0\)

\(d\left( 2 \right) = 2 + 2 - {\left( 2 \right)^2} = 0\)

Thus, the required maximum vertical distance is \(d = \frac{9}{4}\), since \(d\left( {\frac{1}{2}} \right) > d\left( { - 1} \right)\) and \(d\left( {\frac{1}{2}} \right) > d\left( 0 \right)\) .