Optimization of a curve

Question

which point on the curve squareroot(1-x) is closest to the origin

Answer 1

Given:

Curve of y = √( 1 - x )

Distance equation from a coordinate point ( a, b ): s² = ( x - a )² + ( y - b )²

Find:

Point that lies on the curve closest to the origin.

Solution:

Since the point in question is the origin ( 0, 0 ), the distance equation simplifies to:

s² = x² + y²

Next, we have an equation for y, yet, the above equation asks for y²

Thus,

y² = 1 - x

Substituting this in yields,

s² = x² - x + 1

Implicitly differentiating both sides yields,

2s * ds/dx = 2x - 1

The point of this problem is to find when the curve is closing to the point of interest (POI; this time being the origin)

Interpreting the ds/dx term would be as such:

1. ds/dx < 0 means as you changed the increment dx, the increment of distance to your POI, ds, responded in the OPPOSITE way. This is confirmed that the ratio of ds to dx is negative. Say we started from some very large negative number and started moving in the positive direction (the dx term is positive). s will always be positive (the problem wouldn't make sense that DISTANCE is negative, but DISPLACEMENT can be negative, trying to point out a BIG distinction there), and it can be scene that so long as x is less than 1 / 2, ds/dx will evaluate to some negative number. It keeps decreasing the distance, s, as x increases from negative numbers approaching towards x = 1/2. Hinting towards the final answer but not quite there yet.
2. ds/dx > 0 means as you changed the increment dx, the increment of distance to your POI, ds, responded in the SAME way. This is confirmed that the ratio of ds to dx is positive. This means you've moved past the point of closest distance and are making yourself further from the closest point. This can be seen that for values of x greater than 1 / 2, ds/dx will evaluate to some positive number. Again hinting towards the final answer but still not quite there yet.

1. ds/dx = 0 means you increment your dx term and the increment of distance to your POI, ds, responded next to no value, causing the ratio to be very nearly to zero (or in the case of calculus, equal to zero).

Thus, ds/dx = 0 is what we are looking for.

0 = 2x - 1 (thankfully eliminating the s term on the left side of the equation)

x = 1/2

y = √( 1 - 1/2 ) = 1 / √2 y = √(2) / 2

Therefore meaning that the closest point along the curve y = √( 1 - x ) would be the point ( 1 / 2, √(2) / 2 )

Side note: There would actually be 2 points if one were to consider the ± nature of the square-root function.

The reason it wasn't included was because it looks like this problem is dealing with the positive of the ± and not the negative.

This is all to say that the other closest point technically would be ( 1 / 2, -√(2) / 2 )

I hope this helps your understanding of handling problems like this!

Message me in the comments for any further questions you may have!

Answer 2

Let point (x,√(1 - x)). Then we have function f(x) = x² + 1 - x is square of distance to origin.

f'(x) = 2x - 1 = 0; x = 1/2; f''(x) = 2 > 0. So, we have minimum for point (1/2, √1/2)