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How to determine the tangent vector of space curve to find the corresponding tangent equation and normal plane
It is obvious that for the curve represented by the equation system f (x, y, z) = 0 g (x, y, z) = 0, first determine a certain variable as a parameter and convert other variables into a function of this variable. For example, with X as a parameter, the equation system is simplified as: x = x, y = y (x) z = Z (x). Therefore, the tangent vector at any point of the curve is
Some problems about plane normal vector in space vector method In the space rectangular coordinate system, the normal vector formula of a plane can be represented by the linear coordinates of any normal vector perpendicular to the plane I hope you can find the normal vector
The normal vector has only one solution
Coordinate, quantity product is 0
For example, how to find the normal vector of surface ABC
First calculate the coordinates of AB, and then calculate the AC coordinates
Let n be the normal vector of... And the coordinates of n be (x, y, 1)
n*AB=0
n*AC=0
Just calculate n
If you can see it directly on some graphs, you don't have to calculate it. You can directly calculate the coordinates to get the vector
On the solution of tangent and normal plane of space curve I know how to solve ordinary equations However, if the curve is determined by a system of equations (surface equation and plane equation) I don't know how to ask for it I hope you will talk about the idea of solving the problem
Get 2T, 1,1
Therefore, the tangent equation can be obtained as 2T / (X-2) = 1 / (Y-1) = 1 / Z
So the tangent equation is 0 = 2T (X-2) + 1 (Y-1) + 1z
Choose me. I don't know how to ask again. I've been playing for a long time
Is differentiation just derivative?
Almost
Find derivative or differential 1. Y = x ^ 10-10 ^ x + e ^ 3, find y ' 2. Y = arctan (1 / x), find dy 3. Y = LNX, find dy 4. Y = xlnx, find y '' 5. Let x ^ 3 + x ^ 2 · y + y ^ 2 = 1 and find dy / DX
1.y'=10x^9-10^xln10 2.dy=-dx/1+x^2 3.dy=dx/x*lnx*lnlnx 4.y'=lnx+1 y''=1/x 5.3x^2+2xy+x^2*dy/dx+2y*dy/dx=0 dy/dx=-（3x^2+2xy)/(x^2+2y)
Derivative differential TGA = (Y-Y1) / (x-x1) how to find the differential?
The function expression should be:
y=（x+x1）tgA+y1
Where x1, Y1 and a are all constants,
Then the differential of Y over x is easy to find
dy=tgA*dx
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