The Mathematics of Star Trek

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The Mathematics of Star Trek

• Lecture 3 Equations of Motion and Escape
  Velocity

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Topics

• Antiderivatives
• Integration
• Differential Equations
• Equations of Motion
• Escape Velocity

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Antiderivatives

• Weve 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.)

• 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.)

• 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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Integration

• 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.)

• 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 xn1/(n1) C for any rational number
  n ? -1.
• ? ek x dx (1/k)ek x C.

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Integration (cont.)

• For practice, evaluate each integral!
• ? x4 dx
• ? 2u7 du
• ? (t2 3 t 1) dt
• ? 3 e3 x dx

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Differential Equations

• 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.)

• Here are some differential equations

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Differential Equations (cont.)

• 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.)

• 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 objects velocity at time t.
• We can use separation of variables to solve for
  velocity v!

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Differential Equations (cont.)

• ? dv ? a dt
• v a t 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.)

• Recall that velocity is the derivative of the
  objects position function s(t).
• It follows that the objects 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

• 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 t2 v0 t s0
• v a t v0.

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Equations of Motion (cont.)

• One application of these equations of motion is
  projectile motion.
• For example, suppose Commander Siskos 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.)

• Solution
• (a) With initial height s0 0 m and initial
  velocity v0 30 m/sec, the balls equations of
  motion are
• s -4.9 t2 30 t 0 (m)
• v -9.8 t 30 (m/sec)
• At the balls 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.)

• (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.)

• (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.)

• (d) Using the equation for velocity, when the
  ball hits the ground its velocity will be
• v -9.8 (30/4.9) 30 -30 m/sec.

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Escape Velocity

• Another question that we can use differential
  equations to answer is the following
• What initial velocity is required for an object
  to escape the Earths gravitational field?
• To answer this question, we need Newtons Law of
  Universal Gravitation and Newtons Second Law of
  Motion!

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Escape Velocity (cont.)

• Newtons 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.
• Newtons 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.)

• Using these laws, we find that the acceleration
  of an object a distance r from the Earths 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.)

• 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.)

• Substituting v dv/dr for dv/dt in the equation
  dv/dt -gR2/r2, we are led to the following
  model for an objects velocity as a function of
  distance from the Earths center
• v dv/dr -gR2/r2.
• This differential equation can be solved via
  separation of variables!

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Escape Velocity (cont.)

• ? 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 Earths 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.)

• 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 Earths escape velocity, i.e. the
  minimum initial velocity needed for an object to
  escape the Earths force of gravity.

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References

• Calculus Early Transcendentals (5th ed) by
  James Stewart
• Elementary Differential Equations (8th ed) by
  Rainville, Rainville, and Bedient
• Hyper Physics http//hyperphysics.phy-astr.gsu.e
  du/hbase/hph.html
• The Cartoon Guide to Physics by Larry Gonick and
  Art Huffman