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MATH 101 - Section 210
Instructor: Avner Segal ([email protected])
February 28th 2017
Common course page:
http://www.math.ubc.ca/~gerg/teaching/101-Winter2017/
Individual section page:
http://www.math.ubc.ca/~avners/courses/math101-2017.html
Office hours: Thursdays 12:00-14:30 (LSK300)
On Quiz Weeks: Tuesdays 12:00-13:00 (LSK300) - by appointment
only!
Follow announcements on section web page in case of changes.
Center of Mass
Reminder
The center of mass of pointed masses m1 ,...,mn located at x1 ,...,xn
(respectively) is given by
Pn
mi xi
i =1 mi
xc .m. = Pi =n1
In what follows, we use this to compute the center of mass of
regions in the plane with a constant mass density. In this case, the
center of mass, is also called centroid or geometric center.
Example
Calculate the center of mass of the semicircle bounded by the
curves y 2 + x 2 = 1 and y = 0.
It is assumed (implicitly in the question) the mass of the semicircle
is spread evenly; this is called a constant mass density.
Let m (in kg ) be the mass density, i.e. the mass of a unit area (i.e.
the mass of a square with area 1 m2 ). Note that the mass of the
semicircle is given by M = m · A, where A is its area.
Also note that since the semicircle is symmetric about the y -axis, it
is obvious that xc .m. = 0.
Solution
We approximate the center of mass using Riemann sums. We chop
the shape along the y -axispinto rectangles. The width of the
semicircle
at height y is 2 1 − y 2 so the area of each
qrectangle is
q
1
1
∗ 2
∗ 2
n 2 1 − yk ,n and the mass of each rectangle is m n 2 1 − yk ,n .
p
Let f (y ) be the width of the semicircle at y , i.e. f (y ) = 2 1 − y 2 .
The center of mass of the rectangles is given by:
n
X
1
yk∗,n m f (yk∗,n )
n
k =1
n
X
k =1
µ
1
m f (yk∗,n )
n
µ
¶
¶
=
n
1X
y ∗ mf (yk∗,n )
n k =1 k ,n
n
1X
mf (yk∗,n )
n k =1
.
Taking the limit as n → ∞ yields
yc .m. = lim
n→∞
n
1X
y ∗ mf (yk∗,n )
n k =1 k ,n
n
1X
mf (yk∗,n )
n k =1
m
=
1
Z
m
yf (y )dy
0
yf (y )dy
0
=
M
1
Z
.
M
Now we integrate
m
1
Z
0
yf (y )dy = 2m
1
Z
y
0
"
#
3/2 1
(1 − y 2 )
1 − y 2 dy = 2m −
3
q
2
=
0
Now, recalling that M = m · A and A = π·21 = π2 we find that
2m/3
yc .m. = mπ/2 = 34π .
2m
.
3
Alternatively, taking the limit as n → ∞ yields
yc .m. = lim
n→∞
n
1X
y ∗ f (y ∗ )
n k =1 k ,n k ,n
n
1X
f (y ∗ )
n k =1 k ,n
1
Z
=
yf (y )dy
0
A
.
2m/3
Solving the integral as before yields yc .m. = mπ/2 = 34π .
In general, the center of mass of a semicirlce of radius r will be at
4r
3π of its central radius.
A list of centers of mass of simple shapes:
https://en.wikipedia.org/wiki/List_of_centroids, you
should try to do as many of them as you can at home!
Question (Final 2013)
Find the centroid of the region below, which consists of a semicircle
of radius 3 on top of a rectangle of width 6 and height 2.
Solution
The region R consists of a semicircle of radius 3 on top of a
rectangle of width 6 and height 2. The shape is symmetric about
the y -axis and so xc .m. = 0. We now compute yc .m. .
Solution I
The region R consists of a semicircle of radius 3 on top of a
rectangle of width 6 and height 2. The center of mass of the
semicircle is at y = 43·π3 and the center of the rectangle is at y = −1.
2
The mass of the semicircle is π23 m (where m is the mass density)
and the mass ofthe rectangle is 6 · 2 · m. We conclude that
yc .m. =
4
π
· 9π2m + (−1) · 12m
9π m
2
+ 12m
=
4
.
8 + 3π
Solution II
We will now calculate yc .m. by integration again, but instead of
chopping the shape along the y -axis, we will chop it along the x
axis.
Solution II
Here, we will average of the centers of mass of the rectangles. For
each of these rectangles
6
n
³q
´
9 − xk2,n − (−2) .
Ï
Its area is given by
Ï
Its mass is given by m · n6 ·
Ï
Its
(y-coordinate of its) center of mass is located at
q
9−xk2,n +(−2)
2
.
·
³q
´
9 − xk2,n − (−2) .
The center of mass of all these rectangles together is
n
X
q
9 − xk2,n − 2
2
k =1
·m·
´
6 ³q
9 − xk2,n + 2
·
n
n
X
´
6 ³q
m · · 9 − xk2,n + 2
n
k =1
6
n
=
n h
X
i
(9 − xk2,n ) − 4
k =1
.
n q
X
6
2
2· n
9 − xk ,n + 2
k =1
Taking the limit n → ∞, the denominator goes to 2 · A = 24 + 9π,
while the numerator goes to
Z
3
−3
(5 − x 2 )dx = 12.
This yields
yc .m. =
4
.
8 + 3π
More Generally
The center of mass of the region (with area A)
©
¯
(x , y )¯a ≤ x ≤ b, f (x) ≤ y ≤ g (x)
ª
between the curves y = f (x) and y = g (x) (with a constant mass
density) is given by
b
Z
a
xc .m. =
A
b¡
Z
yc .m. =
x (g (x) − f (x))dx
a
Notice that the units here fits!
g (x)2 − f (x)2 dx
¢
2A
.
Example (Last year quiz v.M1)
Express the x–coordinate of the centroid of the triangle with
vertices (−1, −1), (−1, 1), and (1, 0) in terms of a definite integral.
Do not evaluate the integral.
Solution
The equations of the top and bottom of the triangle are
y = f (x) =
1−x
,
2
y = g (x) =
x −1
.
2
The area of the triangle is A = 12 · 2 · 2 = 2 and hence
xc .m. =
1
A
Z
1
−1
x (f (x) − g (x))dx =
1
2
Z
1
1
x(1 − x)dx = − .
3
−1
Separable Differential Equations
Some examples of "real-life" differential equations:
Newton’s Law of Motion
Maxwell’s equations
Navier–Stokes equations
Heat equation
Wave equation
Schrödinger equation
Stress-strain equations
Black–Scholes models
Predator–prey equations
Einstein’s equations
Ludwig–Jones–Holling’s equation
Zeeman’s model
Sherman–Rinzel–Keizer model
Hodgkin–Huxley equations
describes motion of particles
describes electromagnetic radiation
describes fluid motion
describes heat flow
describes wave motion
describes atoms, molecules and crystals
describes elastic materials
used for pricing financial options
describes ecosystem populations
connects gravity and geometry
models spruce budworm/Balsam fir ecosystem
models heart beats and nerve impulses
for electrical activity in Pancreatic β–cells
models nerve action potentials
In this course, we will only look at only the simplest examples of
differential equations. Try reading the optional sections in the CLP
notes.
Example - Introduction I
Consider the differential equation y 0 = 3y 2 . We are looking for
solutions of such equations.
Ï
For which values of C and D is y (x) = Cx D solution of the
equation? Assume it is. Then
!
³
y 0 (x) = C · Dx D −1 = 3y 2 (x) = 3 Cx D
´2
= 3C 2 x 2D .
If C = 0 then this is a solution (admittedly, not an interesting
one but it is important to remember the boring solution...) i.e.
y (x) = 0. Otherwise, it follows that C · D = 3C 2 and
1
.
D − 1 = 2D. Hence D = −1 and C = − 13 . So y (x) = − 3x
¡
¢
2
1
1
0
2
y = 3x 2 = 3 − 3x = 3y .
Example - Introduction II
Consider the differential equation y 0 = 3y 2 .
Ï
Suppose y = f (x) is a solution. Show that y = f (x − a) is also a
solution for any a. What is the solution with f (0) = 1?
The assumption means that f 0 (x) = 3f (x)2 for any x. Hence
[f (x − a)]0 = f 0 (x − a) = 3 · (f (x − a))2
fo any x.
Ï
Applying part I, we see that for any a the following is a solution
f (x) = −
1
3(x − a)
.
We solve f (0) = 1 for a.
1
1
=1⇔a= .
3
3a
We conclude that the solution we look for is y = 1−13x .
Seperable Differential Equations
A separable differential equation is an equation for a function y (x)
of the form
dy
(x) = f (x) · g (y (x)).
dx
We look for solutions y (x) for this equation.
Given such an equation, we write (and this just abstract nonsense)
dy
= f (x)dx .
g (y )
Integrating, we get
Z
dy
=
g (y )
Z
f (x)dx .
Miraculously, this will give all solutions to the equation
dy
dx (x) = f (x) · g (y (x)).
Example
Solve the separable differential equation y 0 = x 3 .
Solution
We write
dy
dx
= x 3 . Using the idea in the previous slide, we write
Z
dy =
Z
x 3 dx
Solving both sides,we have
y=
x4
+C.
4
Example
Solve the separable differential equation y 0 = x 3 such that y (1) = 1.
The condition y (1) = 1 is called initial condition.
4
We consider the general solution we found y (x) = x4 + C . Plugging
in the initial condition yields
1 = y (1) =
1
+C.
4
It follows that C = 34 and conclude that
y (x) =
x4 + 3
.
4
Questions
Solve the following separable differential equations:
(a) y 0 = 5y .
(b) (Final 2012) y 0 = xy , y (0) = e.
(c) (Last years quiz)
dy
dx
= −xy 3 , y (0) = − 41 .
2
(d) (Final 2014) x dy
dx + y = y , y (1) = −1.
Solution (a) - y 0 = 5y
dy
= 5y
dx
Z
dy
=
y
Z
5dx
log |y | = 5x + C
By continuity, either y (x) > 0 for all x or y (x) < 0 for all x.
Ï
If y > 0 then y = e 5x +C = De 5x (here D > 0).
Ï
If y < 0 then y = −e 5x +C = De 5x (here D < 0).
In both cases, we can write the solution as y (x) = De 5x , D (and its
sign) is decided by an initial condition.
Note that y 0 = 5De 5x = 5y .
Solution (b) - y 0 = xy , y (0) = e
dy
= xy
dx
Z
dy
=
y
log |y | =
x2
Z
xdx
x2
+C
2
x2
So y (x) = ±e 2 +C = De 2 , where D = ±e C . Plugging in the initial
condition yields e = y (0) = De 0 = D so
y (x) = e
Indeed, y (0) = e and y 0 = xy .
x2
2 +1
.
Solution (c) -
dy
dx
= −xy 3 , y (0) = − 14
Z
−
dt
=−
y3
Z
xdx
1
x2
+C
=
−
2
2y 2
y2 =
1
x 2 − 2C
1
y (x) = ± p
2
x − 2C
1
1
− = y (0) = ± p
,
4
−2C
1
y (x) = − p
x 2 + 16
Indeed, y (0) = − 14 , y 0 = −xy 3
C = −8
2
Solution (d) - x dy
dx + y = y , y (1) = −1
dy
Z
Z
=
y2 −y
Z
dy
y2 −y
Z
=
dy
−
y −1
Z
dx
x
¯y −1¯
dy
¯+C
= log ¯¯
y
y ¯
¯
¯
¯
¯
¯
¯
¯
¯
¯y −1¯
¯ = log |x | + C ⇒ ¯ y − 1 ¯ = e C |x |
¯
¯
y ¯
y
log ¯¯
y −1
= Dx ⇒ y − 1 = Dxy ⇒ y (1 − Dx) = 1
y
y (x) =
1
1
⇒ −1 = y (1) =
⇒ D =2
1 − Dx
1−D
y (x) =
Check that this is a solution!
1
1 − 2x
Example - Harmonic Osciliator
A physical system satisfies the equation E = 12 mv 2 + 12 kx 2 , where m,
k and E are constants (mass, spring constant, energy, respectively)
and v = dx
dt is the velocity. We solve for x(t).
dx
dt
1. Note that
q
=v =
2. We solve
1
m
dx
Z
q
q
· (2E − kx 2 ) =
2E
m
·
q
Z
2
1 − kx
2E
dz =
k
2E dx,
q
4. z = sin
2E
m
·
µq
q
q
dt
q
k
2E x,
2
1 − kx
2E .
with
to get
dx
Z
·
=
3. We make the change of variables z =
q
2E
m
1−
kx 2
k
mt +C
2E
=q
1
2E
m
·
Z
q
¶
⇒ x(t) =
k
2E
q
2E
k
p
dz
1 − z2
sin
µq
arcsin(z)
= p
k
mt +C
k/m
¶
+C.
Let’s understand this solution
1
1
E = mv 2 + kx 2
2
2
s
x(t) =
s
2E
sin 
k

k
t +C
m
This is called an harmonic oscillator.
Note that this is a periodic
q
m
function with period T = 2π k .
https://upload.wikimedia.org/wikipedia/commons/9/9d/
Simple_harmonic_oscillator.gif
https://universe-review.ca/I15-71-classical.gif
q
q
k
When m
t0 + C = π2 we have x(t0 ) = 2E
k , this is the maximal
distance between the object and the origin. This is when v = 0 and
hence E = 21 kx 2 .
The role of C in the solution is to fix a moment at which the object
passes through the origin.