These problems are done with the CAS. See Hand Solutions for details.

The arc length of a space curve is computed.

syms t

r = [cos(t) sin(t) log(cos(t))]

Dr = diff(r,t)

f = simplify(sum(Dr.^2)) % or sec(t)^2

g = sec(t) % sqrt(f) for 0 <= t <= pi/4

L = int(g, t, 0, pi/4)

L_appx = double(L) % in cm

L_appx = 0.8814

Parameterize the curve of intersection then proceed as in #1.

syms t x y z

elliptical_cylinder = 4*x^2 + y^2 == 4

plane = x + y + z == 2

r = [cos(t) 2*sin(t) 2-cos(t)-2*sin(t)]

Dr = diff(r,t)

g = sqrt(simplify(sum(Dr.^2)))

L = int(g, t, 0, 2*pi)

L_appx = double(L) % in cm, to 4 decimal places

L_appx = 13.5191

The unit tangent, unit normal, and curvature function for a space curve are computed.

syms t positive

r = [t^2 sin(t)-t*cos(t) cos(t)+t*sin(t)]

Dr = diff(r,t)

mag_Dr = sqrt(simplify(sum(Dr.^2)))

T = Dr / mag_Dr % unit tangent vector

DT = diff(T,t)

mag_DT = sqrt(simplify(sum(DT.^2)))

N = DT / mag_DT % unit normal vector

kappa = mag_DT / mag_Dr % curvature

The curvature to a space curve at a point corresponding to is computed using Formula 10 on page 864.

syms t positive

r = [t^2 log(t) t*log(t)]

Dr(t) = diff(r,t), D2r(t) = diff(Dr,t)

v = Dr(1), a = D2r(1)

kappa = simplify(norm(cross(v,a)) / norm(v)^3)

kappa_appx = double(kappa)

kappa_appx = 0.3043

We use Formula 11 on page 865 to analyze the curvature of a curve in the xy-plane.

syms x positive

f = log(x), Df = diff(f,x), D2f = diff(Df,x)

k = simplify(abs(D2f) / (1 + Df^2)^(3/2))

Dk = simplify(diff(k,x))

xM = solve(Dk == 0, x) % Maximum curvature occurs at x = sqrt(2)/2 via FDT for extr vals.

L = limit(k, x, inf) % As x approaches infinity, the curvature approaches zero.

A parametric space curve and its curvature are graphed for on separate plots.

syms t positive

r = [t-sin(t) 1-cos(t) 4*cos(t/2)]

Dr = diff(r,t);

mag_Dr = sqrt(simplify(sum(Dr.^2)));

T = Dr / mag_Dr; % unit tangent vector

DT = diff(T,t);

mag_DT = sqrt(simplify(sum(DT.^2)));

N = DT / mag_DT; % unit normal vector

kappa = mag_DT / mag_Dr % The curvature is obtained. (Intermediate results suppressed.)

%

figure

fplot3(t-sin(t), 1-cos(t), 4*cos(t/2), [0 8*pi], 'LineWidth', 2)

xlabel('x'); ylabel('y'); zlabel('z')

title('SET8, 869/36: space curve, viewed from above')

view(0,90) % View looking down the positive z-axis, straight onto the xy-plane.

figure

fplot(t, kappa, [0 8*pi], 'r', 'LineWidth', 2)

axis([0 8*pi 0 7]); grid on

xlabel('x'); ylabel('y')

title('SET8, 869/36: curvature')

ax = gca;

S = sym(ax.XLim(1):pi:ax.XLim(2));

ax.XTick = double(S);

ax.XTickLabel = arrayfun(@texlabel,S,'UniformOutput',false);

% OBSERVATION:

% The curvature is minimal at odd multiples of pi

% and becomes infinite at even multiples of pi.

Unit tangent, normal, and binormal vectors are computed at a point corresponding to on a space curve.

syms t

r = [cos(t), sin(t), log(cos(t))]

Dr = diff(r,t), Dr = [-sin(t) cos(t) -tan(t)]

mag_Dr = sec(t)

T = simplify(Dr / mag_Dr) % unit tangent vector

DT = simplify(diff(T,t))

mag_DT = sqrt(simplify(sum(DT.^2)))

N = DT / mag_DT % unit normal vector

% The unit binormal vector is B = T x N.

T0 = subs(T, t, 0), N0 = subs(N, t, 0), B0 = cross(T0,N0) % T, N, B for t = 0

A point on a space curve is found whose normal plane is parallel to a given plane.

syms c t x y z

given_plane = 6*x + 6*y - 8*z == 1 % normal vector to given plane: [6 6 -8]

r(t) = [t^3 3*t t^4], Dr(t) = diff(r(t),t)

% The tangent vector to the space curve is perpendicular to the normal plane.

% To be parallel to the given plane, it must be a multiple of [6 6 -8].

%

eqs = Dr(t) == c*[6 6 -8]

s = solve(eqs, [c t])

sc = s.c, st = s.t

P = r(-1), n = 1/2*[6 6 -8] % The desired point is (-1, -3, 1), at which norm vect is [3 3 -4].

syms t real

syms a b positive

r = [a*cos(t) a*sin(t) b*t]

Dr = diff(r,t), D2r = diff(Dr,t), D3r = diff(D2r,t)

mag_Dr = sqrt(simplify(sum(Dr.^2)));

T = Dr / mag_Dr % unit tangent vector

DT = diff(T,t)

mag_DT = sqrt(simplify(sum(DT.^2)));

N = DT / mag_DT % unit normal vector

kappa = mag_DT / mag_Dr % constant curvature

c = cross(Dr,D2r)

tau = simplify(dot(c, D3r) / norm(c)^2) % constant torsion

syms t real

r = [sinh(t) cosh(t) t] % = [0 1 0] when t = 0

Dr = diff(r,t), D2r = diff(Dr,t), D3r = diff(D2r,t)

%

kappa(t) = simplify(norm(cross(Dr,D2r)) / norm(Dr)^3)

kappa0 = kappa(0) % Curvature is 1/2 at (0,1,0).

%

c = cross(Dr,D2r)

tau(t) = simplify(dot(c, D3r) / norm(c)^2)

tau0 = tau(0) % Torsion is -1/2 at (0,1,0).