Max distance between a line and a parabola

Question

I am suppose to use calculus to find the max distance between the line $y = x + 2$ and the parabola $y = x^2$ on the interval $x$ greater then or equal to $-1$ and less then or equal to $2$.

I really have no idea what to do. I found the critical numbers and that didn't help at all. I guess all the integers and that didn't help either.

Answer 1

Note: The following solves the problem of maximizing the distance between two points, one on a line and one on a parabola, subject to a constraint. This is what the question above asks, although comments from the OP suggest that the intent was to maximize the difference of the $y$ values for a given $x$.

Here is another way:

Let $(x_1,x_1+2)$ be a point on the line and $(x_2,x_2^2)$ be a point on the parabola. Maximizing the distance is equivalent to maximizing the square of the distance, and the square is more tractable. So, we want to maximize $f(x_1,x_2) = (x_1-x_2)^2+(x_1+2-x_2^2)^2$, subject to $ x_1, x_2 \in [-1,2]$.

First, notice that the function $x_1 \mapsto f(x_1,x_2)$ is always a convex quadratic (ie, a quadratic in $x_1$, and the square term has a non-negative multiplier), regardless of the value of $x_2$. A convex quadratic on a closed interval takes its extreme value at the boundary of the interval. In this case, that gives, $\max(f(-1,x_2),f(2,x_2)) \geq f(x_1,x_2)$. Hence we may presume that $x_1 \in \{-1,2\}$. To find a solution we can maximize $f$ with $x_1$ set to $-1$, and again with $x_1=2$ and pick the maximum value.

Let $f_1(x_2) = f(-1,x_2), f_2(x_2) = f(2,x_2)$. Expanding these gives $f_1(x_2) = (1-x_2^2)^2+(1+x_2)^2$, $f_2(x_2) = (4-x_2^2)^2+(x_2-2)^2$.

We have $f_1'(x_2) = 2(x_2+1)(2x_2^2-2 x_2+1)$. The latter factor has no real roots, hence $f_1$ is non-negative, and is monotonic on the intervals $(-\infty,-1]$ and $[-1,\infty)$. Hence the maximum of $f_1$ on $[-1,2]$ is $f_1(2) = 18$.

$f_2'(x_2)= 2(x_2-2)(x_2+1-\frac{1}{\sqrt{2}})(x_2+1+\frac{1}{\sqrt{2}})$. Hence $f_2$ is monotonic on the intervals $(-\infty, -1-\frac{1}{\sqrt{2}}]$, $[-1-\frac{1}{\sqrt{2}},-1+\frac{1}{\sqrt{2}}]$, $[-1+\frac{1}{\sqrt{2}}, 2]$ and $[2,\infty)$. A moment's thought shows that $f_2$ is maximized at $x_2=-1+\frac{1}{\sqrt{2}}$ on the interval $[-1-\frac{1}{\sqrt{2}}, 2]$, and $f_2(-1+\frac{1}{\sqrt{2}}) = \frac{71+\sqrt{128}}{4} > 18$.

Hence the maximum distance is $\sqrt{\frac{71+\sqrt{128}}{4}}$ and it occurs at $x_1 = 2, x_2 = -1+\frac{1}{\sqrt{2}}$.

Addendum: Here is a solution to the actual problem I think the OP is trying to ask (with apologies to @Git Gud):

First, you should draw a picture.

Solve $\max_{x \in [-1,2]} |x+2-x^2|$. First note that $x+2 \geq x^2$ if and only if $x \in [-1,2]$, so the problem becomes $\max x+2-x^2$. Setting the derivative to zero gives $x=\frac{1}{2}$, hence the maximum value is $\frac{9}{4}$ which occurs at $x=\frac{1}{2}$.

Answer 2

Clearly, $P(t,t^2)$ is any point of the parabola.

The distance of the line: $y=x+2$ from $P(t,t^2)$ is $$\left|\frac{t-t^2+2}{\sqrt{1^2+1^2} }\right|=\frac{|t-t^2+2|}{\sqrt2}$$

Let $f(t)=t-t^2+2$

so, $f'(t)=1-2t$

For the extreme values of $f(t), f'(t)=0\implies t=\frac12$

Now, $f''(t)=-2<0$

So, $f(t)$ will be maximum at $t=\frac12$

$$f(t)_{max}=\frac12-\left(\frac12\right)^2+2=\frac94$$

So, the maximum distance is $\frac{\frac94}{\sqrt 2}$ unit $=\frac 9{4\sqrt2}$ unit and $P$ being $(\frac12, \left(\frac12\right)^2)$ or $(\frac12,\frac14)$

The maximum value of $f(t)$ can also be derived in the following way.

Let $y=t-t^2+2\implies t^2-t+2-y=0$ which is a quadratic equation in $t$.

As $t$ is real, the discriminant $(-1)^2-4(y-2)\ge 0\implies y\le \frac94$

Answer 3

What is 'distance'? My guess is you don't understand what's being asked, so let me clarify:

Hint: You're looking for the number, if it exists, $\max \limits_{x\in[-1,2]}\left (|f(x)|\right)$ where $f(x)=x^2-(x+2)$, for all $x\in [-1,2]$. Notice that $[-1,2]$ is closed and bounded, therefore any continuous function defined on that interval will attain a maximum and a minimum. This proves that the number exists. Now just look at the critical points of $f$ and take their absolute values (for both maxima and minima).

Answer 4

Hint. See the Math Java Applet in

http://www.ies.co.jp/math/java/conics/draw_parabola/draw_parabola.html

for a geometry ideia.