MATH 251: Calculus 3, SET8

13: Vectors Functions

13.4: Motion in Space; Velocity and Acceleration

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

1. [878/4]

Plane curve with velocity and acceleration vectors
syms t positive % wlog
r(t) = [t^2 1/t^2], v(t) = diff(r(t),t), a(t) = diff(v(t), t)
r(t) = 
v(t) = 
a(t) = 
r1 = r(1), v1 = v(1), a1 = a(1)
r1 = 
v1 = 
a1 = 
speed = norm(v(t))
speed = 
%
figure
fplot(t^2, 1/t^2, [-0.5 2.5], 'LineWidth', 2); grid on; hold on
plot(1, 1, 'go', 'MarkerFaceColor', 'g', 'MarkerSize', 10)
quiver(1,1, 2,-2, 'r', 'LineWidth', 2, 'AutoScale', 'off', 'MaxHeadSize', 0.5) % velocity vector
quiver(1,1, 2,6, 'm', 'LineWidth', 2, 'AutoScale', 'off', 'MaxHeadSize', 0.2) % acceleration vector
axis equal; axis([-2 6 -2 8])
xlabel('x'); ylabel('y')
title('SET8, 878/4')

2. [878/8]

Space curve with velocity and acceleration vectors
syms t
r(t) = [t 2*cos(t) sin(t)], v(t) = diff(r(t),t), a(t) = diff(v(t), t)
r(t) = 
v(t) = 
a(t) = 
r0 = r(0), v0 = v(0), a0 = a(0)
r0 = 
v0 = 
a0 = 
speed = norm(v(t))
speed = 
%
figure
fplot3(t, 2*cos(t), sin(t), [-2*pi 2*pi], 'LineWidth', 2); grid on; hold on
plot3(0, 2, 0, 'go', 'MarkerFaceColor', 'g', 'MarkerSize', 10)
quiver3(0,2,0, 1,0,1, 'r', 'LineWidth', 2, 'AutoScale', 'off', 'MaxHeadSize', 0.8) % velocity vector
quiver3(0,2,0, 0,-2,0, 'm', 'LineWidth', 2, 'AutoScale', 'off', 'MaxHeadSize', 0.8) % acceleration vector
axis equal; axis([-2 6 -2 2 -1 1])
xlabel('x'); ylabel('y'); zlabel('z')
title('SET8, 878/8')
view(-26,26)

3. [878/12]

Given the position function, we compute velocity, acceleration, and speed once again.
syms t positive
r = [t^2 2*t log(t)]
r = 
v = diff(r,t), a = diff(v, t)
v = 
a = 
speed = simplify(norm(v))
speed = 

4. [878/16]

Given acceleration plus initial velocity and position, we obtain the position function.
See the Hand Solutions for how to do this step by step. (Or look at 861/42 in Section 13.2.)
Here we dispatch the problem in fully automatically by solving a system of differential equations with initial conditions, as in Math 308.
syms x(t) y(t) z(t)
Dx = diff(x,t), Dy = diff(y,t), Dz = diff(z,t)
Dx(t) = 
Dy(t) = 
Dz(t) = 
deqs = [diff(x,t,2) == sin(t), ... % components of acceleration
diff(y,t,2) == 2*cos(t), ...
diff(z,t,2) == 6*t]
deqs(t) = 
conds = [Dx(0)==0, Dy(0)==0, Dz(0)==-1,... % initial velocity components
x(0)==0, y(0)==1, z(0)==-4] % initial position components
conds = 
soln = dsolve(deqs, conds) % Solve the initial value problem.
soln = struct with fields:
y: [1×1 sym] x: [1×1 sym] z: [1×1 sym]
x(t) = soln.x, y(t) = soln.y, z(t) = soln.z % Assign position components.
x(t) = 
y(t) = 
z(t) = 
r(t)= [x(t) y(t) z(t)] % Aggregate to obtain position vector function.
r(t) = 

5. [878/18]

Same drill as 878/16 along with a graph of the position function, a space curve.
syms x(t) y(t) z(t)
Dx = diff(x,t), Dy = diff(y,t), Dz = diff(z,t)
Dx(t) = 
Dy(t) = 
Dz(t) = 
deqs = [diff(x,t,2) == t, ... % components of acceleration
diff(y,t,2) == exp(t), ...
diff(z,t,2) == exp(-t)]
deqs(t) = 
conds = [Dx(0)==0, Dy(0)==0, Dz(0)==1, ... % initial velocity components
x(0)==0, y(0)==1, z(0)==1] % initial position components
conds = 
soln = dsolve(deqs, conds) % Solve the initial value problem.
soln = struct with fields:
y: [1×1 sym] x: [1×1 sym] z: [1×1 sym]
x(t) = soln.x, y(t) = soln.y, z(t) = soln.z % Assign position components.
x(t) = 
y(t) = 
z(t) = 
r(t)= [x(t) y(t) z(t)] % Aggregate to obtain position vector function.
r(t) = 
%
figure
fplot3(t^3/6, exp(t)-t, 2*t+exp(-t), [-3.5 3.5], 'LineWidth', 2)
axis equal; % axis([-3 3 1 9 1 7])
xlabel('x'); ylabel('y'); zlabel('z')
title('SET8, 878/18')
% view(-26,26)

6. [878/26]

A formula on page 873 relates distance with initial velocity and angle of elevation.
syms d g v0 alpha
assume(0 < alpha < pi/2)
eq1 = d == v0^2 * sin(2*alpha) / g
eq1 = 
eq2 = subs(eq1, [d g v0], [3000 98/10 400])
eq2 = 
ang = solve(eq2, alpha)
ang = 
ang_rad_appx = double(ang)
ang_rad_appx = 2×1
0.0924 1.4784
deg = round(rad2deg(ang_rad_appx), 2) % angles of elevation in degrees
deg = 2×1
5.2900 84.7100
% The latter is preferable unless shooting over a hill.

7. [879/36]

8. [879/38]

Tangential and normal components of acceleration are computed.
syms t
r = [2*t^2 2/3*t^3 - 2*t 0]
r = 
v = diff(r,t), a = diff(v,t)
v = 
a = 
aT = simplify(dot(v,a) / norm(v))
aT = 
aN = simplify(norm(cross(v,a)) / norm(v))
aN = 4

9. [879/40]

Tangential and normal components of acceleration are computed.
syms t
r = [t 2*exp(t) exp(2*t)]
r = 
v = diff(r,t), a = diff(v,t)
v = 
a = 
aT = simplify(dot(v,a) / norm(v))
aT = 
aN = simplify(norm(cross(v,a)) / norm(v))
aN = 

10. [879/42]

Tangential and normal components of acceleration are computed.
syms t positive
r = [1/t 1/t^2 1/t^3]
r = 
v = diff(r,t), a = diff(v,t)
v = 
a = 
aT(t) = simplify(dot(v,a) / norm(v))
aT(t) = 
aN(t) = simplify(norm(cross(v,a)) / norm(v))
aN(t) = 
aT1 = simplify(aT(1)), aN1 = simplify(aN(1))
aT1 = 
aN1 = 
chk1 = simplify(aT1 == sym(-50/sqrt(14))) % formats
chk1 = TRUE
chk2 = simplify(aN1 == sym(sqrt(38/7))) % alternative
chk2 = TRUE