Nerdy is the new black
Patrick Bardsley
January 20, 2014
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Weather forecasting
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Weather forecasting
Optimizing material design
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Weather forecasting
Optimizing material design
Applications of integrals
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Weather forecasting
Optimizing material design
Applications of integrals
Fourier analysis
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Weather forecasting
Optimizing material design
Applications of integrals
Fourier analysis
Spectrum analysis of acoustic signals
Patrick Bardsley
Nerdy is the new black
Overview
Applications of derivatives
Weather forecasting
Optimizing material design
Applications of integrals
Fourier analysis
Spectrum analysis of acoustic signals
The Fundamental Theorem of Calculus
Patrick Bardsley
Nerdy is the new black
Weather forecasting
The temperature T (t) in a given area (SLC) can be described as
dT
= f (t, T , wind, cloud cover, humidity, . . .)
dt
where f is some function of all of those variables. As a simple
example consider
dT
1 11
10
= 6 sin
t
− 0.001t,
T (0) = 40
dt
5
Patrick Bardsley
Nerdy is the new black
dT
= 6 sin
dt
Patrick Bardsley
1 11
t 10
5
− 0.001t
Nerdy is the new black
dT
1 11
10
= 6 sin
t
− 0.001t
dt
5
1 11
t 10 − 0.001t dt
dT = 6 sin
5
Patrick Bardsley
Nerdy is the new black
dT
1 11
10
= 6 sin
t
− 0.001t
dt
5
1 11
t 10 − 0.001t dt
dT = 6 sin
5
Z
Z 1 11
t 10 − 0.001t dt
dT =
6 sin
5
Patrick Bardsley
Nerdy is the new black
dT
1 11
10
= 6 sin
t
− 0.001t
dt
5
1 11
t 10 − 0.001t dt
dT = 6 sin
5
Z
Z 1 11
t 10 − 0.001t dt
dT =
6 sin
5
Z 1 11
T (t) =
6 sin
t 10 − 0.001t dt + C
5
Patrick Bardsley
Nerdy is the new black
dT
1 11
10
= 6 sin
t
− 0.001t
dt
5
1 11
t 10 − 0.001t dt
dT = 6 sin
5
Z
Z 1 11
t 10 − 0.001t dt
dT =
6 sin
5
Z 1 11
T (t) =
6 sin
t 10 − 0.001t dt + C
5
Problem: We can’t integrate this! But... we could use an
approximation of this function, which we find from a linearization.
Patrick Bardsley
Nerdy is the new black
Linearized differential equation
Using the linearization formula
f (t) ≈ f (a) + f 0 (a)(t − a) = L(t)
Patrick Bardsley
Nerdy is the new black
Linearized differential equation
Using the linearization formula
f (t) ≈ f (a) + f 0 (a)(t − a) = L(t)
we find the linearization L(t) of the function
1 11
t 10 − 0.001t.
f (t) = 6 sin
5
Patrick Bardsley
Nerdy is the new black
Linearized differential equation
Using the linearization formula
f (t) ≈ f (a) + f 0 (a)(t − a) = L(t)
we find the linearization L(t) of the function
1 11
t 10 − 0.001t.
f (t) = 6 sin
5
Then we could use the approximation
dT
= f (t) ≈ L(t)
dt
Patrick Bardsley
Nerdy is the new black
Linearized differential equation
Using the linearization formula
f (t) ≈ f (a) + f 0 (a)(t − a) = L(t)
we find the linearization L(t) of the function
1 11
t 10 − 0.001t.
f (t) = 6 sin
5
Then we could use the approximation
dT
= f (t) ≈ L(t)
dt
and we can solve the approximate differential equation to find
Z
Z
T (t) = f (t)dt + C ≈ L(t)dt + C .
Patrick Bardsley
Nerdy is the new black
The linearization near time equal to zero of
1 11
10
t
− 0.001t.
f (t) = 6 sin
5
is given by
L(t) = f (0) + f 0 (0)(t) = −0.001t.
Patrick Bardsley
Nerdy is the new black
So solving the linearized differential equation we have
dT
= −0.001t
dt
Patrick Bardsley
Nerdy is the new black
So solving the linearized differential equation we have
dT
= −0.001t
dt
dT = −0.001tdt
Patrick Bardsley
Nerdy is the new black
So solving the linearized differential equation we have
dT
= −0.001t
dt
dT = −0.001tdt
Z
Z
dT =
Patrick Bardsley
−0.001tdt + C
Nerdy is the new black
So solving the linearized differential equation we have
dT
= −0.001t
dt
dT = −0.001tdt
Z
Z
dT =
−0.001tdt + C
T (t) = −0.001
Patrick Bardsley
t2
+C
2
Nerdy is the new black
So solving the linearized differential equation we have
dT
= −0.001t
dt
dT = −0.001tdt
Z
Z
dT =
−0.001tdt + C
T (t) = −0.001
t2
+C
2
T (t) = −0.0005t 2 + C
Patrick Bardsley
Nerdy is the new black
So solving the linearized differential equation we have
dT
= −0.001t
dt
dT = −0.001tdt
Z
Z
dT =
−0.001tdt + C
T (t) = −0.001
t2
+C
2
T (t) = −0.0005t 2 + C
and using the fact that T (0) = 40 we find a temperature model
T (t) = −0.0005t 2 + 40.
Patrick Bardsley
Nerdy is the new black
45
True Temperature
Approximate Temperature
44
43
Temperature
42
41
40
39
38
37
36
35
0
0.5
1
1.5
2
2.5
Hours
Why can’t we trust the 7-day forecast?
This linearized approximation is great for short times (within 2
hours in this case).
Patrick Bardsley
Nerdy is the new black
80
True Temperature
Approximate Temperature
75
70
Temperature
65
60
55
50
45
40
35
0
5
10
15
20
Hours
Why can’t we trust the 7-day forecast?
The linearized approximation breaks down as we move further away
from time zero. So in general forecasting is hard for long periods
of time because of the approximations to the model.
Patrick Bardsley
Nerdy is the new black
Optimal design of a climbing cam
The curved shape of a climbing cam
Patrick Bardsley
Nerdy is the new black
Optimal design of a climbing cam
The curved shape of a climbing cam is described by a sector of
what is known as a logarithmic spiral r = e µθ where r is the radius
on the spiral, θ is the angle in radians, and µ describes how fast
the spiral opens up.
Patrick Bardsley
Nerdy is the new black
If this camming unit is placed inside a parallel crack we can look at
a force diagram to understand what keeps climbers alive.
Patrick Bardsley
Nerdy is the new black
If this camming unit is placed inside a parallel crack we can look at
a force diagram to understand what keeps climbers alive. Here the
contact angle is what is most important for our purposes since it
helps describe how much force we can expect on the center pin of
the cam for a given load T .
Patrick Bardsley
Nerdy is the new black
The angle a is given by tan(a) = y /x where x describes the
horizontal distance of the center pin to the wall, and y describes
the vertical distance of the contact point below the center pin.
Patrick Bardsley
Nerdy is the new black
The angle a is given by tan(a) = y /x where x describes the
horizontal distance of the center pin to the wall, and y describes
the vertical distance of the contact point below the center pin. To
find the angle for a given choice of parameter µ we need to find
where the curve has a vertical tangent line.
Patrick Bardsley
Nerdy is the new black
The angle a is given by tan(a) = y /x where x describes the
horizontal distance of the center pin to the wall, and y describes
the vertical distance of the contact point below the center pin. To
find the angle for a given choice of parameter µ we need to find
where the curve has a vertical tangent line. The curved shape of
the cam is given in polar coordinates by
r = e µθ
and we have y = r sin(a) and x = r cos(a).
Patrick Bardsley
Nerdy is the new black
Now instead of thinking of the spiral curve in polar coordinates
r = e µθ , we can think of the curve being described in Euclidean
(standard) coordinates as a function y (x).
Patrick Bardsley
Nerdy is the new black
Now instead of thinking of the spiral curve in polar coordinates
r = e µθ , we can think of the curve being described in Euclidean
(standard) coordinates as a function y (x). By the relationship of
polar coordinates we have
r 2 = e 2µθ
−1 (y /x)
x 2 + y 2 = e 2µ tan
Patrick Bardsley
.
Nerdy is the new black
Implicit differentiation gives
2x + 2yy 0 = e 2µ tan
−1 (y /x)
Patrick Bardsley
x2
2µ
(xy 0 − y )
+ y2
Nerdy is the new black
Implicit differentiation gives
2x + 2yy 0 = e 2µ tan
−1 (y /x)
x2
2µ
(xy 0 − y )
+ y2
Gross algebra gives
0
y =
−2x − ye 2µ tan
2y −
Patrick Bardsley
−1 (y /x)
−1
2µe 2µ tan (y /x)
x 2 +y 2
Nerdy is the new black
Implicit differentiation gives
2x + 2yy 0 = e 2µ tan
−1 (y /x)
x2
2µ
(xy 0 − y )
+ y2
Gross algebra gives
0
y =
−2x − ye 2µ tan
2y −
−1 (y /x)
−1
2µe 2µ tan (y /x)
x 2 +y 2
Set the bottom to zero to find a vertical tangent line! Everything
simplifies to
µ = tan(a) ⇐⇒ a = tan−1 (µ).
Patrick Bardsley
Nerdy is the new black
Now we can relate the total force applied to the center pin from
the camming action of the device to find
s
1
T
1 + 2.
|F | =
2
µ
Patrick Bardsley
Nerdy is the new black
Now we can relate the total force applied to the center pin from
the camming action of the device to find
s
1
T
1 + 2.
|F | =
2
µ
If we know µ, and we know the amount of force |F | which will
make the cam fail, we can find out how much of load the device
can handle (i.e. how far we can fall before the unit fails).
Patrick Bardsley
Nerdy is the new black
Fourier analysis
What if we could build a complicated function f (x) out of a bunch
of other relatively simple functions? For example:
f (x) = a0 +a1 cos(πx)+b1 sin(πx)+a2 cos(2πx)+b2 sin(2πx)+· · ·
Patrick Bardsley
Nerdy is the new black
Example: f (x) = x 2 on the interval 0 ≤ x ≤ 2
Patrick Bardsley
Nerdy is the new black
Example: f (x) = x 2 on the interval 0 ≤ x ≤ 2
x 2 = a0 + a1 cos(πx) + b1 sin(πx) + a2 cos(2πx) + b2 sin(2πx) + · · ·
where
Patrick Bardsley
Nerdy is the new black
Example: f (x) = x 2 on the interval 0 ≤ x ≤ 2
x 2 = a0 + a1 cos(πx) + b1 sin(πx) + a2 cos(2πx) + b2 sin(2πx) + · · ·
where
Z
1 2 2
4
a0 =
x dx =
2 0
3
Z 2
4
an =
x 2 cos(nπx)dx = 2 2
n π
0
Z 2
−4
bn =
x 2 sin(nπx)dx =
nπ
0
Patrick Bardsley
Nerdy is the new black
So in a very general form we can write x 2 as
x2 =
∞
X
(an cos(nπx) + bn sin(nπx))
n=0
=
4
4
−4
+ 2 cos(πx) +
sin(πx)+
3 π
π
1
−2
cos(2πx) +
sin(2πx) + · · ·
2
π
π
Fourier series
This function is easy already (f (x) = x 2 ), but this will be very
useful for more difficult functions!
Patrick Bardsley
Nerdy is the new black
Spectrum analysis
A recording of a musical instrument is a time signal which may
look like
Time signal
0.1
0.05
0
−0.05
−0.1
−0.15
0
0.5
Patrick Bardsley
1
time
1.5
Nerdy is the new black
2
5
x 10
Spectrum analysis
A recording of a musical instrument is a time signal which may
look like
Time signal
0.1
0.05
0
−0.05
−0.1
−0.15
0
0.5
1
time
1.5
2
This is just a function f (t) - Let’s use Fourier series!
Patrick Bardsley
Nerdy is the new black
5
x 10
If T denotes the final time of the acoustic signal we are going to
reconstruct the time function f (t) as
f (t) =
∞
X
an cos(nπt) + bn sin(nπt),
n=0
where
1
a0 =
T
T
Z
f (t)dt
0
Z
T
an = C1
f (t) cos(nπt)dt
0
Z
T
bn = C2
f (t) sin(nπt)dt
0
where we know exactly what C1 and C2 are.
Patrick Bardsley
Nerdy is the new black
Obviously, we can’t use infinitely many terms or we’d be working
on one problem forever... literally.
f (t) =
∞
X
an cos(nπt) + bn sin(nπt)
n=0
Patrick Bardsley
Nerdy is the new black
Obviously, we can’t use infinitely many terms or we’d be working
on one problem forever... literally.
f (t) =
∞
X
an cos(nπt) + bn sin(nπt)
n=0
So we use a finite amount N ∗ of them as an approximation:
∗
f (t) ≈
N
X
an cos(nπt) + bn sin(nπt)
n=0
and we obtain the following sounds.
Patrick Bardsley
Nerdy is the new black
The Fundamental Theorem of Calculus
Recall if
Z
d(t) =
t
v (τ )dτ
0
then
Patrick Bardsley
Nerdy is the new black
The Fundamental Theorem of Calculus
Recall if
Z
d(t) =
t
v (τ )dτ
0
then
d 0 (t) = v (t).
If v represents velocity (mph), then d(t) represents the distance
traveled (mi) over the time interval [0, t].
Patrick Bardsley
Nerdy is the new black
Let’s say v (t) = 80 mph, then in 1 hour we have traveled
Z
d(1) =
1
80dt
0
Patrick Bardsley
Nerdy is the new black
Let’s say v (t) = 80 mph, then in 1 hour we have traveled
Z
1
d(1) =
80dt
Z 1
= 80
dt
0
0
Patrick Bardsley
Nerdy is the new black
Let’s say v (t) = 80 mph, then in 1 hour we have traveled
Z
1
d(1) =
80dt
Z 1
= 80
dt
0
1
= 80t 0
0
Patrick Bardsley
Nerdy is the new black
Let’s say v (t) = 80 mph, then in 1 hour we have traveled
Z
1
d(1) =
80dt
Z 1
= 80
dt
0
1
= 80t 0
0
= 80 (1 − 0)
Patrick Bardsley
Nerdy is the new black
Let’s say v (t) = 80 mph, then in 1 hour we have traveled
Z
1
d(1) =
80dt
Z 1
= 80
dt
0
1
= 80t 0
0
= 80 (1 − 0)
= 80.
Patrick Bardsley
Nerdy is the new black
Let’s say v (t) = 80 mph, then in 1 hour we have traveled
Z
1
d(1) =
80dt
Z 1
= 80
dt
0
1
= 80t 0
0
= 80 (1 − 0)
= 80.
Wow, if we are traveling 80mph, in one hour we will have traveled
80 miles! Who knew?!
Patrick Bardsley
Nerdy is the new black
Thanks for a great semester!
Good luck on your finals and
have a great winter break!
Patrick Bardsley
Nerdy is the new black