Name:
Math 1320 - Lab 10
uNiD:
Due: 11/14
READ BEFORE PROCEEDING: Today we will continue with some application problems in multivariate calculus. Don’t forget to write your reasoning (in COMPLETE and CONCISE sentences) behind
your approach to each problem so that someone who has taken Math 1310 would understand what you are
doing. Present your work in such a way that someone without access to the questions can read through your
solutions and understand it. To achieve this, set-up the problem in your own words and describe NEATLY
and LEGIBLY the procedures you take in solving the problem. Disjoint or unclear explanations may result
in lost points.
1. Design of train tracks. Suppose a train track weaves through a valley on a sinusoidal center-path
r(s) = hs, A sin(2πs), 0i, where s is in kilometers. On each turn, the track must be banked on an angle
θ appropriately so that passengers in the dining car won’t experience lateral forces from cornering that
would cause passengers to sway sideways, and their coffee cups to slosh (lawsuits!). That is, the force
from each turn that is experienced by passengers (and their beverages) must be matched by the lateral
component of the normal force from the floor of the train. The train’s speed v must remain constant at
all times.
In order to determine the banking angle, θ, consider the plane normal to r0 (s) for each s. We can
create local coordinate unit vectors u (horizontal) and w (vertical). Each passenger car experiences a
gravitational force Fg = −mgw. The normal force vector Fn from the track onto the passenger car vector
balances Fg so that there is no vertical motion on the track center. The normal vector Fn is oriented
normal to the bank of the track at an angle θ so Fn = h−mg tan(θ), mgi. The train’s acceleration is
characterized as
a = v 2 κN
because dv/dt = 0. The normal N is pointing in the u direction because there is no vertical acceleration
and no forward-backward track acceleration. To achieve no-slosh, the w normal force must match the
centripetal force, that is:
mv 2 κw = mg tan(θ)w
(a) Introduce and restate the problem in your own words.
(b) Determine the train track angle θ(s) for the track for every location s on the track. Note that
r(s) describes the physical location of the path of the train track center, but does NOT describe
the motion of the train (the derivative of r(s) with respect to s is not the velocity of the train—
||r0 || =
6 v).
(c) Suppose a train is moving at 100 km/h, and A = 1/32 [km]. What will be the peak bank angle?
Where does it occur?
(d) If the train is 3 meters wide, then passengers in the window seats, 1.5 meters from the center, will
experience some up-down motion as they are carried through each banked turn. Make a rough
approximation for the angular motion of the train to gauge how much vertical acceleration exists
for window seat passengers. What order of magnitude is it relative to a single g = 9.8? Can it be
ignored for the parameters in (c), or should additional constraints be put on the track?
HINTS: Solve for θ, for each s, by computing the net force from gravity Fg and the normal force
from the floor Fn and equating it to the centripetal force obtained from the curve N. Note also the
acceleration vector of the train lies in the osculating plane, which is decomposed into forward acceleration
2
T, and centripetal acceleration N according to a(t) = dv
dt T + v κN. One way to think about the
physical situation is that the bank angle would make the coffee cup slide off the table angled at θ,
but the turning train is also constantly moving to perfectly position the table underneath cup, thereby
making the cup stationary on the table. Also recall that curvature is in units of inverse distance.
Also, it will be useful to use a Taylor approximation to estimate window seat vertical motion to be
zvert (t) = 1.5 sin(θ(t)) ≈ 1.5 tan(θ(t)) for small angles, once θ(t) is found.