Download The Mathematics of Star Trek

The Mathematics of Star
Trek
Lecture 3: Equations of
Motion and Escape Velocity
Topics
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Antiderivatives
Integration
Differential Equations
Equations of Motion
Escape Velocity
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Antiderivatives
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We’ve already seen the idea of the
derivative of a function.
A related idea is the following:
Given a function f(x), find a function F(x)
such that F’(x) = f(x).
If such a function F(x) exists, we call F
an antiderivative of f.
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Antiderivatives (cont.)
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For example, given f(x) = 3x2, an antiderivative
of f is F(x) = x3, since F’(x) = 3x2.
Question: Can you think of any other
antiderivatives of f(x) = 3x2?
Possible answers: G(x) = x3 + 1, H(x) = x3 + 4,
K(x) = x3 - 5, etc.
Notice that all these antiderivatives of f(x) = 3x2
differ by a constant!
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Antiderivatives (cont.)
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Graphically, this should
make sense, since for a
fixed x-value, all of the
antiderivatives given
above have the same
slope!
This is true in general!
If F(x) and G(x) are
antiderivatives of f(x) on
an interval, then
F(x) = G(x) + C, for
some constant C.
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5
0
-5
-10
-2
-1
0
1
2
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Integration
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The process of finding an antiderivative of a
given function f(x) is called antidifferentiation
or integration.
If F’(x) = f(x), then we denote this by writing
∫ f(x) dx = F(x) + C.
We call the symbol ∫ an integral sign, f(x) the
integrand, and C the constant of integration.
Thus, we can write ∫ 3x2 dx = x3 + C.
This notation is due to Leibnitz!
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Integration (cont.)
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Using the fact that integration is
differentiation “backwards”, we can apply
derivative shortcuts to get integration
shortcuts!
∫ k f(x) dx = k ∫ f(x) dx for any constant k.
∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx.
∫ xn dx = xn+1/(n+1) + C for any rational
number n  -1.
∫ ek x dx = (1/k)ek x + C.
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Integration (cont.)
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For practice, evaluate each integral!
∫ x4 dx
∫ 2u7 du
∫ (t2 + 3 t + 1) dt
∫ 3 e3 x dx
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Differential Equations
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One branch of mathematics is differential
equations.
Many applications that involve rates of change
can be modeled with differential equations.
A differential equation is an equation involving
one or more derivatives of an unknown
function.
When solving a differential equation, the goal
is to find the unknown function(s) that satisfy
the given equation.
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Differential Equations (cont.)
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Here are some differential equations:
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Differential Equations (cont.)
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We can use integration to help solve certain
differential equations!
A differential equation is said to be separable if
it can be put into the form:
In this case, rewrite the equation in the form
s 1/g(y) dy = s f(x) dx and integrate each side
with respect to the appropriate variable.
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Differential Equations (cont.)
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Here is an example!
An object moving along a straight line with
under the influence of a constant acceleration
a is described by the differential equation:
dv/dt = a,
where v is the object’s velocity at time t.
We can use separation of variables to solve for
velocity v!
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Differential Equations (cont.)
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∫ dv = ∫ a dt
v=at+C
(general solution)
If the object has initial velocity v0 at time
t = 0, then we can find C.
v0 = a (0) + C
v0 = C
Thus, v = a t + v0 (particular solution).
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Differential Equations (cont.)
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Recall that velocity is the derivative of the
object’s position function s(t).
It follows that the object’s position function
satisfies the differential equation:
ds/dt = a t + v0.
If the object has initial position s0 at time t = 0,
then separation of variables can be used to
show that:
s = ½ a t2 + v0 t + s0.
(HW: Show this!)
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Equations of Motion
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In this last example, we have derived the
Equations of Motion for an object moving
along a straight line, under the influence
of a constant acceleration a, with initial
position s0 and initial velocity v0:
s = ½ a t 2 + v 0 t + s0
v = a t + v0.
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Equations of Motion (cont.)
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One application of these equations of motion is
projectile motion.
For example, suppose Commander Sisko’s baseball is
thrown straight up into the air with an initial velocity of
30 m/sec.
Assuming that the acceleration due to gravity is -9.8
m/sec2, find each of the following:
(a) The time at which the ball reaches its maximum
height.
(b) The maximum height that the ball reaches.
(c) The time at which the ball hits the ground.
(d) The velocity with which the ball hits the ground.
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Equations of Motion (cont.)
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Solution:
(a) With initial height s0 = 0 m and initial velocity v0 =
30 m/sec, the ball’s equations of motion are:
s = -4.9 t2 + 30 t + 0 (m)
v = -9.8 t + 30 (m/sec)
At the ball’s maximum height, the velocity is zero, so
using the second equation:
0 = -9.8 t + 30
t = 30/9.8 = 3.06 seconds is time at which maximum
height is reached.
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Equations of Motion (cont.)
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(b) To find the maximum height the ball
reaches, use the first equation with t =
30/9.8 sec:
s = -4.9 (30/9.8)2 + 30 (30/9.8) = 45.9
m.
Maximum height of the ball reaches is
approximately 45.9 meters.
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Equations of Motion (cont.)
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(c) When the ball hits the ground, its
height will be zero, so using the first
equation,
0 = -4.9 t2 +30 t = t(-4.9 t + 30),
Thus t = 0 sec or t = 30/4.9 = 6.1 sec.
The ball hits the ground approximately
6.1 seconds after it is thrown into the
air.
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Equations of Motion (cont.)
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(d) Using the equation for velocity, when
the ball hits the ground it’s velocity will
be:
v = -9.8 (30/4.9) +30 = -30 m/sec.
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Escape Velocity
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Another question that we
can use differential
equations to answer is the
following:
What initial velocity is
required for an object to
escape the Earth’s
gravitational field?
To answer this question,
we need Newton’s Law of
Universal Gravitation and
Newton’s Second Law of
Motion!
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Escape Velocity (cont.)
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Newton’s Law of Universal Gravitation: The
gravitational force between two masses M and
m is proportional to the product of the masses
and inversely proportional to the square of the
distance between them, i.e. F = GMm/r2, where
G is a constant.
Newton’s Second Law: The net external
force on an object is equal to its mass times
acceleration, i.e. F = ma.
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Escape Velocity (cont.)
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Using these laws, we
find that the
acceleration of an
object a distance r
from the Earth’s
center is given by the
equation:
a = dv/dt = -k/r2,
where k is a constant
of proportionality.
r
R
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Escape Velocity (cont.)
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When r = R, then a = -g, the acceleration at the
surface of the Earth, so:
-g = -k/R2,
which yields k = gR2.
Thus, a = dv/dt = -gR2/r2.
Now, since v is a function of position r and r is
a function of time t, we can write:
dv/dt = dv/dr dr/rt = v dv/dt. (This is an
application of the Chain Rule from Calculus.)
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Escape Velocity (cont.)
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Substituting v dv/dr for dv/dt in the
equation dv/dt = -gR2/r2, we are led to
the following model for an object’s
velocity as a function of distance from
the Earth’s center:
v dv/dr = -gR2/r2.
This differential equation can be solved
via separation of variables!
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Escape Velocity (cont.)
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∫ v dv = ∫ -gR2 r-2 dr
½ v2 = g R2 r-1 + C
v2 = (2g R2)/r + 2 C (general solution)
If the object leaving the Earth’s surface
has an initial velocity of v0, then we can
find constant C!
v02 = (2g R2)/R + 2 C
C = ½ (v02 - 2g R)
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Escape Velocity (cont.)
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Thus, our solution to this differential equation is:
v2 = (2g R2)/r + v02 - 2g R.
In order for the velocity v to stay positive, we need
v02 - g R ≥ 0, which means that
We call the right-hand side of this last expression Earth’s
escape velocity, i.e. the minimum initial velocity needed
for an object to escape the Earth’s force of gravity.
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References
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Calculus: Early Transcendentals (5th ed) by
James Stewart
Elementary Differential Equations (8th ed) by
Rainville, Rainville, and Bedient
Hyper Physics: http://hyperphysics.phyastr.gsu.edu/hbase/hph.html
The Cartoon Guide to Physics by Larry
Gonick and Art Huffman
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