Question

Optimization. What is the minimum vertical distance between y=x^2+1 and y=x-x^2?
I'm semi familiar with optimization problems and the hardest part is figuring out how to begin. How do I start this?

Answer 1

Let's write the curves as (x,y) = (t^2+1, t) and (x,y) = (s-s^2, s). The distance between them is given by (t-s)^2 + (t^2+1 - (s-s^2))^2; now we just have to find the minimum of that expression.
Are you familiar with finding the extrema of multivariate functions? If not, I'm pretty sure there has to be another way to do this.

Answer 2

No. I don't understand that method. I'm in Calc1, a beginning calc student.
I have worked through other optimization problems, so I know in general you usually find the formula you need to work with, simplify it, solve for one of the variables in the original equation, find the derivative, find critical points.

Answer 3

Okay, then we'll need to put the distance as a function of one parameter, not two. The shortest line between two curves is always perpendicular to said curves; we can hopefully get a relationship between t and s from there, and substitute to get a single variable function you can work with.

Answer 4

I did one that involved minimum distance to a point, but there was just one equation. Then I used a distance formula \[\sqrt{(x _{2}-x_{1})^{2} + (y_{2} }-y _{1})^{2}\]
I'm thinking I use it here, too. But There are two fucnctions/equations in this problem and I'm not sure how to begin.

Answer 5

Damn, Firefox crashed on me and ate what I was writing. Think I'm gonna write on notepad, it's less annoying. Give me a second here.

Answer 6

damn, I hate when that happens. It's especially frustrating when inputting equations on this chat. Thanks for replying, no rush.

Answer 7

We can write all the points in the curve y=x^2+1 as a function of a parameter t,
this way: (x = t^2+1, y = t). Doing the same for y=x-x^2, we get (x = s-s^2, y = s).
Notice that the parameters have to be different, because there's no
guarantee that the values of x that minimize the distance will be the same for
both curves. Still with me so far?

Answer 8

kind of. Will we be setting the equations equal to each other?

Answer 9

Setting the equations equal to each other would give us the values of t and s for which the curves intersect; if that happened the distance would be zero. Of course, it's not going to be so easy; the curves never touch each other. You can try solving the equation, you'll find it has no solutions.

Answer 10

However we can use the distance formula on those sets of points, to get the distance
between any point in the first curve and any point in the second curve. The result
will be sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) =
= sqrt( (t^2+1 - (s-s^2) )^2 + (t-s)^2 )
, which is the equation I wrote in the first post.
We have to find the smallest value that function takes. It can be done directly, but it's certainly out of the scope of Calc 1. So what I'm trying to do is find a relationship between s and t, so that we'll get the distance as a function of only one variable.

Answer 11

Lemme try make that a bit fancier:\[\sqrt{(x2-x1)^2+(y2-y1)^2} = \sqrt{(t^2-1 - (s-s^2)^2 + (t-s)^2}\]

Answer 12

t^2+1, that is

Answer 13

so that is setting distance formulas of each equation equal to each other?

Answer 14

I'm using any point from the first curve, x1 = t^2-1, y1 = t

and any point from the second curve, x2 = s-s^2, y2 = s

and substituting those values into the distance equation to get the distance between any point from each curve.

Answer 15

oh, the subscripts cleared up a bit

Answer 16

Man the drawing tools in here are atrocious. Anyway. The shortest line connecting two curves is always perpendicular to those curves.

So, at the points A and B, the tangent to the curves will be perpendicular to the line connecting them.