Proof of curvature of plane curve K | DX / DS | = | D (dy / DS) / DS| S is the length of an arc from a (a, f (a)) to P (x, y) on the curve y = f (x), K is the curvature of the curve at point P

Proof of curvature of plane curve K | DX / DS | = | D (dy / DS) / DS| S is the length of an arc from a (a, f (a)) to P (x, y) on the curve y = f (x), K is the curvature of the curve at point P

Proof of curvature of plane curve K | DX / DS | = | D (dy / DS) / DS|
S is the length of an arc from a (a, f (a)) to P (x, y) on the curve y = f (x). K is the curvature of the curve at P!
∵︱ds/dx︱=︱√(1+y′²)︱,∴︱dx/ds︱=︱1/√(1+y′²)︱
k=︱y″/(1+y′²)^(3/2)︱
So K DX / DS = ｜ y ″ / (1 + y ′ & # 178;), # 178; (1)
︱dy/ds︱=︱y′dx/ds︱=︱y′/√(1+y′²)︱
︱d(dy/ds)/ds|=︱{d[y′/√(1+y′²)]/dx}(dx/ds)︱=︱{[y″√(1+y′²)]-y′²y″/√(1+y′²)]/(1+y′²)}[1/√(1+y′²)]
=︱{[y″(1+y′²)-y′²y″]/(1+y′²)^(3/2)}[1/√(1+y′²)]=︱y″/(1+y′²)²︱.(2)
According to (1) (2), K | DX / DS | = | D (dy / DS) / DS|

The straight line y = - x + m and hyperbola y = 4 △ x intersect at two points c and D, and CD = 3 root sign 2, find m.ps: given the whole process, find a detailed explanation
Substituting y = - x + m into y = 4 / x, we get
-x+m=4/x,
It is reduced to x ^ 2-mx + 4 = 0,
△=m^2-16,
|CD|=√[2△]=3√2,
∴△=9,m^2=25,
｜ M = soil 5
Why | CD | = √ [2 △] = 3 √ 2?

Let C (x1, Y1), D (X2, Y2), then y1-y2 = - X1 + x2 = - (x1-x2), | x1-x2 | = √ △ so, | CD | = √ [(x1-x2) ^ 2 + (y1-y2) ^ 2] = √ [(x1-x2) ^ 2 + (x1-x2) ^ 2] = √ 2 × | x1-x2 | = √ 2 × △
This result can basically be used as a formula. Knowing the difference D of abscissa of two intersections and the slope k of the line where the intersection is, the distance between two intersections can be expressed as √ (1 + K ^ 2) × D