Knowing the curvature of the curve, find the curve equation

This involves differential equations.

Curvature k(x) = |y''|/[1+(y')^2]^(3/2);

When y > 0, it can be seen from the graph that the curve is convex, then y at this time''<0;

then k(x) = -y''/[1+(y')^2]^(3/2);

y'' = -k(x)·[1+(y')^2]^(3/2);

Let p=y', then y''=dp/dx;

Then: dp dx = -k(x)· (1+p 2) (3 2);

Separation variables: dp (1+p 2) (3 2) = -k(x)·dx;

Integral on both sides: 1 (1+p 2) (3 2) dp = - 0,x)k(x)·dx;

Calculate the integral on the left: let p=tan u, then dp=du cos 2 u; 1/(1+p^2)^(3/2)=1/(sec^2 u)^(3/2) = 1/sec^3 u = cos^3 u;

then 1 (1+p 2) (3 2) dp

cos^3 u ·du/cos^2 u

cos u ·du

sin u -c1

sin arctan p -c1

i.e.: sin arctan p = - k(x)·dx +c1; ①

arctan p= arcsin[-∫k(x)·dx +c1];

p = tan

∫k(x)·dx +c1]/√

If the curve is tangent to the y-axis at point o, the limit lim(x 0)y can be known', i.e., lim(x 0)p= , is obtained by .

0,0) k(x)·dx +c1 =sin(π/2)=1;

c1 = 1;

Rule. y'=p=[-∫k(x)·dx +1]/√

Points. y=∫(0,x) [k(x)·dx +1]/√ dx +c2;

and y(0)=0, substituting c2=0;

Rule. y=∫(0,x) [k(x)·dx +1]/√ dx

Give me k(x) and I'll try it first.

2/(l*x^2 - 2*l^2*x) -x + 2*x)/(2*l^3*x - l^2*x^2) -2*i*atan((i*x)/l - i))/l^3

matlab results, I don't know if it's right...

Suppose the curve is y=f(x), the center of the curvature circle (a, b), and the radius r;

The essence of the curvature circle is to require the tangent of the curve and the circle at this point.

Same as the depression.

First, the circular equation for curvature is obtained: (x-a) 2 + y-b) 2 = r 2;

Assuming that the curve is concave at this point, then b > y, giving y = b - r 2 - x-a) 2) (1 2) ;

y' =1/2)[(r^2 - x-a)^2)^(1/2) ]2)(x-a) =x-a) (r^2 - x-a)^2)^(1/2) ；Type A.

y'' r^2 - x-a)^2)^(1/2) +x-a)*(1/2)(r^2 - x-a)^2)^(3/2)*(2)(x-a)

r^2 - x-a)^2)^(1/2) +x-a)^2(r^2 - x-a)^2)^(3/2)

Type B. According to the reasons a and b, (x-a) can be eliminated to give an expression of radius r.

by y'with y''Denote;

However, it is more troublesome to directly substitute for the envy line into the elimination element, which can be replaced as follows:

Knows by a (r 2 - x-a) 2) (1 2) = y'(x-a) substitution of the b formula has:

y'' y’/(x-a) +x-a)^2 (y'/(x-a))^3 = y'/(x-a) +y'^3 / x-a) =y' +y'^3) /x-a)

x-a) =y' +y'^3) /y''In this formula, there is a : in the A formula

y' =y' +y'^3) /y'')r^2 - y' +y'^3) /y'')2)^(1/2)

r^2 = 1 + y'^2) /y'')2 + y' +y'^3)

y'')2(1 + y'^2)^3) /y''^2)

r = 1 + y'^2)^(3/2)

y''The curvature is 1 r;

There is a radius r, a normal.

Slope (-1 y.)'It is easy to find the center of the curvature circle, and then find the curvature circle of the square brother to make an uproar.

I don't know if it helps you.

Suppose the curved line is y=f(x), the center of the curvature circle (a, b), and the radius is r.

The essence of the curvature circle is to require the tangent of the curve and the circle at this point.

Same as the depression.

First, the circular equation for curvature is obtained: (x-a) 2 + y-b) 2 = r 2;

Assuming that the curve is concave at this point, then b > y, giving y = b - r 2 - x-a) 2) (1 2) ;

y' =1/2)[(r^2 - x-a)^2)^(1/2) ]2)(x-a) =x-a) (r^2 - x-a)^2)^(1/2) ；Type A.

y'' r^2 - x-a)^2)^(1/2) +x-a)*(1/2)(r^2 - x-a)^2)^(3/2)*(2)(x-a)

r^2 - x-a)^2)^(1/2) +x-a)^2(r^2 - x-a)^2)^(3/2)

Type B. According to the reasons a and b, (x-a) can be eliminated, and an expression for the half-brother rr can be obtained.

by y'with y''Denote;

However, it is more troublesome to directly substitute the elimination element, which can be replaced as follows:

Knows by a (r 2 - x-a) 2) (1 2) = y'(x-a) substitution of the b formula has:

y'' y’/(x-a) +x-a)^2 (y'/(x-a))^3 = y'/(x-a) +y'^3 / x-a) =y' +y'^3) /x-a)

x-a) =y' +y'^3) /y''In this formula, there is a : in the A formula

y' =y' +y'^3) /y'')r^2 - y' +y'^3) /y'')2)^(1/2)

r^2 = 1 + y'^2) /y'')2 + y' +y'^3)

y'')2(1 + y'^2)^3) /y''^2)

r = 1 + y'^2)^(3/2)

y''Bend.

The curvature is 1 r;

There is a radius r, a normal.

Slope (-1 y.)'), it is easy to find the center of the circle of curvature, and then find the equation of the circle of curvature.

I don't know if it helps you.

In the orange belt projection, it is the round family reed 0

x 2 + y 2 = ax is projected on xoz, and spike is projected on xoz, in other words, on the plane y=0.

Substituting y=0 into the above equation yields the projection on xoz as x 2=ax

Subtract y from the two equations, and get z 2 = a 2 - ax, which is a parabola on the xoz plane.

The tangent slope of a point is the derivative y of that point'Blind orange limbs, by the title, y'=3y+2，y'-3y=2, which is a first-order linear differential equation that uses the Wu Zao formula to find y

y=ce (-3dx)+e (-3dx) 2e (3dx)dx=ce (3x)+e (3x)(-2 3)e (-3x)=ce (3x)-2 3,c is an arbitrary constant.

Because y is over (0,1), substituting it to obtain c=5 3, milling, the curve equation is y=5 3*e (3x)-2 3

This question p(x)=-3, q(x)=2

Let the square of the curve be y=f(x), then the tangent slope of the curve at the point (x,y) is f'(x)=3y+2。

According to the known conditions, the point (0,1) is on the curve, i.e., f(0)=1, so it is possible to perform a definite integration of the curve equation to obtain:

f(x) =f(0) +0,x] f'(t) dt

where [0,x] denotes the definite integral from 0 to x. Replace f'(x)=3y+2 Substituting the above equation yields:

f(x) =1 + 0,x] (3f(t)+2) dt

Solve its integral to the right side of the above equation and obtain:

f(x) =1 + 3∫[0,x] f(t) dt + 2x

This is a first-order linear ordinary differential equation that can be solved using the constant variable method of excitation. Suppose its general solution is f(x)=a*e (3x) -2 3, where a is a constant to be determined. Substituting it into the expression of f(x) gives :

a*e^(3x) -2/3 = 1 + 3∫[0,x] (a*e^(3t) -2/3) dt + 2x

a*e^(3x) =3∫[0,x] a*e^(3t) dt + 2x + 5/3

The integral of both sides of the above equation can be obtained by taking the integral at the same time

a*e^(3x) =a*e^(3x) -a + 2x + 5/3

The move yields: a = 5 9

Therefore, f(x)=5 9*e (3x)-2 3

The slope of the tangent at any point (x,y) on the curve is equal to 3 times the ordinate of the point plus 2, Biling.

So y'=3y+2, which variable to separate Sanhui gets dy (3y+2)=dx, and the integral punching code gets ln|3y+2|=3x+c, the curve passes through the point (0,1), so c=ln5, so |3y+2|=5e^(3x).

y=2^(1-3x)

y'=-3ln2*2^(1-3x)

y'Messy (0) = -6ln2

The tangent equation for the sum of the shed leakage is y=-6ln2*x+2

Curvature k=y''/1+(y'2) (3 2)], where y',y"They are the first and second derivatives of the function y against x, respectively.

1. Let the curve r(t) = x(t), y(t)), curvature k=(x'y" -x"y')/x'Suspicious band) 2 + y')^2)^(3/2).

2. Let the curve r(t) be a three-dimensional vector function, and the curvature k=|r'×r"|/r'|)3/2),|x|Represents the length of the vector x.

3. The outer pants of the vector a, b are vertical, if a = (a1, a2, a3), b = (b1, b2, b3), a b = (a2b3-a3b2, a3b1-a1b3, a1b2-a2b1).