Curvature radius formula

Curvature radius formula

ρ=|(1+y'^2)^(3/2)]/y"|

How to calculate the radius of curvature

Radius of curvature = 1 / curvature
The analytic formula of known curve y = f (x)
Curvature = (the second derivative of F / (the square of 1 + the first derivative of F) ^ (3 / 2))

How to prove that a curve with constant radius of curvature is a circle?

You can search Baidu Encyclopedia of "radius of curvature", which has the formula of radius of curvature. You can make it a constant. That's what I do

Given that point a (1, - K + 2) is on hyperbola y = K / x, find the value of constant K

X = 1, y = - K + 2 is substituted into hyperbolic equation
-k+2=k/1
2k=2
k=1

Given the point a (1, - K + 2) on the hyperbola y = x / K, find the value of constant K

Because point a is on the hyperbola, we take x = 1 into the hyperbolic equation and get y = K and y = 2-k, so k = 2-k and K = 1

Ln (secx) derivation

Let secx = t
Int'=1/t*t'
=1/sec*tanxsecx
=tanx

What is the period of y = ln (secx + TaNx)?

Obviously, the period of y = ln (secx + TaNx) depends on secx + TaNx = cosx / SiNx + SiNx / cosx = [(SiNx) ^ 2 + (cosx) ^ 2] / (SiNx * cosx) = 2 / sin2x, so the period is 2 Π / 2 = Π

Y = 5 + ln x, then dy = what

Solution
y=5+lnx
∴
y‘=dy/dx=(5+lnx)=1/x
∴
dy=(1/x)dx

Y = secx / (1 + TaNx) derivation

Reduce secx = 1 / cosxtanx = SiNx / cosx multiply cosxy = 1 / (cosx + SiNx) y '= [1' (cosx + SiNx) - 1 * (cosx + SiNx)] / (cosx + SiNx) ^ 2 = - (- SiNx + cosx) / (cosx + SiNx) ^ 2 = (SiNx cosx) / (cosx + SiNx) ^ 2 = (SiNx cosx) / (cosx + SiNx) ^ 2

Derivation: y = in (secx + TaNx)

=[1/(secx+tanx)]*(secxtanx+sec²x)
=(secxtanx+sec²x)/(secx+tanx)
=secx(secx+tanx)/(secx+tanx)
=secx