How to find tangent vectors at any point on a spatial curve

If it's curvilinear

Parametric equation. Then the coordinate components pair the parameters.

Derivation. The resulting vector is at that point.

Tangent vectors. If a curve is given as a curve in the form of a surface intersection, then find the normal vector of the two surfaces at that point.

The product of the crosses of both.

This is the tangent vector of the curve.

For example, y=x 2, consider x as a variable, y as the dependent variable, and then find the partial derivative of y versus x.

Take a system of equations. f（x，y，z）＝0

g（x，y，z）＝0

The curve is represented by a variable that is determined to be a parameter first, and the other variables are reduced to a function of this variable, for example, with x as a parameter, the system of equations is reduced to :

x＝xy＝y（x)

z＝z（x）

So, the tangent vector known at any point on the curve is.

1，dy/dx，dz/dx

Extended Materials. Basic property 1: The same number or the same algebraic formula is added (or subtracted) to both sides of the equation at the same time.

The result is still the equation.

It is expressed in letters as: if a=b, c is a number or an algebraic formula. Then:

1)a+c=b+c

2)a-c=b-c

Fundamental property 2: Both sides of the equation are multiplied or divided by the same way.

The result of a number that is not 0 is still an equation.

3) If a=b, then b=a (symmetry of the equation).

First, the tangent vector of any point (x,y) is obtained according to the parametric equation, and then the tangent vector in the problem is obtained by the parametric equation at (x0,y0), and finally the tangent equation is obtained by the point formula.

The tangent vector of the curve at a point can be understood as that pointTangents(with a directional arrow).

Description: The tangent vector of a surface can be thought of as a vector in the tangent plane.

For the more general manifold m, the tangent vector of m at the point p is the tangent vector at p in m by the curve at the point p.

The concept of tangential high reputation is a geometric concept, i.e. its definition has nothing to do with the selection of coordinates.

Because it is a geometric quantity. This is the most basic concept in differential geometry.

Real-world application: The space we come into contact with, from the universe to the cells, is full of colorful and changing curves. Such as the planets of the solar system.

The orbit, the airplane's fairway, the winding mountain.

Winding roads, springs in sofas, fabric patterns, silhouettes of gears and cams, double helix structures of DNA of living genetic material, and so on.

Among the curves that people come into contact with, the simplest ones are straight lines and circles. These curves are the objects discussed in elementary plane geometry. The next more complex curve is the quadratic curve.

i.e. elliptical, hyperbola.

and parabola. These have been studied in plane analytic geometry, and the methods discussed are using coordinates and unary quadratic algebraic equations.

For more complex curves, just using the first algebra is generally not enough. The study of the geometric properties of smooth curves in general, calculus.

It's a powerful tool. We can use calculus to derive three basic geometric quantities that characterize the geometric properties of a spatial curve, namely arc length, curvature, and deflection.

If it's curvilinear

Parametric equation. Then the coordinate components pair the parameters.

Derivation. The resulting vector is at that point.

Tangent vectors. If a curve is given in the form of a curved surface intersection, then first find the normal vectors of the two surfaces at that point, and the cross product of the two is the tangent vector of the curve.

For example, y=x 2, treat x as a variable, y as the dependent variable state, and then find the partial derivative of y versus x. Take a system of equations.

f（x，y，z）＝0

g(x,y,balance slow z) 0

The curve is represented by a variable that is determined to be a parameter first, and the other variables are reduced to a function of this variable, for example, with x as a parameter, the system of equations is reduced to :

x＝xy＝y（x)

z＝z（x）

So, the tangent vector known at any point on the curve is.

1，dy/dx，dz/dx

Extended Materials. Basic property 1: Add (or modulus) the same number or the same algebraic formula to both sides of the equation at the same time, and the result is still the equation.

It is expressed in letters as: if a=b, c is a number or an algebraic formula. Then:

1)a+c=b+c

2)a-c=b-c

Basic property 2: Multiplying or dividing both sides of the equation by the same non-0 number is still an equation.

3) If a=b, then b=a (symmetry of the equation).

4) If a=b, b=c, then a=c (transitivity of the equation).

The parametric equation of the channel curve, then the vector derived by the coordinate component to the parameter is the tangent vector at the point. If a curve is given in the form of a surface intersection, then first find the normal vectors of the two surfaces at that point, and the cross product of the two is the tangent vector of the curve.

A vector tangent to the curve, given a point p on the curve c, q is the point of proximity to p on c, and when the q point approaches p along the curve, the limit position of the secant pq is called the tangent of the curve c at the p point.

One of the characteristics of manifold is that one of its local domains can establish a one-to-one mapping relationship between points with an n-dimensional Euclidean space, and each of its local domains can establish a one-to-one mapping relationship between points with its own n-dimensional Euclidean space, and on this basis, a local coordinate system of manifolds can be established for each locality, so as to become measurable.

.Do you have to use vectors?

Assuming MA OA, then AM is the tangent. No problem.

Other words. ma=（（x1-x0),（y1-y0））；

oa=（x1,y1).

Two verticals: x1*(x1-x0)+y1*(y1-y0)=0;

The generation of the circular equation is r2-x1x0-y1y0=0;That's the equation you said!

So it seems that your answer is not right.

Do you understand that? Your tangent equation should contain m. Do you substitute m for the equation?

I won't count the correct answer, and then discuss whether it is perpendicular to the coordinate system is your business!

Upstairs is a disguised slope, which has little to do with the vector, but the result must be right, just add the limit solution that y can tend to infinity

Written as a standard, the coefficient ratio below the denominator is the tangent vector.

Easy to obtain, x=y 2

z=y^4+y^2-y

Select y as the parameter, and DX dy=2y

dz/dy=4y^3+2y-1

Substituting y=1 can be obtained.

dx/dy=2

dz/dy=5

So, the tangent vector is.

t=(2，1，5)

Take a system of equations. f(x, then late y,z) 0

g(x,y,wuvolz) 0

The curve is represented by a variable that is determined to be a parameter first, and the other variables are reduced to a function of this variable, for example, with x as a parameter, the system of equations is reduced to :

x＝xy＝y（x)

z＝z（x）

So, the tangent vector at any point on the curve is.

1，dy/dx，dz/dx

This part of the content belongs: the equation composed of 2 equations containing 3 variables can determine two unary implicit functions, and the derivatives of the 2 implicit functions can be expressed by formulas, and the specific expressions can be seen in the textbook.