Parametric Equations of Lines and Curves in the Plane

Parametric equations of 2D curves in the plane: [pc2; x(t); y(t); itv = (a, b)]

Parametric equations provide a useful tool for describing curves and graphs not only for functions and equations, but also for paths traced out by moving particles. The location of a particle's motion in a plane can be expressed as points (x(t), y(t)), where x(t) and y(t) are parametric equations of time t over an interval I = [a, b].

Use "pc2; x(t); y(t); itv=(a,b)" to plot curves of parametric equations, where "pc2" (parametric curve 2d) is the operation name, 'x(t)' corresponds to the expression of x(t), 'y(t)' to y(t), and "itv" to the interval I (default interval is [-10, 10]).

Examples
(1) pc2;t;2*(t-1)+3 || pc2;t;-(t+2)-3 || pc2;t;0.8*(2*t-7)-5 ||.
(2) Horizontal lines y = b parametrized by x = t, y = b and graphed by pc2;t;2 || pc2;t;-3 || pc2;2*t;-5 || pc2;-t;3.4 ||.
(3) Vertical lines x = a can be parametrized by x = a, y = t and graphed by pc2;1;t || pc2;-2;t || pc2;-4;-t ||.

Parametric equations of lines (point-slope form) A line through a point P(a, b) with a slope m has an equation of the form y = m(x - a) + b.
(1) A natural parametrization of the line is x = t, y = m(t - a) + b for all t.
(2) Or the parametric equations can be x = a + t, y = b + mt.
(3) Or a parametrization can be x = a + ht, y = b + kt for m = k/h.

Examples A line of slope 0.8 through a point (3, -2) has an equation y = 0.8(x - 3) + 2. Plot it by plt;0.8*(x-3)+2 ||. Graph a natural parametrization by pc2;t;0.8*(t-3)+2 ||, or by pc2;3+t;-2+0.8*t ||, or pc2;3+5*t;-2+4*t ||.

Parametric equations of lines (two-points form) A parametrization of the line through two points P(a, b) and Q(c, d) is x = a + (c - a)t, y = b + (d - b)t for all t. The line segment PQ has the same set of parametric equations, but the parameter t is restricted at 0 ≤ t ≤ 1.

Example A parametrization of the line through P(1, -2) and Q(-3, 5) is x(t) = 1 - 4t, y(t) = -2 - 7t. Graph it by pc2;1-4*t;-2-7*t ||.

Examples Find a parametrization for a line 2x - 3y = 5 through the point (1, -1). You need the slope or another point.
(1) Since the line y = 2x/3 - 5/3, its slope is 2/3, and a parametrization of point-slope form is x = 1 + 3s, y = -1 + 2s by pc2;1+3*s;2*s-1 ||.
(2) Since x = 7, y = 3, or the point (7, 3) is also on the line, a parametrization (two-points form) can be x = 1 + 6t, y = -1 + 4t by pc2;1+6*t;-1+4*t ||.

Parametric equations of curves in planes: [pc2; x(t); y(t)], [pc2; t; f(t)], [plt; f(x)]; [imf; f(x, y)]

The set of points (x, f(x)) defines the graph of y = f(x) for all x in the domain of f, and the set of points (x(t), y(t)) defines a parametric curve in a direction as t increases. Parametric curves in the plane include graphs for which the vertical line test fails, or graphs (level curves) that are defined implicitly in the equation f(x, y) = c for some constant c.

Parametrization of explicit functions y = f(x) An explicit function y = f(x) can be parametrized as (t, f(t)), so you can use both "plt" and "pc2" modules to graph it. To use "pc2" module, let x = t, y = f(t), and plot it by "pc2; t; f(t)".

Examples pc2;t;t**2 || plt;x**2 || produce the same graph of y = x². Similarly, pc2;t;1/(1+t**2) || plt;1/(1+x**2) || yield the same graph.

Graphs of implicit functions For implicit curves f(x, y) = 0 for which the vertical line test fails, graph it by "pc2" with proper parametric equations.

Examples
(1) The curve x² = y can be parametrized by x = t², y = t by pc2;t**2;t ||.
(2) The parametric equations of a circle of radius 5 can be x = 5cos(t), y = 5sin(t). Graph by pc2;5*cos(t);5*sin(t) ||.
(3) The parametric equations of x³ = y² can be x = t², y = t³. Graph by pc2;t**2;t**3 ||.

Examples pc2;t**2-4;t**2+2*t || pc2;(1-t**2)/(1+t**2);2*t/(1+t**2) || pc2;4*sin(t)**3;4*cos(t)**3 || pc2;t**3;t**2-3 || pc2;t-sin(t);1-cos(t) || pc2;2*t-2*sin(t);2-2*cos(t) || pc2;2*t-sin(t);2-cos(t) || pc2;cos(t);exp(sin(t)) || pc2;-2*t**3+5*t**2-1;2*t**3-2*t**2+5*t-1 ||.

For curves that involve functions of nth roots (e.g., \(x^{\frac{m}{n}}\)), the "plt" module only plots values in positive domains. In this case, use "pc2" or "imf" modules to graph the curves in both positive and negative domains.

Examples
(1) Graph \(y=\sqrt[3]{x}\) by plt;x**(1/3) ||, pc2;t**3;t || or by imf;x-y**3 ||.
(2) Plot \(y=\sqrt[3]{x^2} ≥0\) by plt;x**(2/3) ||, pc2;t**3;t**2 || or imf;x**2-y**3 ||. Notice that if you parametrize \(y=x^{\frac{1}{3}}\) by \((t,t^{\frac{1}{3}})\) and use the "pc2" module, you will only get the positive portion of the graph. But parametrizing it by \((t^3,t)\), which implies \(y=x^{\frac{1}{3}}\), and you will see the entire graph in both positive and negative domains.

Examples Check pc2;t**4;t**3 || pc2;t**5;t**2 || pc2;t;exp(sin(t)) || pc2;t**3;t**(-2) || pc2;t**3;t**2-t || pc2;t**4+t**3;t**3-1 || pc2;t**2;t || pc2;(t-2)**2;t+1 || pc2;-(t+3)**2;t-2 || pc2;cos(u)**3;sin(u)**3;itv=(0,6.3) || pc2;3*cos(t);2*exp(sin(t));0;6.3 ||.

Conic Sections

Conic sections The general quadratic equation of a conic section is \(ax^2+bxy+cy^2+dx+ey+f=0\), where \(a,b,c,d,e,f\) are constants and \(a,b,c\) are not all 0. This equation may have degenerate cases, or the equation does not define a conic section, because the set of solutions may be empty, a single point, a single line, a pair of parallel lines, or a pair of intersecting lines.

For example, \(x^2+y^2=0\) is just a point (0, 0), there is no solution to \(x^2+y^2=-7\), and \(y^2-y=0\) defines two horizontal lines \(y=0\) and \(y=1\).

Conic sections defined by this equation are not necessarily in standard position. They may not be centered at the origin, and their focal and minor axis may be rotated relative to the coordinate axes. If the equation is non-degenerate and the cross term \(bxy=0\) or \(b=0\), then we can write the equation in standard form (translated) by completing the square, which means the axes of a translate of a conic section are parallel to the coordinate axes.

Suppose the general equation is non-degenerate and \(D=b^2-4ac\). Then the equation defines an ellipse or circle if \(D < 0\), a hyperbola if \(D>0\), and a parabola if \(D=0\).

Standard form of conic sections If the cross product term \(bxy\) is not shown in the general equation, then \(Ax^2 + By^2 + Cx + Dy + E = 0\) becomes a stand form of conic sections, where \(A, B, C, D, E\) are constants and \(A, B\) are not both 0. The graph of the equation, which is in the standard position, may be a circle, ellipse, parabola or hyperbola except some degenerate cases.

Circle \(A = B\): General form is \(Ax^2 + Ay^2 + Cx + Dy + E = 0\) or \(x^2 + y^2 + C_1x + D_1y + E_1 = 0\). Standard form is \(x^2 + y^2 = r^2\) or \((x - h)^2 + (y - k)^2 = r^2\) with center \((h,k)\).

Ellipse \(AB > 0, A ≠ B\): Coefficients A and B have the same signs and standard form is \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\).

Parabola \(AB = 0\): If \(A = 0\) and \(B ≠ 0\), general form is \(By^2 + Cx + Dy + E = 0\), and standard form is \((x - h)^2 = 4p(y - k)\). If \(B = 0\) and \(A ≠ 0\), general form is \(Ax^2 + Cx + Dy + E = 0\), and standard form is \((y - k)^2 = 4p(x - h)\). If \(A=B=0\), it becomes a line.

Hyperbola \(AB < 0\): \(A\) and \(B\) have different signs and standard form is \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\) or \(\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1\).

The standard equations of conic sections are implicitly defined in a bivariate quadratic equation, which is difficult to graph by explicit functions. But their parametric equations are straightforward to obtain and graph by "pc2" module. Or use "imf" module to graph a conic by its defining equation.

Parametrizations of Conic Sections

circles The parametric equations of a circle of radius \(r\) centered at \((h, k)\) are \(x(t) = h + r\cos(t)\), \(y(t) = k + r\sin(t)\).

Examples pc2;cos(t);sin(t) || pc2;3*cos(t);3*sin(t) || pc2;5*sin(t);5*cos(t) || pc2;2+3*cos(t);3+3*sin(t);pt=(2,3) || pc2;-4+2*cos(t);-1-2*sin(t);pt=(-4,-1) || pc2;3*sin(x);3*cos(x);itv=(0,6.3) || pc2;t+1/t;t-1/t || pc2;2+3*t;1-4*t ||.

ellipses The parametric equations of an ellipse centered at \((h, k)\) are in form of \(x(t) = h + a\cos(t)\), \(y(t) = k + b\sin(t)\).

Examples pc2;2*cos(t);3*sin(t) || pc2;4*sin(t);2*cos(t) || pc2;3*sin(u);3-2*cos(u);itv=(0,6.3) || pc2;-2+5*cos(s);3+2*sin(s);itv=(0,6.3) || pc2;3*cos(z);5*sin(z);pt=[(0,0),(0,4),(0,-4)] || pc2;4*cos(x);3*sin(x);pt=[(-2.646,0),(2.646,0)] || pc2;4*cos(x);3*sin(x);pt=[(-2.6,0),(2.6,0)];ln=[(6,-8),(6,8)];ln=[(-6,-8),(-6,8)] ||.

hyperbolas The parametric equations of a hyperbola centered at \((h, k)\) are in form of \(x(t) = h + a\sec(t)\), \(y(t) = k + b\tan(t)\), or \(x(t)=at+\frac{a}{t},y(t)=bt-\frac{b}{t}\). You may also parametrize one branch of hyperbola by \((h+a\cosh(x),k+b\sinh(x))\) and the other by \((h-a\cosh(x),k+b\sinh(x))\).

Examples imf;x**2/4-y**2/5-1 || imf;x**2-2*y**2+x-y || imf;x**2/3-y**2-2*x+3*y-1 || imf;y**2/4-x**2/2-1 || imf;y**2-x**2-y-4 ||.

Examples pc2;2*sec(y);2.2361*tan(y);pt=[(3,0),(-3,0)] || pc2;3*sec(y);4*tan(y);itv=(-1.56,1.56);pt=(5,0);ln=[(1.8,-10),(1.8,10)] || pc2;3*sec(y);4*tan(y);pt=[(5,0),(-5,0)];ln=[(1.8,-10),(1.8,10)];ln=[(-1.8,-10),(-1.8,10)] || pc2;3*sec(y);4*tan(y);ln=[(-9,-12),(9,12)];ln=[(-9,12),(9,-12)] || pc2;2+tan(x);-1-sec(x) || pc2;3*sec(y);2*tan(y) || pc2;-sec(s);tan(s) || pc2;2*tan(t);-sec(t) || pc2;t+1/t;t-1/t || pc2;2*t-2/t;3*t+3/t || pc2;1+t+1/t;2+t-1/t ||.

parabolas Use "plt" to graph parabolas with vertices at \((h, k)\) if they have an explicit form of \(y=f(x)=\frac{1}{4p}(x-h)^2+k\). Or use "pc2" module to graph other parabolas in form of \((x-h)^2=4p(y-k)\). The general parametric equations for this form are \(x(t)=h+4pt^2\), \(y(t)=k+t\).

Examples
(1) Plot \(y=0.5(x-1)^2-3\) by plt;0.5*(x-1)**2-3;pt=(1,-2.5) ||, or pc2;t;0.5*(t-1)**2-3;pt=(1,-2.5) ||.
(2) Graph y² = - 3x by pc2;-t**2/3;t ||.

Examples Use "imf" module to graph the equation (x-1)² - 2y = 0 by plt;0.5*(x-1)**2 || imf;(x-1)**2-2*y ||. Or use the "pc2" module to generate their graphs and foci by pc2;t**2;t-2;pt=(0.25,-2) || pc2;3-0.2*t**2;t-1;pt=(1.75,-1);ln=[(4.25,-10),(4.25,10)] || pc2;t;0.25*t**2;pt=(0,1);ln=[(-8,-1),(8,-1)] || pc2;1+0.5*t**2;t+2;pt=(1.5,2);ln=[(0.5,-4),(0.5,10)] ||.

Examples Graph the equation (x-1)² - 3y² = 4 by imf;(x-1)**2-3*y**2-4 ||, or by pc2;1+2*(t+1/t);2/3**(1/2)*(t-1/t) ||.

Derivatives of parametric equations: [tnv; x(t)i + y(t)j; t], [dif(y(t), t)/dif(x(t), t)], [dif(dif(y(t), t)/dif(x(t), t), t)/dif(x(t), t)]

For a parametric curve \((x(t),y(t))\), if both \(x(t)\) and \(y(t)\) are differentiable at \(t\) and \(x'(t)≠0\), then the derivative \(y'=\frac{dy}{dx}=\frac{y'(t)}{x'(t)}\) and \(y''=\frac{d}{dx}\frac{dy}{dx}=\frac{\frac{d}{dt}y'}{x'(t)}=\frac{x'(t)y''(t)-y'(t)x''(t)}{x'(t)^3}\), where \(x'(t)=\frac{dx}{dt},y'(t)=\frac{dy}{dt}\). If the curve is a path traced by a moving particle, \(y'\) is ratio of vertical and horizontal velocities.

Use "tnv; x(t)i + y(t)j; t" to find derivatives y' and y'' of parametric equations. First obtain the tangent vector ⟨x'(t), y'(t)⟩ of the parametric equations, and then substitute these two values in the formulas (ratio) for y' and y''. The "tnv" function requires the parametric equations to be written as a vector expression in form of x(t)i + y(t)j with the parameter t (or any other name but not conflict with names of basis vector i, j, k). Refer to vectors and tangent vectors for more information.

Attention Also use the "dif" function to get y' by "dif(y(t), t)/dif(x(t), t)" and y'' by "dif(dif(y(t), t)/dif(x(t), t), t)/dif(x(t), t)".

Note that the magnitude (length) \(\sqrt{x'(t)^2+y'(t)^2}\) of a tangent vector serves as an integrand for finding arc length of parametric curves. See more details in the next section.

Examples
(1) Find y' for x(t) = cos³(t), y(t) = sin³(t) by tnv;cos(t)**3*i+sin(t)**3*j;t ||. Substituting the results in the ratio, you will have y' = -tan(t). Or dif(sin(t)**3,t)/dif(cos(t)**3,t) ||.
(2) To find y'', check tnv;-3*sin(t)*cos(t)**2*i+3*cos(t)*sin(t)**2*j;t ||, and substitute the results in the ratio for y'' = sec⁴(t)csc(t)/3. Or by dif(-tan(t),t)/dif(cos(t)**3,t) || dif(dif(sin(t)**3,t)/dif(cos(t)**3,t),t)/dif(cos(t)**3,t) ||.

Examples If x = 2s² + 4, y = 3s³-5s, find y' by tnv;(2*s**2+4)*i+(3*s**3-5*s)*j;s ||, which is (9s² - 5)/4s, and y'' by tnv;4*s*i+(9*s**2-5)*j;s || (4*s*18*s-(9*s**2-5)*4)/(4*s)**3 ||, which is (9s² + 5)/(16s³).

Arc length: [itg; \(\sqrt{x'(t)^2+y'(t)^2}\); t; a; b], [lit; 1; x(t)i + y(t)j; t; a; b]

The arc length is given by \(s=\int_a^b\sqrt{x'(t)^2+y'(t)^2}dt\), and the different \(ds=\sqrt{x'(t)^2+y'(t)^2}dt\) is called arc length differential. In most cases, the integral need to be evaluated numerically. The Cartesian formula of the arc length \(L=\int_a^b\sqrt{1+f'(x)}dx\) is a special case of the formula for a continuously differentiable function \(y=f(x)\). So finding arc length is to integrate the length of a tangent vector over a parameter domain.

Example To find the arc length of an astroid x(t) = cos³(t), y(t) = sin³(t) for t from 0 to π/2, obtain the derivatives by tnv;cos(t)**3*i+sin(t)**3*j;t ||, substitute results in the formula and use itg;3*sin(t)*cos(t);t;0;pi/2 ||, or approximate the integral by nit;3*sin(t)*cos(t);t;0;pi/2;20 ||, and you will see the result (arc length) is 3/2.

Examples It is hard to evaluate exact arc length due to radical forms involved in integrals, but use numerical integration to approximate it by "nit".
(1) Find the length of y = x³ for x ∈ [0, 3] nit;(1+9*x**4)**(1/2);x;0;3;50 ||.
(2) Find the length of the curve ⟨\(e^t,t^2\)⟩ for \(0≤t≤2\) by nit;(exp(2*t)+4*t^2)^(1/2);t;0;2;50 || pc2;exp(t);t^2 ||.

Cycloid The parametric equations of a cycloid generated by a circle of radius R are x(t) = R(t - sin(t)), y(t) = R(1 - cos(t)), and the length for t from 0 to 2π is 8R by the formula. To see this result, x'(t)² + y'(t)² = 2R²(1 - cos(t)) = 4R²sin(t/2)². For 0 ≤ t ≤ 2π, the square root of 4R²sin(t/2)² gives 2Rsin(t/2) and \(∫_0^{2π}2R\sin\frac{t}{2}dt\) = 8R.

Examples
(1) If R = 1, tnv;(t-sin(t))*i+(1-cos(t))*j;t ||, the arc length, which is 8, can be obtained by itg;(2-2*cos(t))**(1/2);t;0;2*pi ||. But this integral is hard to evaluate. Use numerical approximation by nit;(2-2*cos(t))**(1/2);t;0;2*pi;20 ||, and you will see the results will be very close to 8.
(2) Similarly, R = 2, itg;2*(2-2*cos(t))**(1/2);t;0;2*pi || nit;2*(2-2*cos(t))**(1/2);t;0;2*pi;20 ||.
(3) If R = 3, check itg;3*(2-2*cos(t))**(1/2);t;0;2*pi || nit;3*(2-2*cos(t))**(1/2);t;0;2*pi; 20 ||.

Attention Use scalar line integral to calculate arc length. The parametrization ⟨x(t), y(t)⟩ for the curve is the same, but the integrand is equal to 1. So the code pattern is "lit; 1; x(t)*i+y(t)*j; t; a; b", where t is the parameter, and (a, b) is the integration limits, or the start and end points of the arc.

Examples
(1) The length of a parametric curve x(t) = 3t - 2, y(t) = 5 - 4t from t = 0 to t = 3 by lit;1;(3*t-2)*i+(5-4*t)*j;t;0;3 ||.
(2) The length of (3t² - 2, 4t² + 5) from t = 1 to t = 4 by lit;1;(3*t**2-2)*i+(4*t**2+5)*j;t;1;4 ||.
(3) The length of the unit circle (cos(t), sin(t)) can be got by lit;1;cos(t)*i+sin(t)*j;t;0;2*pi ||.
(4) The length of (e^tsin(t), e^tcos(t)) for 0 ≤ t ≤ 2 by lit;1;exp(t)*sin(t)*i+exp(t)*cos(t)*j;t;0;2 ||.
(5) The length of (e^-tsin(t), e^-tcos(t)) for 0 ≤t ≤ 1 by lit;1;exp(-t)*sin(t)*i+exp(-t)*cos(t)*j;t;0;1 ||.
(6) The length of (2(cos(t)-tsin(t)), 2(sin(t)-tcos(t))) for 0 ≤ t ≤ 2π, lit;1;2*(cos(t)+t*sin(t))*i+2*(sin(t)-t*cos(t))*j;t;0;2*pi ||.

If \((x(t),y(t))\) is a path of motion, \(\frac{ds}{dt}=\sqrt{x'(t)^2+y'(t)^2}\) is the speed of a particle moving along the path. To find the maximum or minimum speed is to find extreme values of \(f(t)=x'(t)^2+y'(t)^2\).

Examples If the path is (2cos(t), 3sin(t)) for 0 ≤ t ≤ 2π, f(t) = 4sin²(t) + 9cos²(t), f'(t) = -10sin(t)cos(t), and t = 0, π/2, π, 3π/2, 2π are critical numbers. So the maximum speed is 3, and minimum 2. Check dif;4*sin(t)**2+9*cos(t)**2;t;1;cp ||.

Areas enclosed by parametric curves: [itg; y(t)x'(t); t; a; b], [itg; x(t)y'(t); t; c; d]

The area bounded by a parametric curve between \(t=a\) and \(t=b\) and the x-axis is \(\int_a^by(t)x'(t)dt\), and the area enclosed by the curve between \(t=c\) and \(t=d\) and the y-axis is \(\int_c^dx(t)y'(t)dt\).

Examples Find the area of an astroid x(t) = cos³(t), y(t) = sin³(t) for t from 0 to 2π. Graph pc2;cos(t)**3;sin(t)**3 ||, and area can be got by itg;-12*sin(t)**4*cos(t)**2;t;pi/2;0 ||.

Surface area of revolving a curve: [itg; \(2πy\sqrt{x'(t)^2+y'(t)^2}\); t;a;b], [itg; \(2πx\sqrt{x'(t)^2+y'(t)^2}\); t;c;d]
Let \((x(t),y(t))\) be a parametric curve for \(y(t)\ge 0\) and \(x(t)\) is increasing. Assume \(x'(t)\) and \(y'(t)\) are continuous and \(x'(t)^2+y'(t)^2>0\) for all \(t\in [a,b]\). Then the area of the surface obtained by rotating the curve about the x-axis for \(a\le t\le b\) is given by \(2\pi\int_a^by(t)\sqrt{x'(t)^2+y'(t)^2}dt\)\(=\int_a^b2\pi yds\), where \(ds=\sqrt{x'(t)^2+y'(t)^2}dt\) is the arc length differential. The surface area of revolving a parametric curve about the y-axis is \(\int_c^d2\pi xds\).

Attention Use scalar line integral to calculate the surface area obtained by resolving the parametric curve C(t) = ⟨x(t), y(t)⟩ about x-axis or y-axis by "lit; 2*pi*y; x(t)*i+y(t)*j; t; a; b; x; y" or "lit; 2*pi*x; x(t)*i+y(t)*j; t; c; d; x; y".

Examples
(1) The surface area of the cone generated by revolving the parametric curve (3t, 4t) from t = 0 to t = 2 about the x-axis is \(\int_0^22π(4t)5dt\) = 80π by lit;2*pi*y;3*t*i+4*t*j;t;0;2;x;y ||.
(2) The surface area of revolving the cycloid (t - sin(t), 1 - cos(t)) about the x-axis for t from 0 to 2π can be obtained by itg;2*pi*(1-cos(t))*2*sin(t/2);t;0;2*pi || because \(ds=\sqrt{2-2\cos(t)}dt=2\sin\frac{t}{2}dt\), and tangent vector by tnv;(t-sin(t))*i+(1-cos(t))*j;t ||.

Volume of a solid formed by revolving a parametric curve: [itg; πy(t)²x'(t); t; a; b], [itg; πx(t)²y'(t); t; c; d]
The volume obtained by rotating \(c(t)\) about the x-axis for \(a\le t\le b\) is given by \(V=\pi\int_a^by^2(t)dx=\pi\int_a^by^2(t)x'(t)dt\). The volume of a solid formed by revolving a parametric curve \(c(t)\) about the y-axis for \(c\le t\le d\) is given by \(V=\pi\int_c^d x^2(t)y'(t)dt\).

Examples Find the volume of ball formed by a circle of radius 3 revolving about the x-axis. It is known the volume is 36π. Apply the above formula to verify it. The parametric equations are (3cos(t), 3sin(t)). The volume is by itg;-27*pi*sin(t)**3;t;pi;0 ||.

Examples Find the volume of x³ = y², or ⟨t², t³⟩ revolving about the y-axis for 0 ≤ t ≤ 2. Check the graph by pc2;t**2;t**3;itv=(0,2) ||. The volume is \(\int_0^23t^6dt\) = 384/7 by itg;3*t**6;t;0;2 ||.

Polar coordinates system The polar coordinates \(P(r,\theta)\) are related to the Cartesian coordinates \((x, y)\) in the equations \(x=r\cos\theta, y=r\sin\theta,r^2=x^2+y^2\) and \(\tan\theta=\frac{y}{x}\).

Symmetry By symmetry of polar graphs, if a point (x, y) corresponds to (r, θ), then
(1) symmetric about x-axis: (x, -y) corresponds to (r, 2π - θ)
(2) symmetric about y-axis: (-x, y) to (r, π - θ)
(3) symmetric about origin or pole: (-x, -y) to (r, π + θ)
(4) symmetric about line y = x: (y, x) corresponds to (r, π/2 - θ)

Polar and Cartesian coordinates conversions: [plt;f(x)], [pc2; x(t); y(t); itv], [imf; f(x,y)]; [plt; f(x)]

One can convert points, functions and equations in polar coordinates to Cartesian coordinates and vice versa using these relations.

Examples
(1) The point (0, -1) corresponds to (1, 3π/2), and the point (-1, 1) corresponds to (\(\sqrt{2}\), 5π/4).
(2) The equation θ = π/4 converts to a line y = x, and θ = π/3 to y = \(\sqrt{3}\)x.
(3) The polar equation r = 2/(sin(θ) - 3cos(θ)) corresponds to the equation y - 3x = 2, and the circle r = 2 to x² + y² = 4.
(4) r = 2sin(θ) ⇒ r² = 2rsin(θ) converts to the circle x² + (y - 1)² = 1.

Examples
(1) After conversion, plot these curves by "plt", "imf" and "pc2" modules on the xy-plane. plt;x || plt;3**(1/2)*x || plt;2*x+3 || imf;x**2-y**2-4 || imf;x**2+(y-1)**2-1.
(2) Also graph the polar equation rcos(θ) = 5 ⇒ x = 5 by pc2;5;t ||.
(3) Graph rsin(θ) = -3 ⇒ y = -3 by pc2;t;-3 ||.
(4) Graph r² = 4rsin(θ) by imf;x**2+y**2-4*y || or pc2;2*cos(t);2+2*sin(t) || by first converting it to x² + y² - 4y.

Graph polar functions: [pol; f(θ), itv = (a, b)]

Explicit polar functions are in form of r = f(θ), where r is the radial coordinate, and θ the angular coordinate measured by radians. For explicit polar functions, use "pol; f(r); f(s); f(t)" to graph up to three polar curves. Also change the default θ interval to (a, b) by adding the keyword "itv = (a, b)" to the end.

Examples The constant polar function is r = c for some constant c. The graph of r = 1 is a unit circle centered at the pole (origin), and r = 2 is a circle of radius equal to 2. Check pol;1 || pol;2 || pol;-3 || pol;2;3;4 || pol;-1;-2;-3 ||.

Lines in polar coordinates The ray θ = c represents a line in polar system, and we convert it to a rectangular equation y = tan(c)x. But the equation θ = c is not in form of r = f(θ), and thus is hard to plot it by the "pol" module.

Horizontal and vertical lines A horizontal line y = b converts to a polar function rsin(θ) = b ⇒ r = bcsc(θ); the vertical line x = d corresponds to rcos(θ) = d ⇒ r = dsec(θ). So in polar system the graphs of polar functions r = bcsc(θ) and r = dsec(θ) are horizontal and vertical lines, respectively.

Examples
(1) The vertical lines x = 1, x = 2, and x = -3 can be obtained by pol;sec(t);itv=(-1.2,1.2) || pol;2*sec(t);itv=(-1.2,1.2) || pol;-3*sec(t);itv=(-1.2,1.2) ||.
(2) The horizontal lines y = 1, y = -3, y = 2 can be graphed by pol;csc(t);itv=(0.1,3) || pol;-3*csc(t);itv=(1,2) || pol;2*csc(t);itv=(0.1,3) ||.
(3) pol;2*csc(t);-2*csc(t); itv=(0.2,3) || pol;sec(t);-sec(t);itv=(-1.5,1.5) || pol;csc(t);2*csc(t);3*csc(t);itv=(0.2,3) || pol;sec(t);2*sec(t);3*sec(t);itv=(-1.5,1.5) ||. Be sure to provide a reasonable interval for your graph.

In general, the polar function r = d/(bsin(θ)-ccos(θ)) converts to the line by - cx = d. Using the module "pol", you can graph these lines in polar system. It is important to specify a proper interval or scale so that the features of a polar curve can be well displayed.

Examples
(1) The polar function r = sec(θ - π/4) represents a line that has distance 1 to the pole and rotates counterclockwise an angle π/4 from r = sec(θ), and can be graphed by pol;sec(t-pi/4);itv=(-0.5,2) ||.
(2) Graph r = 2sec(θ + pi/3) by pol;2*sec(t+pi/3);itv=(-2,0.5) ||.
(3) pol;2/(sin(t)-cos(t));itv=(-1.5,0.2) || pol;4/(2*sin(t)-3*cos(t));itv=(-1.3,0.7) || pol;3/(2*sin(t)-cos(t));itv=(-1.5,0.2) || pol;2*sin(t)*tan(t);itv=(-1.4,1.4) || pol;2+cos(2*t);-sin(t);2+sin(t);itv=(-7,7) ||.

Polar curves The function r = f(θ) can be any polar curve including lines, and the vertical line test does not apply to polar functions and graphs.

Examples
(1) circles pol;2 || pol;-3 || pol;5 || pol;2*sin(t) || pol;-2*cos(t) || pol;cos(t)+sin(t) || pol;2*cos(t)+3*sin(t) || pol;3*sin(t)-5*cos(t) || pol;cos(t)+2*sin(t) ||.
(2) roses the codes pol;sin(2*t) || pol;cos(3*t) || pol;sin(t/2) || pol;cos(2*t/3) || pol;2*cos(pi*t) ||.
(3) lines pol;2*sec(t) || pol;csc(t) || pol;1/(2*sin(t)+3*cos(t)) || pol;1/(sin(t)-2*cos(t)) || pol;1/(3*sin(t)-4*cos(t));itv=(-6,6).
(4) limacons/cardioids pol;1+cos(t) || pol;1+2*cos(t) || pol;1-2*sin(t) || pol;3+cos(t) || pol;1+sin(t)**2 || pol;2+2*cos(t)**2 || pol;2+3*cos(t) || pol;0.5 + cos(t) || pol;2+4*sin(t) ||.
(5) loops/spirals pol;exp(0.5*t);itv=(0,6.3) || pol;t;itv=(0,6.3) || pol;exp(0.5*t)*sin(t);itv=(0,6.3) || pol;t**(1/2)*cos(t);itv=(0,6.3) || pol;abs(cos(t))**(1/2)*(exp(sin(t)))**(1/2);itv=(0,6.3) || pol;0.5*cos(t)*exp(sin(t));itv=(-1.56,1.58) || pol;t*sin(t);itv=(-12.57,12.57) ||.

Examples If possible, solve an implicit polar equation for two polar functions, and plot them into two separate polar curves.
(1) Graph r² = 1+sin(θ) by pol;(1+sin(t))**(1/2);-(1+sin(t))**(1/2) ||.
(2) Graph r² = tan²(θ) by pol;tan(t);-tan(t);itv=(-1.5,1.5) ||.

Graph conic sections in polar functions: [pol; f(t); itv = (a, b)]

The unified definition of conic sections in polar functions is \(r=\frac{ed}{1\pm e\cos\theta}\) or \(r=\frac{ed}{1\pm e\sin\theta}\), where \(e>0\) is the eccentricity and \(d>0\) is related to the directrix x = ±d or y = ±d. It is important to provide an appropriate θ interval for each conics, because the fraction can be very large when the denominate is close to 0.

Examples
Ellipses The codes pol;0.5*2/(1+0.5*cos(t)) || pol;0.7*3/(1-0.7*cos(t)) || pol;0.3*5/(1+0.3*sin(t)) pol;0.8*2/(1-0.8*sin(t)) || pol;6/(3+2*sin(t)) || pol;1/(2-cos(t)) || give graphs of ellipses \(e < 1\).

Parabolas The codes pol;1/(1+cos(t));itv=(-2,2) || pol;1/(1-cos(t));itv=(1.3,5) || pol;2/(1-sin(t));itv=(3,6.4) || pol;3/(1+sin(t));itv=(0,3.2) || pol;5/(2+2*sin(t));itv=(0,3.2) || yield graphs of parabolas \(e=1\).

Hyperbolas The codes pol;2/(1+1.3*cos(t));itv=(-2,2) || pol;3/(1-1.5*cos(t));itv=(1.58,4.7) || pol;4/(1+1.6*sin(t));itv=(0,3.2) || pol;6/(1-1.4*sin(t));itv=(3.2,6.3) || produce the graphs of hyperbolas \(e> 1\).

When rotating conic sections, it is convenient to use polar equations, because the graph of r = f(θ - α) is easy to obtain by rotating the graph of r = f(θ) an angle of α counterclockwise about the origin.

Examples The resulting ellipse has a polar equation \(r=\frac{2}{2+\cos(\theta-\frac{\pi}{4})}\) after rotating the graph of ellipse \(r=\frac{2}{2+\cos(\theta)}\) counterclockwise about its focus an angle of \(\frac{\pi}{4}\) by pol;2/(2+cos(t));2/(2+cos(t-pi/4)) || pol;2/(2+cos(t));2/(2+cos(t-pi/4));2/(2+cos(t+pi/4)) ||.

Arc length for polar curves: [itg; \(\sqrt{r^2+(r')^2}\); θ; a; b]

The arc length of a smooth polar curve \(r=f(θ)\) for \(θ\) from \(a\) to \(b\) is \(\int ds = \int_a^b\sqrt{r^2+(r')^2}d\theta\), where \(r'=f'(\theta)\).
Examples
(1) The length of r = cos(t) from t = 0 to 2π can be obtained by itg;1;t;0;2*pi ||.
(2) The length of r = e^t from 0 to log(2) can be got by itg;exp(t)*2**(1/2);t;0;log(2) ||.
(3) The length of the cardioid r = 1 + cos(θ) from t = 0 to π is by itg;2*2*cos(t/2);t;0;pi ||.

Area enclosed by polar curves: [itg; r²/2; θ; a; b]

The area enclosed by a polar curve \(r=f(θ)\) for \(θ\) from \(a\) to \(b\) and two rays through the pole is \(\frac{1}{2}\int_a^br^2d\theta\).

Examples
(1) The codes itg;0.5*(1+cos(t))**2;t;0;2*pi || gives the area of the cardioid r = 1 + cos(θ).
(2) The codes itg;2*sin(t)**2;t;0;2*pi || yields the area 2π of the circle r = 2sin(θ).
(3) The area of \(r^2=\sin(θ)e^{\cos(θ)}\) for \(θ\) from 0 to \(π\) can be got by itg;0.5*sin(t)*exp(cos(t));t;0;pi || nit;0.5*sin(t)*exp(cos(t));t;0;pi;20 ||.
(4) The area of one petal of r = sin(2θ) is by itg;0.5*sin(2*t)**2;t;0;pi/2 ||, the area of one petal of r = sin(4θ) is by itg;0.5*sin(4*t)**2;t;0;pi/4 ||.
(5) The area enclosed by r = 2 + cos(θ) by pol;2+cos(t);itv=(0,6.3) || itg;0.5*(2+cos(t))**2;t;0;2*pi || nit;0.5*(2+cos(t))**2;t;0;2*pi;50 ||.

Surface area formed by revolving a polar curve: [itg; \(\pi r\sin\theta\sqrt{r+(r')^2}\); θ; a; b]

The surface area formed by resolving a polar curve \(r=f(θ)\) for \(θ\) from \(a\) to \(b\) about the polar x-axis is \(\int 2\pi yds= \int_a^b2\pi r\sin\theta\sqrt{r+(r')^2}d\theta\).

Example Find the surface area formed by revolving \(r = 1 + \cosθ\) about polar x-axis. The surface area involves an integral \(\int_0^{π}2π(1+\cosθ)\sinθ\sqrt{1+\cosθ+\sin^2θ}dθ\), which is hard to integrate. Use numerical integration by nit;2*pi*(1+cos(t))*sin(t)*(1+cos(t)+sin(t)**2)**(1/2);t;0;pi;40 ||, which is about 17.67.

Derivatives and equations of tangent lines for polar curves: [tnv; f(θ)cos(θ)i + f(θ)sin(θ)j; θ]

A polar curve \(r = f(θ)\) has parametric equations \(x(θ)=f(θ)\cos(θ), y(θ)=f(θ)\sin(θ)\), so the derivative of \(f(θ)\) at a point \((r, θ)\) is given by \(\frac{dy}{dx}=\frac{y'(θ)}{x'(θ)}=\frac{f(θ)\cos\theta+f'(θ)\sin\theta}{f(θ)'\cos\theta- f(θ)\sin\theta}\). Use 'tnv; x(θ)i + y(θ)j; θ' to first find the tangent vector (x'(θ), y'(θ)), and then substitute the results into the ratio.

Attention Use function call to get y' by dif(f(θ)sin(θ), θ)/dif(f(θ)cos(θ), θ), so you don't need enter the results of y'(θ) and x'(θ) for a ratio.

Examples
(1) The derivative of the polar curve r = sin(θ) can be got by tnv;sin(t)*cos(t)*i+sin(t)**2*j;t || 2*sin(t)*cos(t)/(cos(t)**2-sin(t)**2) ||. Note that the parametric equations of r = sin(θ) are x(θ) = sin(θ)cos(θ), y(θ) = sin²(θ). Use function call for y' by dif(sin(t)**2,t)/dif(sin(t)*cos(t),t) ||.
(2) The derivative of polar curve r = e^θ is by tnv;exp(t)*cos(t)*i+exp(t)*sin(t)*j;t || (sin(t)+cos(t))/(cos(t)-sin(t)). Or use function call dif(exp(t)*sin(t),t)/dif(exp(t)*cos(t),t) ||.
(3) The derivative of the polar curve r = cos(t) + sin(t) can be got by tnv;(cos(t)+sin(t))*cos(t)*i+(cos(t)+sin(t))*sin(t)*j;t || (cos(2*t)+sin(2*t))/(cos(2*t)-sin(2*t)) ||, or dif((sin(t)+cos(t))*sin(t),t)/dif((sin(t)+cos(t))*cos(t),t) ||.

Curvature of polar curves: [dif; f(θ); θ]

The curvature of a smooth polar curve \(r = f(θ)\) at a point \((r,θ)\) can be calculated by \(\kappa =\frac{r^2-rr''+2(r')^2}{[r^2+(r')^2]^{\frac{3}{2}}}\). Use "dif" module to find r' = f'(θ) and r'' = f''(θ), and substitute r, r' and r'' in the formula for curvature.

Examples
(1) Verify that a line r = csc(θ) has curvature 0 everywhere by dif;-csc(t)*cot(t);t || csc(t)**2-csc(t)*(csc(t)*(1+2*cot(t)**2))+2*csc(t)**2*cot(t)**2 ||.
(2) If r = sin(t), get the curvature by (sin(t)**2+sin(t)**2+2*cos(t)**2)/(sin(t)**2+cos(t)**2)**(3/2) ||.
(3) If r = θ, get the curvature by (t**2+2)/(t**2+1)**(3/2) ||.

Parametrizations of 3D curves and surfaces [ps3; x(u, v), y(u, v), z(u, v)], [pc3; x(t), y(t), z(t)]

Lines in spaces Find a parametrization for a line through two points, or a line through one point and parallel to a direction.

Two-point form A parametrization of a line through the points (m, n, k) and (a, b, c) is (1 - t)⟨m, n, k⟩ + t⟨a, b, c⟩.
Examples Parametrize a line through P(1, 0, -2) and Q(5, -3, 4) by (1 - t)⟨1, 0, -2⟩ + t⟨5, -3, 4⟩ = ⟨ 1 + 4t, -3t, 6t - 2⟩.

Point-direction form A line through the point (m, n, k) and parallel to the direction of the vector ⟨a, b, c⟩ has parametrizations x(t) = m + at, y = n + bt, z = k + ct.

Examples
(1) A parametrization of the line through (2, -6, 4) and parallel to ⟨2, 2, 1⟩ is x = 2 + 2t, y = 2t - 6, z = 4 + t. Use the "pc3" module to graph this lines by pc3;2+2*t;2*t-6;4+t ||.
(2) A line passes through the points (2, 1, 0) and (-3, 4, 7) can be graphed by pc3;2-5*t;1+3*t;7*t ||.
(3) pc3;2*t;3*t;t || 1+2*t;2-t;3-2*t || pc3;t;1;2 || pc3;1;t;3 || pc3;2;-2;t || pc3;2-3*t;4+2*t;5-3*t;0;1 ||.

Curves in spaces: [pc3; x(t); y(t); z(t)]

The parametric equations for space curves are (x(t), y(t), z(t)), and they can be graphed by the "pc3" module, which requires a function of one parameter for each coordinate. Use "pc3; x(t); y(t); z(t)" to plot space curves, where "pc3" (parametric curve 3d) is the operation name, and "x(t); y(t); z(t)" are the expressions representing each coordinate function. To change the default interval [-10, 10] for t, just add two numbers "a; b" for your interval [a, b] to the end.

Examples
(1) The helix can be got by pc3;cos(t);sin(t);t || pc3;t;cos(t);sin(t) || pc3;cos(t);2*t;3*sin(t) ||.
(2) pc3;t;t**2;t**3 || pc3;cos(t); sin(t);1-t || pc3;t;t**2;t || pc3;cosh(x);1;sinh(x) || pc3;cosh(x);sinh(x);3 || pc3;t;t**2;t**3 || pc3;sin(t);cos(t);cos(3*t) || pc3;sin(t);cos(t);sin(2*t);0;6.3 || pc3;cos(t);sin(t);cos(4*t);0;6.3 || pc3;cos(2*t);sin(t);cos(3*t);0;6.3 ||.

Planes in spaces Planes and surfaces are two dimensional graphs, so they needs to be parametrized by two parameters. The general equation for a plane is ax + by +cz = d, and they can be graphed by the "sf3" and "ps3" modules. However, the "sf3" module cannot be used to plot vertical planes such as y = 1 or x = 2. You need to find their parametrizations (x, 1, z) or (2, y, z) and graph them by the "ps3" module.

To use "ps3" (parametric surface 3d), enter "ps3; x(u, v); y(u, v), z(u, v)", where "x(u, v); y(u, v), z(u, v)" are the parametric equations for each coordinate function, and "u; v" are two distinct parameters. To change the default intervals of u and v, add "a; b" for u interval and "c; d" for v interval to the end.

Examples
(1) A plane x + 3y - z = 4 can be got by ps3;x;y;x+3*y-4 || sf3;x+3*y-4 ||.
(2) The codes ps3;x;y;(2*x+3*y-6)/5 || sf3;(2*x+3*y-6)/5 || plots the plane 2x + 3y - 5z = 6.
(3) The vertical planes x = 1, x = 0, x = 3, x = -4 can be got by ps3;1;y;z || ps3;0;y;z || ps3;3;y;z || ps3;-4;y;z ||. The vertical planes y = -2, y = 0, y = 3 can be got by ps3;x;-2;z || ps3;a;0;b || ps3;m;3;n ||.
(4) The horizontal planes z = -2, z = 0, z = 5 can be produced by ps3;x;y;-2 || ps3;u;v;0 || ps3;x;y;5 ||.

Surfaces in spaces Space surfaces are two-dimensional, so parametrizations of surfaces involve two parameters. The graph of z = f(x, y) is a 3d surface, and the set of points (x, y, f(x, y)) consists of the surface above or below the xy-plane, so a natural parametrization of the function z = f(x, y) is (x, y, f(x,y)).

Examples Check ps3;(2+cos(s))*cos(t);(2+cos(s))*sin(t);2+sin(s) || ps3;u*v;u**2+v**2;u**2-v**2 || ps3;2*u*sin(v);2*u*cos(v);u || ps3;x;y;x*y || ps3;2*cos(t);3*sin(t);z || ps3;u;v;u**2-v**2 ||.

We cannot visualize a 4d surface defined by w = f(x, y, z), but we can parametrize and graph a level surface implicitly defined by the equation f(x, y, z) = c for some constant c. Assume z = g(x, y) is a function of x and y on a domain in the xy-plane, then (x, y , g(x, y)) is a natural parametrization of the level surface. Similarly, (h(y, z), y, z) or (x, u(x, z), z) is a parametrization of an implicit surface defined on the yz-plane or xz-plane, respectively. For more complicated surfaces like quadratic surfaces, we usually describe them in cylindrical and spherical coordinates system.

Examples ps3;x+y**2;x;y || ps3;x;x**2*z;z || ps3;x;y;(x+y)/(x**2+y**2)**(1/2) ||.

Cylindrical and spherical coordinates A point (x, y, z) in Cartesian coordinates system corresponds to a point (rcos(θ), rsin(θ), z) in the cylindrical coordinates system, and to (ρsinφcosθ,ρsinφsinθ,ρcosφ) in the spherical coordinates system, where (r, θ) are polar coordinates. So cylindrical and spherical coordinates are an extension of polar coordinates to 3d space.

Using the relations r² = x² + y², tanθ = y/x, x = rcosθ = ρsinφcosθ, y = rsinθ = ρsinφsinθ, z = ρcosφ, r = ρsinφ, and ρ² = r² + z² = x² + y² + z², you can convert coordinates and equations from one system to another.

Examples
(1) If r = k > 0 is a sphere of radius k centered at the origin, θ = k is a half plane containing z-axis and making an angle k radian with the xz-plane, φ = k is a cone with vertex at the origin and making an angle k radian with positive z-axis.
(2) If x² + y² + 2z² = 3, the cylindrical equation is r² + 2z² = 3; the spherical equation is ρ² + z² = 3.

To graph cylindrical and spherical equations, convert them to rectangular equations and use the "sf3" or "ps3" module.

Examples
(1) To graph z = 3r²cosθsinθ, convert z = 3xy, and graph the saddle surface by sf3;3*x*y || ps3;x;y;x*y ||.
(2) To graph r = 2cosθ, convert it to x² + y² - 2x = 0, and graph the circular cylinder by ps3;1+cos(t);sin(t);z ||.
(3) Graph z² = 1 - r²sin²θ. Convert it to z² = 1 - y², and graph the cylinder by ps3;x;cos(t);sin(t) ||.

General equation for quadratic surfaces Quadratic surfaces are related to quadratic curves and equations, and parametrizations of quadratic surfaces involve cylindrical and spherical coordinates. In this section, we focus on the following groups of quadratic surfaces: spheres, ellipsoids, hyperboloids, elliptic cones, paraboloids, and quadratic cylinders.

Quadratic curves are related to conic sections, whose curves in R² are defined by quadratic equations in two variables. A quadratic surface in R³ is defined by a quadratic equation in three variables. The general equation of a quadratic surface in R³ is \(Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\), assuming the degree-2 coefficients \(A,B,C,D,E,F\) are not all 0.

Parametric equations for quadratic surfaces: [ps3; x(u, v), y(u, v), z(u, v)]
1. Quadratic cylinders: [ps3;a*cos(θ), b*sin(θ), z]

The following are various equations for quadratic cylinders in standard position.
Elliptic cylinder: \(\frac{x^2}{a^2} + \frac{y^2}{b^2}\) = 1 or \(\frac{x^2}{a^2} + \frac{z^2}{c^2}\) = 1 or \(\frac{y^2}{b^2} + \frac{z^2}{c^2}\) = 1.
Circular cylinder: \(x^2+y^2 = r^2\) or \(x^2+z^2 = r^2\) or \(z^2+y^2 = r^2\).
Hyperbolic cylinder: \(\frac{x^2}{a^2} - \frac{y^2}{b^2}\) = 1 or \(\frac{x^2}{a^2} - \frac{z^2}{c^2}\) = 1 or \(\frac{y^2}{b^2} - \frac{z^2}{c^2}\) = 1.
Parabolic cylinder: \(x^2=4pz\) or \(x^2=4py\); \(y^2=4px\) or \(y^2=4pz\); \(z^2=4px\) or \(z^2=4py\).

We use cylindrical coordinates to parametrize some quadratic cylinders, and then graph them by "ps3". Elliptic and circular cylinders can be parametrized by (acos(θ), bsin(θ), z), (acos(θ), y, csin(θ)), (x, bcos(θ), csin(θ)), and then use the "ps3" module to graph these elliptic cylinders.

Examples ps3;cos(t);sin(t);z || ps3;3*cos(t);0.5*sin(t);z || ps3;cos(t);y;2*sin(t) || ps3;x;2*cos(t);5*sin(t) ||.

Examples Parabolic cylinders can be parametrized by (t, t²/(4p), z), (t²/(4p), t, z), (x, t²/(4p), t) and graphs by ps3;t;t**2/4;z || ps3;t;y;t**2/8 || ps3;t**2/6;y;2*t || ps3;x;t**2/10;t ||.

Examples Parametrize hyperbolic cylinders by (acosh(t), bsinh(t), z), (-acosh(t), bsinh(t), z), (x, acosh(t), bsinh(t)), (acosh(t), y, bsinh(t)), (bsinh(t), acosh(t), z), and graphs by ps3;2*cosh(t);sinh(t);z || ps3;x;2*cosh(t);4*sinh(t) || ps3;cosh(t);y;5*sinh(t) || ps3;-4*cosh(t);sinh(t);z ||.

Examples Also parametrize hyperbolic cylinders by (asec(t), btan(t), z), (asec(t), y, btan(t)), and (x,asec(t), btan(t)), and graphs by ps3;2*sec(t);3*tan(t);z;-1;1.5 || ps3;x;sec(t);3*tan(t);-1;1.5 || ps3;4*sec(t);y;tan(t);-1;1.5 ||.

2. Ellipsoids and spheres: [ps3; a*cos(t)*sin(s); b*sin(t)*sin(s); c*cos(s)]
The standard equation for ellipsoid is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\). The trace is an ellipse in the plane \(z=z_0\) or \(y=y_0\) or \(x=x_0\). A parametrization for an ellipsoid is \((a\sin\phi\cos\theta, b\sin\phi\sin\theta,c\cos\phi)\). A sphere \(x^2+y^2+z^2=r^2\) is a special ellipsoid for \(a=b=c\).

Examples A sphere of radius 2 by ps3;2*sin(s)*cos(t);2*sin(s)*sin(t);2*cos(s) ||, and an ellipsoid by ps3;sin(s)*cos(t);2*sin(s)*sin(t);3*cos(s) ||.

3. Elliptic and hyperbolic paraboloids: [ps3; x; y; (x/a)^2 + (y/b)^2],[ps3; x; y; (x/a)^2 - (y/b)^2]
The equation of an elliptic paraboloid is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=z\), or \(\frac{x^2}{a^2}+\frac{z^2}{c^2}=y\), \(\frac{z^2}{c^2}+\frac{y^2}{b^2}=x\). The trace is an ellipse in the plane \(z=z_0\) and a parabola in \(x=x_0\) or \(y=y_0\). Elliptic paraboloids can be parametrized by \((x, y,\frac{x^2}{a^2}+\frac{y^2}{b^2})\), \((\frac{y^2}{b^2}+\frac{z^2}{c^2}, y, z), (x, \frac{x^2}{a^2}+\frac{z^2}{c^2}, z)\).

Examples Graph various elliptic paraboloids by ps3;x;y;x**2+y**2 || ps3;x;y;x**2+2*y**2 || ps3;x;3*x**2+2*z**2;z || ps3;3*y**2+z**2;y;z ||.

Remember parametrization of a curve or surface is not unique. So you can also parametrize elliptic paraboloids by \((a\sqrt{z}\cosθ,b\sqrt{z}\sinθ,z), (a\sqrt{y}\cosθ,y,b\sqrt{y}\sinθ), (x,a\sqrt{x}\cosθ,b\sqrt{x}\sinθ)\).

Examples ps3;z**(1/2)*cos(t);z**(1/2)*sin(t);z;0;4;0;6.3 || ps3;2*y**(1/2)*cos(t);y;y**(1/2)*sin(t);0;4;0;6.3 || ps3;x;2*x**(1/2)*cos(t);3*x**(1/2)*sin(t);0;4;0;6.3 ||.

The equation of a hyperbolic paraboloid is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=z\), or \(\frac{x^2}{a^2}-\frac{z^2}{c^2}=y\), or \(\frac{z^2}{c^2}-\frac{y^2}{b^2}=x\)). The trace is a hyperbola in plane \(z=z_0\) and a parabola in \(x=x_0\) or \(y=y_0\). Hyperbolic paraboloids can be parametrized by \((x, y,\frac{x^2}{a^2}-\frac{y^2}{b^2})\), \((\frac{y^2}{b^2}-\frac{z^2}{c^2}, y, z), (x, \frac{x^2}{a^2}-\frac{z^2}{c^2}, z)\).

Examples Graphs by ps3;x;y;x**2-y**2/4 || ps3;x;x**2/5-z**2/3;z || ps3;y**2/2-z**2/1.5;y;z ||.

4. Circular and elliptic cones: [azsin(θ); bzsin(θ); cz], [zsin(θ); zsin(θ); z]
The equation of an elliptic cone is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}\), or \(\frac{x^2}{a^2}+\frac{z^2}{c^2} =\frac{y^2}{b^2}\), or \(\frac{z^2}{c^2}+\frac{y^2}{b^2}=\frac{x^2}{a^2}\)). The trace is an ellipse in the plane \(z=z_0\) and a hyperbola in \(x=x_0\) or \(y=y_0\). Elliptic cones can be parametrized by \((az\cos\theta, bz\sin\theta,cz)\), \((ax, bx\sin\theta, cx\cos\theta), (ay\cos\theta, by, cy\sin\theta)\).

A circular cone is a special case of an elliptic cone with an equation similar to \(x^2 + y^2 = z^2\) and can be parametrized by ⟨\(z\cos(θ), z\sin(θ),z\)⟩.

Examples Graphs by ps3;0.8*z*cos(t);0.6*z*sin(t);z || ps3;0.6*z*cos(t);0.5*z*sin(t);z || ps3;1.5*y*cos(t);y;2*y*sin(t);-2;2 || ps3;x;x*cos(t);x*sin(t) ||.

5. Hyperboloids: [ps3;\(a\sqrt{z^2+1}\cos\theta; b\sqrt{z^2+1}\sin\theta; cz\)], [ps3; \(a\sqrt{z^2-1}\cos\theta; b\sqrt{z^2-1}\sin\theta; cz\)]
The equation of a hyperboloids (one sheet) is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1\), or \(\frac{x^2}{a^2}+ \frac{z^2}{c^2}-\frac{y^2}{b^2}=1\), or \(\frac{z^2}{c^2}+\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\). The trace is an ellipse in the plane \(z=z_0\) and a hyperbola in \(x=x_0\) or \(y=y_0\). Parametrize hyperboloids (one sheet) by \((a\sqrt{z^2+1}\cos\theta, b\sqrt{z^2+1}\sin\theta, cz),(ax, b\sqrt{x^2+1}\sin\theta,c\sqrt{x^2+1}\cos\theta), (a\sqrt{y^2+1}\cos\theta\), \(c\sqrt{y^2+1}\sin\theta)\).

Examples Graph hyperboloids (one-sheet) by ps3;0.5*(1+z**2)**(1/2)*cos(t);0.5*(1+z**2)**(1/2)*sin(t);z;-2;2;0;6.3 || ps3;0.5*(1+y**2)**(1/2)*cos(t);y;0.4*(1+y**2)**(1/2)*sin(t);-2;2;0;6.3 || ps3;x;0.6*(1+x**2)**(1/2)*cos(t);0.4*(1+x**2)**(1/2)*sin(t);-1.5;1.5;0;6.3 ||.

The equation of a hyperboloids (two sheets) is \(\frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), or \(\frac{x^2}{a^2}-\frac{z^2}{c^2}- \frac{y^2}{b^2}=1\), or \(\frac{y^2}{b^2}-\frac{x^2}{a^2}-\frac{z^2}{c^2}=1\)). The trace is an ellipse (or no trace) in the plane \(z=z_0\) and a hyperbola in \(x=x_0\) or \(y=y_0\). Parametrize hyperboloid (two sheets) by \((a\sqrt{z^2-1}\cos\theta, b\sqrt{z^2-1}\sin\theta, cz), (ax, b\sqrt{x^2-1}\sin\theta\), \( c\sqrt{x^2-1}\cos\theta),(a\sqrt{y^2-1}\cos\theta, by, c\sqrt{y^2-1}\sin\theta)\).

Examples Graph two sheets by ps3;0.4*(z**2-1)**(1/2)*cos(t);0.5*(z**2-1)**(1/2)*sin(t);z || ps3;0.5*(y**2-1)**(1/2)*cos(t);y;0.3*(y**2-1)**(1/2)*sin(t) || ps3;x;0.4*(x**2-1)**(1/2)*cos(t);0.6*(x**2-1)**(1/2)*sin(t) ||.