Tutorial 5

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Navigation problems: adding velocity vectors given as compass headings
• Work and torque: the dot product in engineering
• Directional derivatives and the gradient
• Tangent planes, normal lines and unit normals
• Divergence and curl and their physical meaning

Introduction

Headings are measured clockwise from North. A velocity of speed at compass heading hdg therefore has components

$(\text{East},\text{North})=\big(s\sin(\text{hdg}),\;s\cos(\text{hdg})\big)$

SymPy's atan2(E, N) inverts this and returns a bearing that we shift into $[0^\circ,360^\circ)$.

import sympy as sy
from sympy.abc import x, y, z, t, r
from sympy import sqrt, sin, cos, tan, pi, atan, atan2, acos, exp, ln, Matrix, simplify
sy.init_printing(use_latex=True)

def velocity(speed, hdg):
    """Velocity vector (East, North) for a given speed and compass heading in degrees."""
    return Matrix([speed*sin(hdg*pi/180), speed*cos(hdg*pi/180)])

def gradient(f, vars=(x, y, z)):
    return Matrix([sy.diff(f, v) for v in vars])

def divergence(F, vars=(x, y, z)):
    return sum(sy.diff(F[i], vars[i]) for i in range(len(vars)))

def curl(F):
    return Matrix([sy.diff(F[2], y) - sy.diff(F[1], z),
                   sy.diff(F[0], z) - sy.diff(F[2], x),
                   sy.diff(F[1], x) - sy.diff(F[0], y)])

Question 1

A boat heads for Dunes Beach. Part (a) gives the heading $250^\circ$; in part (b) a wind of $0.8$ knot at $200^\circ$ acts while the skipper keeps the same heading.

boat = velocity(2.4, 250)                  # intended course, 2.4 knots
wind = velocity(0.8, 200)                  # wind acting on the boat
display([sy.N(v, 4) for v in boat])        # [-2.255, -0.821]  (East, North)
display([sy.N(v, 4) for v in wind])        # [-0.274, -0.752]

res = boat + wind
display([sy.N(v, 4) for v in res])         # [-2.529, -1.573]

speed = sqrt(res.dot(res))
display(speed, sy.N(speed, 4))             # 2.978 knots

hdg = atan2(res[0], res[1]) * 180/pi       # compass heading, may be negative
display(sy.N(hdg, 6), sy.N(hdg + 360, 6))  # -121.88 ... 238.12 degrees

a. The intended heading is $250^\circ$.

c. The altered speed is $2.978$ knots at a heading of $238.12^\circ$.

Question 2

Work done by a constant force, $W=\mathbf{F}\cdot\overrightarrow{PQ}=|\mathbf{F}||\overrightarrow{PQ}|\cos\theta$.

a. $12000\ \text{J}$ of work to pull a sled $200\ \text{m}$ with a $150\ \text{N}$ force.

theta = acos(12000/(150*200))
display(sy.N(theta*180/pi, 6))             # 66.42 degrees

b. $\mathbf{F}=5i$ moving an object from the origin to $(1,1)$.

F = Matrix([5, 0]); d = Matrix([1, 1])
display(F.dot(d))                          # 5 J

c. A 1000 lb force on a sail moves a boat 1 mile.

W = 1000 * 5280 * cos(60*pi/180)
display(W)                                 # 2640000 ft.lb

note

The downloadable solution PDF writes $\mathbf{F}=5\mathbf{j}$, the vertical field. The displacement is purely horizontal, so the correct field is $\mathbf{F}=5\mathbf{i}$ and the work is the $x$-component alone: $5\ \text{J}$.

Question 3

A bolt is tightened with a $20\ \text{N}$ force at $60^\circ$ to a $30\ \text{cm}$ wrench.

tau = 0.3 * 20 * sin(60*pi/180)
display(tau, sy.N(tau, 4))                 # 5.196 N.m

The magnitude of the torque is $5.196\ \text{N}\cdot\text{m}$ (about $5.2$). The two pictures in part (b) have the same magnitude but opposite direction — one tightens the bolt, the other loosens it.

note

Torque is measured in $\text{N}\cdot\text{m}$, so the magnitude $5.196$ rounds to $5.2\ \text{N}\cdot\text{m}$ (not joules).

Question 4

a. Is there a direction in which the rate of change of $T=2xy-yz$ at $P(1,-1,1)$ equals $-3\,^\circ\text{C/ft}$?

T = 2*x*y - y*z
grad_T = gradient(T)
display(grad_T)                            # <2y, 2x - z, -y>

g = grad_T.subs({x: 1, y: -1, z: 1})
display(g)                                 # <-2, 1, 1>
display(sqrt(g.dot(g)), sy.N(-sqrt(g.dot(g)), 6))     # sqrt(6), -2.449

The directional derivative lies in $\left[-|\nabla T|,\,|\nabla T|\right]=\left[-\sqrt{6},\sqrt{6}\right]\approx[-2.449,2.449]$. Since $-3$ is outside that interval, no such direction exists.

b. For the paraboloid $2z=x^{2}+y^{2}$ at $(1,3,5)$, write the surface as $f=x^{2}+y^{2}-2z=0$.

f = x**2 + y**2 - 2*z
grad_f = gradient(f)
display(grad_f)                            # <2x, 2y, -2>

n = grad_f.subs({x: 1, y: 3, z: 5})
display(n)                                 # <2, 6, -2>

# tangent plane:  n . (X - P) = 0
plane = n[0]*(x - 1) + n[1]*(y - 3) + n[2]*(z - 5)
display(simplify(plane))                   # 2x + 6y - 2z - 10
display(simplify(plane/2))                 # x + 3y - z - 5

# unit normal vector
display(n / sqrt(n.dot(n)))                # <1, 3, -1>/sqrt(11)

So $\nabla f=2x\,\boldsymbol{i}+2y\,\boldsymbol{j}-2\,\boldsymbol{k}$, $\nabla f(1,3,5)=\langle2,6,-2\rangle$, the tangent plane is $2x+6y-2z=10$, the unit normal is $\dfrac{\langle1,3,-1\rangle}{\sqrt{11}}$, and the normal line is $x=1+2t,\ y=3+6t,\ z=5-2t$.

note

The gradient of $f=x^{2}+y^{2}-2z$ is $\langle 2x,\,2y,\,-2\rangle$, so the $z$-component of $\nabla f$ is the constant $-2$, not $-2z$.

c.i. $2x^{2}+y^{2}-z^{2}=-3$ at $(1,2,3)$.

f = 2*x**2 + y**2 - z**2 + 3
g = gradient(f).subs({x: 1, y: 2, z: 3})
display(g)                                        # <4, 4, -6>
display(simplify(g[0]*(x-1) + g[1]*(y-2) + g[2]*(z-3)))   # 4x + 4y - 6z + 6

Tangent plane $4x+4y-6z+6=0$; normal line $x=1+4t,\ y=2+4t,\ z=3-6t$.

c.ii. $30-y^{2}-z^{2}=x^{2}$ at $(1,-2,5)$.

f = x**2 + y**2 + z**2 - 30
g = gradient(f).subs({x: 1, y: -2, z: 5})
display(g)                                        # <2, -4, 10>
display(simplify(g[0]*(x-1) + g[1]*(y+2) + g[2]*(z-5)))   # 2x - 4y + 10z - 60

Tangent plane $2x-4y+10z-60=0$; normal line $x=1+2t,\ y=-2-4t,\ z=5+10t$.

Question 5

a. Divergence of $F=\dfrac{x}{y}\boldsymbol{i}+(2x-3y)\boldsymbol{j}$

display(simplify(divergence(Matrix([x/y, 2*x - 3*y]), (x, y))))   # 1/y - 3

The divergence is a scalar, so it describes whether the field is expanding (positive) or compressing (negative) at a point.

b. Curl of $F=x\boldsymbol{i}-y\boldsymbol{j}+z\boldsymbol{k}$

display(curl(Matrix([x, -y, z])))          # <0, 0, 0>

The curl is the zero vector, so this field is irrotational.

Question 6

A $2.5\ \text{N}$ force is applied perpendicular to a $15\ \text{cm}$ spanner.

tau = 2.5 * 0.15 * sin(pi/2)               # 15 cm = 0.15 m
display(tau, tau*100)                      # 0.375 N.m = 37.5e-2 N.m

The torque is $0.375\ \text{N}\cdot\text{m}=37.5\times10^{-2}\ \text{N}\cdot\text{m}$, acting anticlockwise.

note

The spanner is $15\ \text{cm}=0.15\ \text{m}$, which is what gives the quoted answer $37.5\times10^{-2}\ \text{N}\cdot\text{m}$.

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Extra Learning Resources

Key Concepts to Master

1. Relative velocity: add the vectors tip-to-tail, then convert back to speed and heading
2. Work: $W=\mathbf{F}\cdot\mathbf{d}=|\mathbf{F}||\mathbf{d}|\cos\theta$
3. Torque: $\tau=|\mathbf{r}||\mathbf{F}|\sin\theta$, in $\text{N}\cdot\text{m}$
4. Gradient: points in the direction of steepest increase, with magnitude equal to that maximum rate
5. Divergence vs curl: divergence is a scalar (source/sink), curl is a vector (rotation)

SymPy Resources

• SymPy trigonometric functions — atan2, acos, sin, cos
• SymPy Matrices — dot, cross, subs
• SymPy subs — substituting the evaluation point into a gradient

Common Pitfalls

• Compass headings run clockwise from North, so the East component uses $\sin$ and the North component uses $\cos$ — the opposite of the usual maths convention
• atan2(E, N) returns angles in $(-180^\circ,180^\circ]$; add $360^\circ$ if you want a bearing
• A directional derivative can never exceed $|\nabla f|$ in magnitude
• Torque uses $\sin\theta$ (cross product) while work uses $\cos\theta$ (dot product)