Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of...
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Transcript of Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of...
![Page 1: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/1.jpg)
Integration Along a Curve:
Kicking it up a notch
Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000
![Page 2: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/2.jpg)
Motivation: Why do we want to integrate a function along a curve?
![Page 3: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/3.jpg)
A Real Mathematician’s Answer:
Because we can. That’s why!
![Page 4: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/4.jpg)
No, but really...
Motivation: A Massachusetts Dilemma
In Boston, we freeze during the wintah.
School is often cancelled due to the hazids of snow and ice.
During a snowball fight, we notice the ice coating the telephone wiahs
My buddy Maak the physics major says, “I bet you 10 bucks you can’t figure out the total mass of the ice on that wiah!”
![Page 5: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/5.jpg)
You’re on.
But first I need to develop the theory of line integrals.
![Page 6: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/6.jpg)
The thickness of the ice varies as one moves along the wire
The confused person
(an annoying recurring character)
Setting up the Problem…...
r = radius of wire
r
R = radius of wire + ice coating
R
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Recall the Ole Physics formula:
Mass = density * volume
Density of ice = 0.92 g/cm3
Area of a cross section of ice = R- r2) cm2
Linear density f of ice on wire = 0.92 * R- r2) g/cm
So...
The total mass of ice on wire = total accumulation of the linear density function along the wire
![Page 8: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/8.jpg)
Parameterize the wire using a continuously differentiable (i.e. smooth) function
2],[: Rba ))(),(()( tytxt
x(t)
y(t)
Real-valued, continuously differentiable functions
![Page 9: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/9.jpg)
Now we can at least approximate the total mass of the ice along the wire by these 4 easy steps:
1. Partition2. Sample
3. Scale
4. Sum
![Page 10: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/10.jpg)
1. Partition
Partition the arc into n subarcs
How? By partitioning [a,b] into n subintervals, we induce a partition of into n subarcs
x
y
t
tx(t),y(t))
…
… …
… t1
t1
t1
ti
ti
ti
a = t0
a = t0
t0
1
tn = b
tn = b
tnn
ti-1
ti-1
ti-1
i
![Page 11: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/11.jpg)
2. SampleRecall f is the density function defined on [a,b]), our “frozen wire”.
On each subarc i, choose a point ixi*,yi*) and sample f at those points.
x
y
t
a = t0
tx(t),y(t))
a = t0 tn = b
tn = b
ti-1
ti-1
ti
ti
t1
t1
…
… …
…
t0
t1
tn
ti-1
ti
n
i
0x0*,y0
*)
i-1xi-1*,yi-
1*)
nxn*,yn*)
![Page 12: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/12.jpg)
3. Scale
x
y
t
tx(t),y(t))
a = t0 tn = b ti-1 ti t1 … …
t0
t1
tn
ti-1 ti
0
n-1
i-
1
Now we scale those sampled values f(ifxi*,yi*) by the length of the subarc denoted by si
si
![Page 13: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/13.jpg)
4. SumNow we sum those scaled sampled values to get what looks like a Riemann sum.
i
n
iii syxf
)**,(1
But we want a way to actually calculate this mutha.
So we need to bring it down to the case we know: the one variable case.
![Page 14: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/14.jpg)
It seems so pointless.
Should I just give him the money right now?
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No Way!!!!!
We gotta show him up!
Math majors, represent!
Let’s do this.
(The Encouragement Slide)
![Page 16: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/16.jpg)
How, do you ask? By relating everything to t.
fxi*,yi*) = fti*)) for some ti*in [ti-1, ti ]
)()( 1 iii tts
almostsi
yi
xi
Notice that for si small, a continuous curve looks locally linear
(t i-1)
(t i)
![Page 17: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/17.jpg)
ttttttt )(')(')()(
Since is continuously differentiable,
So when t is small,
t
tttt
t
)()()(' lim
0
Since fis path integrable (continuity on [a,b]) is sufficient for this), we may sample f((t)) and partition however we choose:
Let ti = (b-a)/n = t
Let ti* = ti-1
![Page 18: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/18.jpg)
So we have
tttfsyxf i
n
iii
n
iii
*)(')*)(()**,(11
which is a Riemann sum of the one variable real-valued function f((t))|| ’(t)) ||
So letting L(P, f) and U(P, f) be our respective lower and upper Riemann sums...
![Page 19: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/19.jpg)
b
a
dtttffds )('))((
)')(,(inf)')(,(sup)('))(( fPUfPLdtttfPP
b
a
where
Thus we define the path integral of f along the curve
Note: If f (x,y)=1 on [a,b]
dttdsb
a )('
= length of the curve
![Page 20: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/20.jpg)
So to answer Maak’s challenge…
b
a
b
a
dttdtttds )2,1())tf((t,)('))(f( R)f(r, 2
the total mass of the ice is ………....
r = Radius of wiah = 10cm
R = Radius of wiah + ice
b
a
dttt )41()(t * 0.92 224
(t) = (t,t2) parametrization of wire (parabola)
f(r,R) = linear density = 0.92*(R2-r2)
![Page 21: Integration Along a Curve: Kicking it up a notch Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000.](https://reader036.fdocuments.net/reader036/viewer/2022081603/56649f165503460f94c2bcd1/html5/thumbnails/21.jpg)
NOW SHOW ME THE MONEY!!!!
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