Vector Parametrizations and Parametric Equations of Lines in Space

point-direction form In the plane, the linear equation ax + by + c = 0 represents a line. But in space, the equation represents a plane. A vector parametrization of a line in R³ through a point \((x_0, y_0, z_0)\) in the direction of v = ⟨\(a, b, c\)⟩ is r\((t) = ⟨x_0, y_0, z_0 ⟩+t⟨a, b, c⟩\) (point-direction form ) for \(-\infty < t < \infty\). Just as parametric equations, a vector parametrization of a space line is not unique.

Also write the vector parametrization as parametric equations \(x=x_0+at,y=y_0+bt,z=z_0+ct\), or write it as a rectangular equation system \(\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\). These three approaches describe the same line.

Examples Use the "vec" module to simplify a vector parametrization of the line through the point (1, 0, -2) in the direction of v = 2i - 3j + 4k by vec;(i-2*k)+t*(2*i-3*j+4*k) || for all t. The parametric equations are x = 1 + 2t, y = -3t, and z = 4t-2, and graph the line by pc3;1+2*t;-3*t;-2+4*t ||.

Two-point form If a line passes through two points \(P(x_0, y_0, z_0)\) and \(Q(x_1, y_1, z_1)\) in the space, then a direction vector of the line is v = ⟨\(x_1-x_0,y_1-y_0,z_1-z_0\)⟩. Substituting a point and direction in the point-direction form, we get a vector parametrization for the line r\((t) = (1-t)⟨x_0, y_0, z_0 ⟩+t⟨x_1, y_1, z_1⟩\) for all \(t\). In particular, the vector parametrization for the line segment \(PQ\) has the same form, but for \(0≤t≤1\).

Examples
(1) A vector parametrization of a line through the points (a, b, c) to (m, n, h) can be obtained by the codes vec;(1-t)*(a*i+b*j+c*k)+t*(m*i+n*j+h*k) ||.
(2) A vector parametrization of a line passes through (1, 2, 3) and (-1, 0, -2) by vec;(1-t)*(i+2*j+3*k)+t*(-i-2*k) ||, and graphed by pc3;1-2*t;2-2*t;3-5*t ||.
(3) A parametrization of the line segment through (1, -3) and (-2, 0) by vec;(1-t)*(i-3*j)+t*(-2*i) ||, and graphed by pc2;1-3*t;-3+3*t;itv=(-2,1) ||.

Vector Parametrizations and Parametric Equations of Planes in Space

Point-normal form equations The equation ax + by + cz = d represents a plane in space, where ⟨a, b, c⟩ is the normal vector of the plane. If a plane passing through a point \((x_0, y_0, z_0)\) and one of its normal vector is n = ⟨\(a, b, c\)⟩, then an equation (in point-normal form) of the plane can be obtained by ⟨\(a, b, c\)⟩ · ⟨\(x, y, z\)⟩ = ⟨\(a, b, c\)⟩·⟨\(x_0, y_0, z_0\)⟩.

Simplify the point-normal equation of a plane with a normal vector ⟨a, b, c⟩ and through point (m, n, p) by "vec;(a*i+b*j+c*k)*((x-m)*i+(y-n)*j+(z-p)*k)". Can graph the resulting plane equation by the "sf3" module.

Examples Find the equation 2x - y + 5z - 46 = 0 of the plane through the point (3, -5, 7) with a normal vector n = ⟨2, -1, 5⟩ by vec;(2*i-j+5*k)*(x*i+y*j+z*k)-(2*i-j+5*k)*(3*i-5*j+7*k). Graph the plane by sf3;(46-2*x+y)/5 ||.

Example A particle moves from P(2, 3, -15) at t = 0 to Q(1, 4, 2). Check if P and Q are on the same side of the plane 2x - 3y - z = 5. Substitute the coordinates of P and Q into the plane equation, d = 10 for P and -13 for Q, so P and Q are on the opposite of the plane. To find when PQ intersect the plane, parametrize the line through PQ by x = 2 - t, y = 3 + t, z = -15 + 17t, substitute the line equation into the plane equaiton and solve for t = 5/24.

Three-point form equations If a plane passes through three points P, Q, R that are not on the same line, the normal vector can be determined by the vector cross product PQ × PR. Substituting the resulting vector in the point-normal form, you can find the plane equation, and graph it by "sf3" module.

Examples Find the equation of a plane through points P(-1, 6, 3), Q(2, 1, 4) and R(3, -1, 5). The vectors PQ = 3i - 5j + k and PR = 4i -7j +2k. The normal vector is PQ × PR by vec;(3*i-5*j+k)**(4*i-7*j+2*k) ||, which is ⟨-3, -2, -1⟩. Thus, the plane equation is by vec;(-3*i-2*j-k)*((x+1)*i+(y-6)*j+(z-3)*k) ||, or z = 12 - 3x - 2y. Graph it by sf3;12-3*x-2*y ||.

Point-line form plane equations If a plane passes a point P and a line \(x=x_0+at,y=y_0+bt,z=z_0+ct\), pick a point Q(\(x_0,y_0,z_0)\) for \(t=0\) on the line to form vector QP, and the normal vector of the sought plane is QP × v, where v = ⟨\(a,b,c\)⟩ is the direction vector of the line. Using the point-normal form, you can get the plane equation.

Examples If a plane passes through a point P(2, 4, -3) and a line x = 1 - 2t, y = 3t, z = -4 + 7t, then Q = (1, 0, -4), vector QP = ⟨1, 4, 1⟩, v = ⟨-2, 3, 7⟩, and the normal vector is QP × v by vec;(i+4*j+k)**(-2*i+3*j+7*k) ||. It is ⟨25, -9, 11⟩. So the plane equation is by vec;(25*i-9*j+11*k)*((x-2)*i+(y-4)*j+(z+3)*k) ||. Graph it by sf3;(9*y-25*x-19)/11 ||.

In particular, if a plane passes through a point and the line of intersection between two other planes, there is a shortcut to determine the equation of the sought plane.

Examples Find the plane through the point (2, 2, 1) and the intersection line between the planes 3x - 4y + 5z = 8 and x + 7y - 3z = 3. Let the equation of the sought plane be (x+7y-3z-3) + k(3x-4y+5z-8) = 0. Substitute the point in this equation and solve for k = 2 by slv;(16-6)-5*k;k ||. The equation of the plane is 7x - y + 7z = 19.

Relations of Points, Lines and Planes in Space

Distance between points and lines in space The distance between two points can be calculated directly using their coordinates. The distance of a point P to a line L is given by \(\frac{||QP × {\bf v}||}{||\bf v||}\), where Q is a point on L and v is the direction of L. The distance can be regarded as the height of a parallelogram span by vector \(QP\) and v, which is its area divided by its base length.

Examples Find the distance of P(2, -2, 3) to line x = 4 - 2t, y = -1 + 5t, z = 6t. Pick a point Q = (4, -1, 0) on the line. v = ⟨-2, 5, 6⟩, QP = ⟨2, 1, -3⟩, the length of the cross product is by vec;lnt((2*i+j-3*k)^(-2*i+5*j+6*k)) ||, and the length of v can be got by vec;lnt(-2*i+5*j +6*k) ||. Thus, the distance is vec;lnt((2*i+j-3*k)^(-2*i+5*j+6*k))/lnt(-2*i+5*j +6*k), which is about 3.091.

Distance between points and planes in space The distance of a point P to a plane ax + by + cz + d = 0 is given by \(\frac{|PQ·{\bf n}|}{||\bf n||}\), the length of the projection of the vector QP on the normal vector n = ⟨a, b, c⟩, where Q is a point on the plane.

Examples To find the distance between P(4, -5, 9) and the plane 2x - 4y + 3z = 8, the normal vector is n = ⟨2, -4, 3⟩ and Q = (0, 1, 4) is a point on the plane. So PQ = ⟨4, -6, 5⟩, and the distance is by vec;(4*i-6*j+5*k)*(2*i-4*j+3*k)/lnt(2*i-4*j+3*k) ||, which is 47/\(\sqrt{29}\).

Distance between two lines in space The distance between two nonparallel and non-intersecting lines can be found in a similar way. Let L1: \(x = x_0 + at, y = y_0 + bt, z = z_0 + ct\) and L2: \(x = x_1 + At, y = y_1 + Bt, z = z_1 + Ct\) be two nonparallel and non-intersecting. Then the normal vector n = u × v, where u = ⟨a, b, c⟩, v = ⟨A, B, C⟩. Pick a point \(P(x_0, y_0, z_0)\) on L1, \(Q(x_1, y_1, z_1)\) on L2, and the vector \(PQ = ⟨x_1-x_0, y_1-y_0, z_1-z_0⟩\). So the distance between L1 and L2 is \(\frac{|PQ·{\bf n}|}{||\bf n||}\).

Examples If L1: x = 1 + 2t, 3 - t, -4 + 2t and L2: x = -2t, y = 7 - 3t, z = 5 - 2t. Then PQ = ⟨1, -4, -9⟩, u = ⟨2, -1, 2⟩ and v = ⟨-2, -3, -2⟩. So the distance can be found by vec;((i-4*j-9*k)*((2*i-j+2*k)^(-2*i-3*j-2*k)))/lnt((2*i-j+2*k)^(-2*i-3*j-2*k)) || vec;w*(u**v)/lnt(u**v);u;2*i-j+2*k;v;-2*i-3*j-2*k;w;i-4*j-9*k ||, which is \(5\sqrt{2}\).

Distance between two parallel planes in space The distance between two parallel planes \(ax+by+cz=d\) and \(ax+by+cz=k\) is \(\frac{|d-k|}{\sqrt{a^2+b^2+c^2}}\).

Examples The distance between planes 7x + 2y - 6z = 25 and 7x + 2y - 6z = 14 by (25-14)/(7**2+2**2+6**2)**(1/2) ||.

Distance between parallel lines and planes in space If a line \(L\) and a plane \(\mathcal{P}\) are parallel, the distance between \(L\) and \(\mathcal{P}\) can be found by first picking a point \(P\) on \(L\) and a point \(Q\) on \(\mathcal{P}\), and then calculating \(\frac{|PQ·{\bf n}|}{||\bf n||}\), the length of the projection of \(PQ\) on n, where n is the normal vector of \(\mathcal{P}\).

Examples Given the line L: x = -6 + 4t, y = 3 + t, z = 5 + 2t and plane P: x + 2y - 3z = 4, they are parallel. Choose a point M = (-6, 3, 5) on L and Q = (4, 0, 0) on P, so MQ = ⟨10, -3, -5⟩. Since n = ⟨1, 2, -3⟩, the distance can be obtained by vec;(10*i-3*j-5*k)*(i+2*j-3*k)/lnt(i+2*j-3*k) ||, which is 19/\(\sqrt{14}\).

Intersection between two lines in space To find the point of intersection of two lines, match their parametric equations, solve the equation system for parameters t and s, and then substitute the values of t and s in the third equation to see if they are compatible. If not, then the two lines do not intersect.

Examples Let line L be x = 1+2t, y = -1+3t, z = 9-5t and line M be x = 2-s, y = -3+2s, z = 6+3s. Use eqs;1+2*t-(2-s);-1+3*t-(-3+2*s) || for solving s = 1, t = 0. Substitute these values in L and M for z = 9. So the point of intersection is (1, -1, 9).

Intersection between lines and planes To find the point of intersection between a line and a plane, substitute the parametric equations for the line in the plane equation, you have a simple linear equation of t. Solve for the parameter t, substitute it back to the parametric equations, and then find the coordinates of the point of intersection.

Examples Find the intersection point of a line x = 1 - 2t, y = 3t, z = 4 + 7t and a plane 2x - 3y + 5z = 4. Substitute x, y, z in the plane equation, and solve the equation by slv;2*(1-2*t)-3*(3*t)+5*(4+7*t)-4;t || for t = -9/11. So the point of intersection is (29/11, -27/11, -19/11).

Lines of intersection between planes If two planes intersect a line, then the line is perpendicular to the normal vectors of both planes. This means the intersection line is parallel to the cross product of the two planes' normal vectors. Choosing a point on the line, you can write the equation of the line in a point-direction form.

Examples Given planes 3x - 4y + z = 11 and x - 3y + 5z = 7, n₁ = ⟨3, -4, 1⟩, n₂ = ⟨1, -3, 5⟩. So v = n₁ × n₁ is the direction of the line of intersection, and can be got by vec;(3*i-4*j+k)**(i-3*j+5*k) ||, which is ⟨-17, -14, -5⟩. To choose a common point, let z = 0 and use eqs;3*x-4*y-11;x-3*y-7 || to get x = 1, y = -2. So the point (1, -2, 0) is on the line. Finally, the parametric equations for the line of intersection are x = 1-17t, y = -2-14t, z = -5t.

Vector-valued functions: [vec; f(u, v, w); u;uexpr; v;vexpr; w;wexpr, ...], [vec; vexpr]

The functions in form of r(t) = x(t)i + y(t)j + z(t)k are called vector-valued functions of one variable t, where the components x(t), y(t) and z(t) are scalar-valued functions. They are often used to analyze curves and motion in R³ by viewing r(t) as the path traced out by a particle at time t. So r(t) is a vector parametrization of the path, where x(t), y(t) and z(t) indicate the coordinates or positions of the particle at time t.

Examples vec;2*i+3*j-4*k || vec;x*i-y*j+z*k || vec;t*i+t**2*j-t**(1/2)*k || vec;(t*i-k+2*t*k)**(t**2*i-3*j+t*k) || vec;(x-y)*i+z**2*j-3*k ||.

Examples For complicated operations of vector functions, write the operation expression as f(u, v, w) in the "vec" module, and define each vector subsequently by "u;uexpr;v;vexpr;w;wexpr". || vec;u*v;u;sin(t)*i+cos(t)*j+t*k;v;sin(t)*i+cos(t)*j+k || vec;2*u**(3*v-w);u;cos(x)*i-sin(y)*j;v;cos(x)*i+sin(y)*j;w;x*i-y*j+z*k ||. Be sure that each vector must be a linear combination of the standard basis vectors i, j, k.

Projection of a vector field onto a plane The projection of r(t) = x(t)i + y(t)j + z(t)k onto a different coordinate plane (yz-, xz- and xy-plane) can be found by setting x(t) = 0, or y(t) = 0, or z(t) = 0. So x(t)i + y(t)j is the projection of r(t) onto the xy-plane; x(t)i + z(t)k is the projection of r(t) onto the xz-plane; y(t)j + z(t)k is the projection of r(t) onto the yz-plane.

Examples vec;prj(cos(t)*i+sin(t)*j+t*k,cos(t)*i+sin(t)*j) || vec;prj(t*sin(t)**2*i+t*cos(t)**2*j+t*k,t*cos(t)**2*j+t*k) || vec;prj(t*i+t**2*j+exp(t)*k,t*i+exp(t)*k) ||.

Limits of vector functions Using the "lim" module, you can find the limit of a vector function, which is computed componentwise as \(\lim\limits_{t\to c}{\bf r}(t)=⟨\lim\limits_{t\to c}x(t),\lim\limits_{t\to c}y(t),\lim\limits_{t\to c}z(t)⟩\).

Attention Use the combination "vec;lim(vexpr)" to find the limit of each component.

Examples vec;lim((x*i+(x**2-1/x)*j+(2-x)*k),x,1) || vec;lim((u,u,cos(x)*i+sin(2*x)*j+log(x)*k),x,pi) ||.

Integrals of vector functions Use the "int" module one at a time to find the integral of a vector function, computed componentwise as \(\int {\bf r}(t)dt=⟨\int x(t)dt,\int y(t)dt,\int z(t)dt⟩\) or \(\int_a^br(t)dt=⟨\int_a^bx(t)dt,\int_a^by(t)dt,\int_a^bz(t)dt⟩\).

Attention Use the combination "vec;itg((vexpr))" to find the integral of each component.

Examples
(1) vec;itg((2*x*i+(x-3)*j+(x-x**2)*k),x,0,1) || vec;itg(cos(x)*i+sin(2*x)*j,x,0,pi/2) ||.
(2) Find the integral \(∫_1^4⟨t^{-2},\sqrt{t},-t^2⟩dt\) by itg;i/t**2+t**(1/2)*j-t**2*k;t;1;4 ||.
(3) If r'(t) = ⟨t², 1 + t, 2t - 3⟩ and r(0) = ⟨1, 2, -1⟩, find r(t). The integral itg;t**2*i+(1+t)*j+(2*t-3)*k;t || gives r(t) + c. Substituting the result in the equation, you get c = ⟨1, 2, -1⟩. So r(t) = ⟨t³/3 + 1, 3t/2 + 2, -2t - 1⟩.
(4) If r'(t) = ⟨t, -t, 5⟩ and r(1) = ⟨2t, 1+t, t⟩, find r(t). First find vec;itg(t*i-t*j+5*k,t) || and c by vec;2*t*i-(1+t)*j+t*k-itg(t*i-t*j+5*k,t) ||, which is ⟨3/2, -3/2, -4⟩. So r(t) = ⟨(t² + 3)/2, -(t² + 3)/2, 5t-4⟩.
(5) If r''(t) = ⟨1, -2, 1⟩, r'(2) = ⟨0, 1, 2⟩ and r(2) = ⟨2, 0, -1⟩, find r(t). For r'(t), itg(i-2*j+k,t)+u ||. Substitute r'(2) for u = -2i + 5j. For r(t), vec;2*i-k-itg(itg(i-2*j+k,t)-2*i+5*j,t) ||, and substitute t = 2 for c = 4i - 6j - 3k. So r(t) = ⟨t²/2 - 2t + 4, -t² + 5t - 6, t²/2 - 3⟩ by vec;itg(itg(i-2*j+k,t)-2*i+5*j,t)+4*i-6*j-3*k ||.

Derivatives of vector functions: [vec;tnv(x(t)i + y(t)j + z(t)k, t)], [vec;dif(x(t)i + y(t)j + z(t)k, t, n)]

Use "vec;tnv(x(t)i + y(t)j + z(t)k, t)" to find the derivatives of a vector function, computed componentwise as \(\frac{d}{dt}{\bf r}(t)={\bf r}'(t)= ⟨x'(t),y'(t),z'(t)⟩\). It is also called the tangent vector of r(t). If r(t) is a trajectory taced by a moving particle, r'(t) = v(t).

Example Cycolid r(t) = ⟨bt - bsin(t), b - bcos(t)⟩, find v'(t) by "vec;tnv(v,t);v;(b*t-b*sin(t))*i+(b-b*cos(t))*j", the speed ||v(t)|| = \(\frac{ds}{dt}\) by "vec;lnt(tnv(v,t));v;(b*t-b*sin(t))*i+(b-b*cos(t))*j", and the unit tangent vector T = \(\frac{\bf v}{||\bf v||}\) by "vec;tnv(v,t)/lnt(tnv(v,t));v;(b*t-b*sin(t))*i+(b-b*cos(t))*j". Note that \({\bf v}(t)=\frac{d\bf r}{dt}=\frac{d\bf r}{ds}\frac{ds}{dt}={\bf T}||{\bf v}(t)||\).

The code pattern "vec;tnv(x(t)*i + y(t)*j + z(t)*k, t)" calculates the tangent vector, where "x(t)i + y(t)j + z(t)k" represents a vector-valued function. Also use "dif; x(t)i + y(t)j + z(t)k; t; n" to find the nth derivative of r(t).

Attention "dif(x(t)*i + y(t)*j + z(t)*k, t)" returns the first derivative of x(t)*i + y(t)*j + z(t)*k, whose result is not a vector even though it is written as a linear combination of i, j, k. The code "vec;tnv(x(t)*i + y(t)*j + z(t)*k, t)" returns a tangent vector, which can be combined with other vector operations.

Examples
(1) The tangent vector of r(t) = cos(t)i + sin(t)j + tsin(2t)k can be obtained by vec;tnv(cos(t)*i+sin(t)*j+t*sin(2*t)*k,t) ||, and a vector parametrization of the tangent line at t = π/4 can be calculated by r(π/4) + tr'(π/4) by vec;u+v;u;2**(1/2)*i+2**(1/2)*j+(pi/4)*k;v;t*(-2**(1/2)*i+2**(1/2)*j+k) ||. Check dif(cos(t)*i+sin(t)*j+t*sin(2*t)*k,t) ||.
(2) The tangent vector of ⟨t, t², t³⟩ can be got by vec;tnv(t*i+t**2*j+t**3*k,t) ||, and a vector parametrization of the tangent line at t = 2 can be calculated by r(2) + tr'(2) by vec;u+v;u;2*i+4*j+8*k;v;t*(i+4*j+12*k) ||.
(3) The tangent vector of ⟨t²-t; t²; t²+t⟩ at the point (0, 1, 2) can be got by vec;tnv((t**2-t)*i+t**2*j+(t**2+t)*k,t) ||, and a parametrization for the tangent line by vec;u+v;u;j+2*k;v;t*(i+2*j+3*k) ||.
(4). Let r(t) = ⟨t³, tsin(2t), log(3t)⟩. Then find r'(t) by dif;t^3*i+t*sin(2*t)*j+log(3*t)*k;t ||, r''(t) by dif;t^3*i+t*sin(2*t)*j+ log(3*t)*k;t;2 ||, and r''(3) dif;t^3*i+t*sin(2*t)*j+log(3*t)*k;t;3;rt;3 ||.

Examples Let r₁(t) = ti + t²j + t³k, r₂(t) = sin(t)i + sin(t²)j + sin(t³)k.
(1) If r(t) = r₁(t) · r₂(t), find r'(t) by vec;dif((t*i+t**2*j+t**3*k)*(sin(t)*i+sin(t**2)*j+sin(t**3)*k),t) ||, and r''(t) by vec;dif((t*i+t**2*j+t**3*k)*(sin(t)*i+sin(t**2)*j+sin(t**3)*k),t,2) ||.
(2) If s(t) = r₁(t) × r₂(t), find s'(t) by vec;dif((t*i+t**2*j+t**3*k)**(sin(t)*i+sin(t**2)*j+sin(t**3)*k),t) ||.
(3) If g(t) = exp(t), find r₁(g(t)) by dif;exp(t)*i+exp(2*t)*j+exp(3*t)*k;t ||.

Example Find a parametrization of the tangent line to r(t) = ⟨1 - t², t + 1, 2t³⟩ at t = 1. First get the tangent vector r'(1) by dif;(1-t**2)*i+(t+1)*j+2*t**3*k;t;1;rt;1 ||, and then the tangent line is r(1) + tr'(1) by vec;u+v;u;2*j+2*k;v;t*(-2*i+j+6*k) ||, which is ⟨0, 2, 2⟩ + t⟨-2, 1, 6⟩. Or x = -2t, y = 2 + t, z = 2 + 6t.

Higher order derivatives of vector functions: [dif(x(t)*i + y(t)*j + z(t)*k, t, n)]
To find the second and third derivative r''(t) and r'''(t), you need to repeat the process twice or three times by the "tnv" module. Use "dif(x(t)i + y(t)j + z(t)k, t, n)" to directly get the nth derivative of a vector-valued function r(t) = x(t)i + y(t)j + z(t)k. To evaluate the tangent vector at t = c, use "dif(x(t)*i + y(t)*j + z(t)*k, t,1,rt,c)".

Examples Given r(t) = ln(t)i + tj + k/t, vec;tnv(log(t)*i+t*j+k/t,t) || returns the first derivative r'(t), then vec;tnv(i/t+j-k/t**2,t) || gets the second r''(t), and vec;tnv(-i/t**2+2*k/t**3,t) || yields the third r'''(t). Use dif(i/t+j-k/t**2,t) || dif(i/t+j-k/t**2,t,2) || dif(i/t+j-k/t**2,t,3) ||.

Examples Let r(t) = ti - t²j + t³k Show D_t(r × r') = r × r'' by vec;dif((t*i-t**2*j+t**3*k)**(tnv(t*i-t**2*j+t**3*k,t)),t) || vec;u**v;u;t*i-t**2*j+t**3*k;v;dif(t*i-t**2*j+t**3*k,t,2) ||. The two resulting vectors are equivalent.

Tangent vectors, normal vectors, binormal vectors

Normalizing a tangent vector r'(t), we get a unit tangent vector denoted by T(t) = r'(t)\(/\)||r'(t)||. Similarly, the unit normal vector is calculated by N(t) = T'(t)\(/\)||T'(t)||, assuming the curve r(t) is smooth and ||r'(t)|| ≠ 0, which implies the tangent vector cannot be a zero vector. The binormal vector B(t) is orthogonal to both T(t) and N(t), and thus B(t) = T(t) × N(t).

Examples Find T(t), N(t), and B(t) for r(t) = ti + cos(t)j + sin(t)k. First, obtain r'(t) and ||r'(t)|| for T(t) by vec;tnv(t*i+cos(t)*j+sin(t)*k,t) || vec;u/lnt(u);u;i-sin(t)*j+cos(t)*k || vec;u;u;i/2**(1/2)-sin(t)*j/2**(1/2)+cos(t)*k/2**(1/2) ||. Second, get T'(t) and ||T'(t)|| for N(t) by vec;tnv(i/2**(1/2)-sin(t)*j/2**(1/2)+cos(t)*k/2**(1/2),t) || vec;u/lnt(u);u;-sin(t)*j/2^(1/2)-cos(t)*k/2^(1/2) ||. Third, calculate the cross product of T(t) × N(t) for B(t) by vec;t**n;t;i/2**(1/2)-sin(t)*j/2**(1/2)+cos(t)*k/2**(1/2);n;-cos(t)*j-sin(t)*k ||.

Examples Find T for r(t) = ⟨t, t², t³⟩ and t = 1 by vec;tnv(t*i+t**2*j+t**3*k,t) || vec;u/lnt(u);u;i+2*t*j+3*t^2*k ||. Substitute t = 1 into the resulting vector to obtain T = \(\frac{1}{\sqrt{14}}\)⟨1, 2, 3⟩.

The three unit vectors are mutually orthogonal to each other in space forming a set (T, N, B) called the Frenet frame, and they determine three planes. The plane spanned by T × N is called osculating plane that is normal to B, the plane by N × B is the normal (or curvature) plane normal to T, and the plane by T × B is normal to N and related to torsion. Note that to find an equation of a normal or osculating plane, it is not necessary to calculate the unit vectors T or B. It suffices to use r' for T and r' × r'' for B because they are parallel.

Examples Let r(t) = ⟨cos(t), sin(t), t⟩.
(1) Find the normal plane at t = π/4 by vec;tnv(cos(t)*i+sin(t)*j+t*k,t) || for T. Since T is normal to the plane spanned by N and B, the normal plane is by vec;(-i/2+j/2+2**(1/2)*k/2)*((x-2**(1/2)/2)*i+(y-2**(1/2)/2)*j+pi*k/4) ||.
(2) Find the osculating plane, first get B = T'/||T'|| by vec;tnv(-sin(t)*i+cos(t)*j+k,t) ||, which is -cos(t)i - sin(t)j. The osculating plane is vec;(-2**(1/2)*i/2-2**(1/2)*j/2)*((x-2**(1/2)/2)*i+(y-2**(1/2)/2)*j+pi*k/4) ||.

Arc length and speed Assume r(t) is differentiable and r'(t) is continuous on [a, b]. Then the length of the curve r(t) is \(\int_a^b||{\bf r}'(t)||dt=\int_a^b\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}dt\) for a ≤ t ≤ b, where || r'(t) ||dt is called arc length differential. If r(t) is the path traced out by a moving particle, then ||r'(t)|| is the speed at time t.

Examples
(1) Find the length and speed of a path parametrization r(t) = cos(t)i + tj + sin(t)k for 0 ≤ t ≤ π by vec;lnt(tnv(cos(t)*i+t*j+sin(t)*k,t)) ||, which is \(\sqrt{2}\), and the arc length by itg;2**(1/2);t;0;pi ||, which is \(\sqrt{2}\)π.
(2) Find an arc length parametrization for r(t) = ⟨2t - 1, 3t, 4-5t⟩. Get ||r'(t)|| by vec;lnt(tnv((2*t-1)*i+3*t*j+(4-5*t)*k,t)) ||, which is \(\sqrt{38}\). So \(s(t)=∫_0^t\sqrt{38}du=\sqrt{38}t\). To find the inverse of \(s(t), s=\sqrt{38}t⇒t=g(s)=\frac{s}{\sqrt{38}}\). So a parametrization r₁(s) = r(g(s)) = ⟨\(\frac{2s}{\sqrt{38}}-1,\frac{3s}{\sqrt{38}},4-\frac{5s}{\sqrt{38}}\)⟩. Verify the unit of ||r'₁(s)|| = 1 by vec;lnt(tnv((2*s/38**(1/2)-1)*i+3*s*j/38**(1/2)+(4-5*s/38**(1/2))*k,s)) ||, which is 1.

Curvature: [vec;tnv(r(t), t)], [vec;dif((u,u, r(t)), t, 2)]

The curvature of r(t) measures how much a curve bends or how fast it changes directions at a given point. It can be calculated by κ(t) = ||T'(t)||\(/\)||r'(t)|| = ||r'(t) × r''(t)||\(/\)||r'(t)||³.

To find curvature, first get the tangent vector r'(t) of a given parametrization r(t), and then enter r'(t) into the module again for r''(t). Submitting these two results into the formula, you can calculate the curvature of r(t). Or use "dif(x(t)*i+y(t)*j+z(t)*k, t)" and "dif(x(t)*i+y(t)*j+z(t)*k,t,2)" to get the first and second derivatives and submit them into the curvature formula.

Examples
(1) Find the curvature of r(t) = ti - tj + exp(t)k. First get r'(t) by dif;t*i-t*j+exp(t)*k;t ||. Then get r''(t) by dif;i-j+exp(t)*k;t;2 ||. Submitting these two vectors in the formula, vec;u**v;u;i-j+exp(t)*k;v;exp(t)*k || vec;lnt(u**v)/lnt(u)^3;u;i-j+exp(t)*k;v;exp(t)*k ||, which gives the curvature 2**(1/2)*exp(t)/(2+exp(2*t))**1.5 ||. If t = 0, the curvature is 2.0**(1/2)*exp(0)/(2+exp(2*0))**1.5 ||, which is about 0.27.

(2) Get the curvature of r(t) = ⟨cos(t), sin(t), t²⟩ by vec;tnv(cos(t)*i+sin(t)*j+t**2*k,t) || vec;tnv(-sin(t)*i+cos(t)*j+2*t*k,t) || vec;lnt(u**v)/lnt(u)^3;u;-sin(t)*i+cos(t)*j+2*t*k;v;-cos(t)*i-sin(t)*j-2*k ||, which gives \(\frac{\sqrt{5+4t^2}}{(1+4t^2)^{\frac{3}{2}}}\). If t = 1, the curvature is about 0.2683.

Examples
(1) In special situations, a straight line has 0 curvature because it does not bend or turn. Verify this fact with any line such as vec;tnv((1+2*t)*i-t*j+(2+2*t)*k,t) ||. Then ||T'(t)|| = 0 by vec;tnv(2*i/3-j/3-2*k/3,t) ||.

(2) A function y = f(x) has curvature \(\frac{|f''(x)|}{[1+f'(x)]^{\frac{3}{2}}}\). Let r(t) = ⟨x, f(x)⟩. Then r'(t) = ⟨1, f'(x)⟩, and r''(t) = ⟨0, f''(x)⟩. Submitting these results in the formula, you get the desired result.

Examples The curvature of \(y = x^2\) is \(\frac{2}{(1+4x^2)^{\frac{3}{2}}}\) by vec;tnv(x*i+x**2*j,x) || vec;tnv(i+2*x*j,x) || vec;lnt(u**v)/lnt(u)^3;u;i+2*x*j;v;2*j ||, and k(0) = 2.

Tangential and normal components of acceleration: [ vec; prj(avec, vvec) ]

Let r(t) be the position function of a moving object. Then the velocity v(t) = r'(t), speed v(t) = ||r'(t)||, and the acceleration vector, a(t) = r''(t) = a_T·T + a_N·N, where T and N are the unit tangent and unit normal vector of r(t). So a(t) lies in the plane spanned by T and N, and a_T and a_N are called tangential and normal component of acceleration.

Compute a_T = a · T = \(\frac{\bf{a · v}}{||\bf{v}||}\), a_N = a · N = \(\frac{||\bf{v × a}||}{||\bf{v}||}\). As a result, a_T·T is the projection of a(t) along v(t), and the normal component a_N·T = a - a_T·T.

Examples Suppose r(t) = ti + t²j + t²k. Then v(t) = r'(t) = i + 2tj + 2tk by vec;tnv(t*i+t**2*j+t**2*k,t) ||, ||v(t)|| = \(\sqrt{1+8t^2}\), and a(t) = 2j + 2k by vec;tnv(i+2*t*j+2*t*k,t) ||, a_T = \(\frac{8t}{\sqrt{1+8t^2}}\) by vec;a*v/lnt(v);a;2*j+2*k;v;(i+2*t*j+2*t*k) ||. Similarly, find a_N by vec;lnt(a**v)/lnt(v);a;2*j+2*k;v;i+2*t*j+2*t*k ||, which gives a_N = \(\frac{\sqrt{8}}{\sqrt{1+8t^2}}\). Get a_TT by vec;prj(2*j+2*k,i+2*t*j+2*t*k) ||, which is \(\frac{8\rm{t}}{1+8\rm{t}^2}\)⟨1; 2t; 2t⟩. Thus, a_N·T is the difference 2j + 2k - \(\frac{8\rm{t}}{1+8\rm{t}^2}\)⟨1; 2t; 2t⟩ by vec;2*j+2*k-prj(2*j+2*k,i+2*t*j+2*t*k) ||. Thus, verify a(t) = 2j + 2k = a_TT + a_NN = \(\frac{8t}{\sqrt{1+8t^2}}\)T + \(\frac{\sqrt{8}}{\sqrt{1+8t^2}}\)N.

Examples Let r(t) = ⟨e^t, e^-2t, t²⟩. Find a_T and a_N at t = 0. First, v(t) and ||v|| by vec;tnv(exp(t)*i+exp(-2*t)*j+t**2*k,t) ||, and a(t) by dif;exp(t)*i+exp(-2*t)*j+t**2*k;t;2 ||. Substituting t = 0 in each result, you get v(0) = ⟨1, -2, 0⟩, ||v(0)|| = \(\sqrt{5}\), and a(0) = ⟨1, 4, 2⟩. So a_T = -7/\(\sqrt{5}\) by vec;a*v/lnt(v);a;i+4*j+2*k;v;(i-2*j) ||, and a_N = \(2\sqrt{70}/5\) by vec;lnt(v**a)/lnt(v);a;i+4*j+2*k;v;i-2*j ||. Find a_T·T = ⟨-7/5, 14/5, 0⟩ by vec;prj(i+4*j+2*k,i-2*j) ||, a_N·N = ⟨4/3, 10/3, 8/3⟩ by vec;i+4*j+2*k-prj(i+4*j+2*k,i-2*j+2*k) ||, and N = a_N·N/a_N = \(\frac{2}{3\sqrt{5}}\)⟨2, 5, 4⟩.

Examples Let r(t) = cos(t)i + sin(t)j + tk. Then v(t) is by vec;tnv(cos(t)*i+sin(t)*j+t*k,t) ||, a(t) is by vec;tnv(-sin(t)*i+cos(t)*j+k,t) ||, the tangential component a_T = 0 is by vec;a*v/2**(1/2);a;-cos(t)*i-sin(t)*j;v;-sin(t)*i+cos(t)*j+k ||, which is 0, and the normal component a_N = 1 is by vec;lnt(a**v)/lnt(v);a;-cos(t)*i-sin(t)*j;v;-sin(t)*i+cos(t)*j+k ||.