I would also compare the exposed method to the one inspired from Descartes (to find the tangent of an explicit curve): From the origine, inflate a circle (keeping its center at the origine), until it meets the curve to one and only one point of the curve. The radius of the circle would then be the (local) minimum distance that we are looking for. In Cartesian coord. the circle is x^2 + y^2 = r^2. We substitute the y value (assuming an explicit curve y = f(x), to find x^2 + f(x)^2 - r^2 = 0 ( x and r are unknown at this time), and we know that we are looking to r being such what r is a DOUBLE ROOT. Indeed, for r less that the distance, the circle does NOT intercept the curve, while for r greater than the wanted distance, the circle will intercept twice the curve. So at the required r value is a double root for that equation. Since the derivative evaluated at a double root is also zero, we fall back to r' , as in your case.
If only I had this 30 yrs ago. I had to find the minimum distance between the mouse and a cubic curve, and because there were three solutions, I broke the curve up into linear segments and iterated over the fragments :( . It worked but this is so much more elegant.
Great video. I had the idea that you could also calculate it as gamma(t)’ • U(t) = 0, which in explicit cases might make calculations easier, since gamma(t)’ might be a simpler expression than U’(t) which is (gamma(t) - O)’. Although upon reflection the O term would always disappear as it is a constant, so the formulas are the same. Oh well, I’m just gonna leave my comment anyway, thanks again for the video
Your "intuitive" explanation involving the hypotenuse of a triangle does not hold water. The *furthest* point P on the curve from the point O is also such that OP is orthogonal to the curve. An identical argument would suggest that this point is instead a local *minimum* for distance.
I think he went through that in the video and showed that multiple points may satisfy the orthogonality condition on the curve, and that the condition is rather a local characterization of the "minimal" distance.
You're exactly right! I've missed that flaw until now. Shortening due to curvature outpaces the lengthening due to following the hypotenuse. In a later video, we derive the criterion that distinguishes local maxima from local minima.
@@MathTheBeautiful Could I turn my last comment into a question. I guess I am struggling to see geometrically in which situations the second or third order terms in the Taylor expansion become more significant than the linear term. It can't only be when subtraction is involved, as then a straight derivative would also have the same dependence on higher order terms.
I would also compare the exposed method to the one inspired from Descartes (to find the tangent of an explicit curve): From the origine, inflate a circle (keeping its center at the origine), until it meets the curve to one and only one point of the curve. The radius of the circle would then be the (local) minimum distance that we are looking for.
In Cartesian coord. the circle is x^2 + y^2 = r^2. We substitute the y value (assuming an explicit curve y = f(x), to find x^2 + f(x)^2 - r^2 = 0 ( x and r are unknown at this time), and we know that we are looking to r being such what r is a DOUBLE ROOT. Indeed, for r less that the distance, the circle does NOT intercept the curve, while for r greater than the wanted distance, the circle will intercept twice the curve. So at the required r value is a double root for that equation. Since the derivative evaluated at a double root is also zero, we fall back to r' , as in your case.
If only I had this 30 yrs ago. I had to find the minimum distance between the mouse and a cubic curve, and because there were three solutions, I broke the curve up into linear segments and iterated over the fragments :( . It worked but this is so much more elegant.
I also have regrets over how I spent the last thirty years
Nice!
Great video. I had the idea that you could also calculate it as gamma(t)’ • U(t) = 0, which in explicit cases might make calculations easier, since gamma(t)’ might be a simpler expression than U’(t) which is (gamma(t) - O)’. Although upon reflection the O term would always disappear as it is a constant, so the formulas are the same. Oh well, I’m just gonna leave my comment anyway, thanks again for the video
Your "intuitive" explanation involving the hypotenuse of a triangle does not hold water. The *furthest* point P on the curve from the point O is also such that OP is orthogonal to the curve. An identical argument would suggest that this point is instead a local *minimum* for distance.
I think he went through that in the video and showed that multiple points may satisfy the orthogonality condition on the curve, and that the condition is rather a local characterization of the "minimal" distance.
You're exactly right! I've missed that flaw until now. Shortening due to curvature outpaces the lengthening due to following the hypotenuse. In a later video, we derive the criterion that distinguishes local maxima from local minima.
@@MathTheBeautiful I've been trying to explain this to myself in terms of the Taylor expansion, but I can't get past generalities.
@@MathTheBeautiful Could I turn my last comment into a question. I guess I am struggling to see geometrically in which situations the second or third order terms in the Taylor expansion become more significant than the linear term. It can't only be when subtraction is involved, as then a straight derivative would also have the same dependence on higher order terms.