Nerdy is the new black

Transcription

1 January 20, 2014

2 Overview Applications of derivatives

3 Overview Applications of derivatives Weather forecasting

4 Overview Applications of derivatives Weather forecasting Optimizing material design

5 Overview Applications of derivatives Weather forecasting Optimizing material design Applications of integrals

6 Overview Applications of derivatives Weather forecasting Optimizing material design Applications of integrals Fourier analysis

7 Overview Applications of derivatives Weather forecasting Optimizing material design Applications of integrals Fourier analysis Spectrum analysis of acoustic signals

8 Overview Applications of derivatives Weather forecasting Optimizing material design Applications of integrals Fourier analysis Spectrum analysis of acoustic signals The Fundamental Theorem of Calculus

9 Weather forecasting The temperature T (t) in a given area (SLC) can be described as dt dt = f (t, T, wind, cloud cover, humidity,...) where f is some function of all of those variables. As a simple example consider ( ) dt 1 dt = 6 sin 5 t t, T (0) = 40

10 ( ) dt 1 dt = 6 sin 5 t t

11 ( dt 1 dt = 6 sin dt = [ 6 sin 5 t ( 1 5 t ) 0.001t ) ] 0.001t dt

12 ( dt 1 dt = 6 sin dt = dt = [ 6 sin 5 t ( 1 5 t ) 0.001t ) ] 0.001t dt [ ( ) ] 1 6 sin 5 t t dt

13 ( dt 1 dt = 6 sin dt = [ 6 sin 5 t ( 1 5 t ) 0.001t ) ] 0.001t dt [ ( ) ] 1 dt = 6 sin 5 t t dt [ ( ) ] 1 T (t) = 6 sin 5 t t dt + C

14 ( dt 1 dt = 6 sin dt = [ 6 sin 5 t ( 1 5 t ) 0.001t ) ] 0.001t dt [ ( ) ] 1 dt = 6 sin 5 t t dt [ ( ) ] 1 T (t) = 6 sin 5 t t dt + C Problem: We can t integrate this! But... we could use an approximation of this function, which we find from a linearization.

15 Linearized differential equation Using the linearization formula f (t) f (a) + f (a)(t a) = L(t)

16 Linearized differential equation Using the linearization formula f (t) f (a) + f (a)(t a) = L(t) we find the linearization L(t) of the function ( ) 1 f (t) = 6 sin 5 t t.

17 Linearized differential equation Using the linearization formula f (t) f (a) + f (a)(t a) = L(t) we find the linearization L(t) of the function ( ) 1 f (t) = 6 sin 5 t t. Then we could use the approximation dt dt = f (t) L(t)

18 Linearized differential equation Using the linearization formula f (t) f (a) + f (a)(t a) = L(t) we find the linearization L(t) of the function ( ) 1 f (t) = 6 sin 5 t t. Then we could use the approximation dt dt = f (t) L(t) and we can solve the approximate differential equation to find T (t) = f (t)dt + C L(t)dt + C.

19 The linearization near time equal to zero of ( ) 1 f (t) = 6 sin 5 t t. is given by L(t) = f (0) + f (0)(t) = 0.001t.

20 So solving the linearized differential equation we have dt dt = 0.001t

21 So solving the linearized differential equation we have dt dt = 0.001t dt = 0.001tdt

22 So solving the linearized differential equation we have dt dt = 0.001t dt = 0.001tdt dt = 0.001tdt + C

23 So solving the linearized differential equation we have dt dt = 0.001t dt = 0.001tdt dt = 0.001tdt + C T (t) = t2 2 + C

24 So solving the linearized differential equation we have dt dt = 0.001t dt = 0.001tdt dt = 0.001tdt + C T (t) = t2 2 + C T (t) = t 2 + C

25 So solving the linearized differential equation we have dt dt = 0.001t dt = 0.001tdt dt = 0.001tdt + C T (t) = t2 2 + C T (t) = t 2 + C and using the fact that T (0) = 40 we find a temperature model T (t) = t

26 45 44 True Temperature Approximate Temperature Temperature Hours Why can t we trust the 7-day forecast? This linearized approximation is great for short times (within 2 hours in this case).

27 80 75 True Temperature Approximate Temperature Temperature 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.

28 Optimal design of a climbing cam The curved shape of a climbing cam

29 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.

30 If this camming unit is placed inside a parallel crack we can look at a force diagram to understand what keeps climbers alive.

31 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.

32 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.

33 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.

34 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).

35 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).

36 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 x 2 + y 2 = e 2µ tan 1 (y/x).

37 Implicit differentiation gives 2x + 2yy = e 2µ tan 1 (y/x) 2µ x 2 + y 2 (xy y)

38 Implicit differentiation gives Gross algebra gives 2x + 2yy = e 2µ tan 1 (y/x) 2µ x 2 + y 2 (xy y) y = 2x ye2µ tan 1 (y/x) 2y 2µe2µ tan 1 (y/x) x 2 +y 2

39 Implicit differentiation gives Gross algebra gives 2x + 2yy = e 2µ tan 1 (y/x) 2µ x 2 + y 2 (xy y) y = 2x ye2µ tan 1 (y/x) 2y 2µe2µ tan 1 (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).

40 Now we can relate the total force applied to the center pin from the camming action of the device to find F = T µ 2.

41 Now we can relate the total force applied to the center pin from the camming action of the device to find F = T µ 2. 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).

42 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) = a 0 +a 1 cos(πx)+b 1 sin(πx)+a 2 cos(2πx)+b 2 sin(2πx)+

43 Example: f (x) = x 2 on the interval 0 x 2

44 Example: f (x) = x 2 on the interval 0 x 2 x 2 = a 0 + a 1 cos(πx) + b 1 sin(πx) + a 2 cos(2πx) + b 2 sin(2πx) + where

45 Example: f (x) = x 2 on the interval 0 x 2 x 2 = a 0 + a 1 cos(πx) + b 1 sin(πx) + a 2 cos(2πx) + b 2 sin(2πx) + where 2 a 0 = 1 2 a n = b n = x 2 dx = 4 3 x 2 cos(nπx)dx = 4 n 2 π 2 x 2 sin(nπx)dx = 4 nπ

46 So in a very general form we can write x 2 as Fourier series x 2 = (a n cos(nπx) + b n sin(nπx)) n=0 = cos(πx) + π2 π sin(πx)+ 1 2 cos(2πx) + π2 π sin(2πx) + This function is easy already (f (x) = x 2 ), but this will be very useful for more difficult functions!

47 Spectrum analysis A recording of a musical instrument is a time signal which may look like 0.1 Time signal time x 10 5

48 Spectrum analysis A recording of a musical instrument is a time signal which may look like 0.1 Time signal time x 10 5 This is just a function f (t) - Let s use Fourier series!

49 If T denotes the final time of the acoustic signal we are going to reconstruct the time function f (t) as where f (t) = a n cos(nπt) + b n sin(nπt), n=0 a 0 = 1 T T 0 f (t)dt T a n = C 1 f (t) cos(nπt)dt 0 T b n = C 2 f (t) sin(nπt)dt where we know exactly what C 1 and C 2 are. 0

50 Obviously, we can t use infinitely many terms or we d be working on one problem forever... literally. f (t) = a n cos(nπt) + b n sin(nπt) n=0

51 Obviously, we can t use infinitely many terms or we d be working on one problem forever... literally. f (t) = a n cos(nπt) + b n sin(nπt) n=0 So we use a finite amount N of them as an approximation: f (t) a n cos(nπt) + b n sin(nπt) N n=0 and we obtain the following sounds.

52 The Fundamental Theorem of Calculus Recall if then d(t) = t 0 v(τ)dτ

53 The Fundamental Theorem of Calculus Recall if then d(t) = t 0 v(τ)dτ d (t) = v(t). If v represents velocity (mph), then d(t) represents the distance traveled (mi) over the time interval [0, t].

54 Let s say v(t) = 80 mph, then in 1 hour we have traveled d(1) = dt

55 Let s say v(t) = 80 mph, then in 1 hour we have traveled d(1) = 1 = 80 80dt dt

56 Let s say v(t) = 80 mph, then in 1 hour we have traveled d(1) = 1 = 80 80dt = 80t 0 dt

57 Let s say v(t) = 80 mph, then in 1 hour we have traveled d(1) = 1 = 80 80dt dt = 80t 0 = 80 (1 0)

58 Let s say v(t) = 80 mph, then in 1 hour we have traveled d(1) = 1 = 80 80dt dt = 80t 0 = 80 (1 0) = 80.

59 Let s say v(t) = 80 mph, then in 1 hour we have traveled d(1) = 1 = 80 80dt dt = 80t 0 = 80 (1 0) = 80. Wow, if we are traveling 80mph, in one hour we will have traveled 80 miles! Who knew?!

60 Thanks for a great semester! Good luck on your finals and have a great winter break!