Finding the normal equation The normal equation of surface z = 2x square + y square at point (1,1,3) is

Z = 2x ^ 2 + y ^ 2, the partial derivative of Z to X is 4x, the partial derivative of Z to y is 2Y, so the direction vector of the normal of the surface at the point (x, y, z) s = (- 4x, - 2Y, 1)
Therefore, the direction vector s of the normal of the surface at points (1,1,3) is = (- 4, - 2,1)
The normal equation is (x-1) / (- 4) = (Y-1) / (- 2) = (Z-3)

Given that a curve passes through a point (1,1), the intercept of its tangent on the longitudinal axis is equal to the abscissa of the tangent point

The tangent of the curve at (a, f (a)) is y = f '(a) (x-a) + F (a)
Intercept is - AF '(a) + F (a)
From the meaning of the title, there is - AF '(a) + F (a) = a
That is to solve the differential equation: - XY '+ y = x, y (1) = 1
dy/dx=y/x-1
Let Y / x = u, then y = Xu, y '= u + Xu'
Substituting into the equation: U + Xu '= U-1
The result is: Xu '= - 1
That is u '= - 1 / X
That is u = - ln | x | + C
Y = - XLN | x | + Cx
Substituting y (1) = 1 = C
So y = - XLN | x | + 1

Given that a curve passes through a point (1,1), the intercept of its tangent on the longitudinal axis is equal to the abscissa of the tangent point, its equation is solved

Let the curve be y = f (x)
The tangent equation of a curve is y-f (X.) = f '(X.) (X-X.), that is, y = f' (X.) (X-X.) + F (X.)
The problem of X. = - X. f '(X.) + F (X.) can be transformed into the solution of differential equation
The differential equation can be written as x = - XDY / DX + y and dy / DX = Y / X - 1
Let z = Y / x, then y = ZX, dy / DX = Z + XDZ / DX = Y / X-1 = Z-1
XDZ / DX = - 1 separate the two sides of the variable DZ = - DX / X to get z = - LNX + C = Y / X
When y = x (c-lnx) curve passes through the point (1,1) generation, y = x (1-lnx)

Finding the normal equation The normal equation of surface z = 2x square + y square at point (1,1,3) is