DERIVATIVES AND DIFFERENTIAL EQUATIONS

1. DERIVATIVESANDDIFFERENTIAL EQUATIONS June 23,’99 i + v -

2. WHAT IS A DERIVATIVE? • Derivative of f(x) at x=a is given by f’(a) y=f(x) f(a+h) f(a) a a+h

3. SIGNIFICANCE OF DERIVATIVE • In the limit, derivative at x=a is equal to the slope m of the tangent line at a. y=f(x) f(a) a a+h

4. APPLICATIONS OF DERIVATIVE • The most widely used application of derivative is in finding the extremum (max or min) points of a function. • If a function has a local extremum at a number c then either f’(c)=0or f’(c) doesn’t exist

5. CRITICAL NUMBERS • A number c in the domain of a function f is a critical number of f if either f’(c)=0 or f’(c) does not exist f’(c)=0 x=c

6. DERIVATIVES IN MATLAB:diff • MATLAB computes two differentials, dy and dx, using diff, to arrive at the derivative • If x and y are input array of numbers dy=diff( y) dx=diff(x) • Then, yprime=diff(y)./diff(x)

7. How diff works • Take the array y=[3 5 7 5 9]; • diff(y) simply is the pairwise differences, i.e. diff(y)=[5-3 7-5 5-7 9-5] which is equal to diff(y)=[2 2 -2 4] • Naturally, there is one fewer term in diff(y) than it is in y

8. Using diff to find derivatives • Let y=[10 25 30 50 10] for x=[1 2 3 4 5] • Then diff(y)=[15 5 20 -40] diff(x)=[1 1 1 1 1] • Dividing term-by-term yprime=[15 5 20 -40] One fewer component

9. WORKING WITH humps • humps is a built-in MATLAB function, like peaks and is given by

10. HOW IT LOOKS

11. Try it!:derivative of humps • First, we must know how humps was created in order to know dx x=0:0.01:1;%dx in this case is 0.01 y=humps(x); • Now use dy=diff(y); dx=diff(x); yprime=dy./dx;

12. PLOT OF dy/dx Function peaks Deriv. goes to 0

13. Finding critical points:theory vs. application • To find max. or min.we should look for instances where y’=0. • This is where the difference between textbook methods and the real world shows up. • To see for yourself look for instances where yprime=0.

14. Where are the zeros? • If you did the search right, you will realize that nowhere in yprime you can actually see a zero • Since there are no points where y’ is exactly zero, we should look for points of transition from positive to negative, i.e. sign changes + +

15. Zero crossing and sign transition • In Matlab, all derivatives are discrete. Some are positive, some negative but no zeros • Since derivative goes from + to -, we can infer that it went to zero somewhere in between Zero crossing

16. FINDING ZERO CROSSINGS • There are two ways to find where a function crosses zero; 1. Numerical 2. Algebraic(polynomial fitting)

17. ZC USING ARRAY OPERATION • To find where a sign transition takes place, multiply two consecutive numbers in the derivative array. If there is a sign change, the product is negative [10 15 16 9 -4 -3 -4 -9] Zero crossing X X X X X X X + + + - + + -

18. MATLAB’s WAY • Let x be an n element array[x1,x2,…,xn]. To generate the zero crossing array, do y=x(1:n-1).*x(2:n) • Algebraically, this is what is happening y=[x1x2 x2x3 ,..., xn-1xn]

19. Try it! • Evaluate humps in the range x=0 to 1 in increments of 0.01 • This gives you a 1x101 array. Find yprime. This is now a 1x100 array • Find where in this 101 positions sign changes occur

20. LOCATING SIGN CHANGES • Just look for instances of negative sign in the zero crossing array 1st=0.29 2nd=0.63 3rd=0.88

21. CLOSER LOOK Sign change

22. SECOND ZC

23. THIRD ZC

24. Differential Equations i + v -

25. Simplest Diff.Eq • A first order differential equation is of the form y ’=dy/dx=g(x,y) • We are looking for a function y such that its derivative equals g(x,y)

26. FEW EXAMPLES • Find y that such that y ’=3x2 y ‘=y y ‘=(2x)cos2y • Answers: y=x3 y=exp(x) y=???

27. SOLVING ODE’s USING MATLAB • MATLAB solves ordinary DE in two ways: 1):numerical and 2):symbolic • For numerical solution use [x,y]=ode23(‘function’,a,b,initial) • See next slide for usage

28. EXPLAINING ode23 • Here are the elements of [x,y]=ode23(‘function’,a,b,initial) function - this is g(x,y) in y ’=g(x,y). Must be written as a separate MATLAB function a - left point of the interval b - right point of the interval initial - initial condition, i.e. y(a)

29. USING ode23 • Let’s solve y ‘ =3x2 • First write a function as follows function yprime=anything(x,y) yprime=3*x.^2; • Then call [x,y]=ode23(‘anything’,2,4,0.5) • y is the solution. Plot x vs.y

30. RC CIRCUIT • Equation describing a source-free RC circuit is vc C R ic

31. SOLVING FOR VOLTAGE • The analytical solution is • where Vo is the initial voltage across the capacitor

32. MATLAB’s WAY • To use ode23, we need to cast the problem in the form of y ‘=g(x,y) • Here, y is vc. . Therefore, vc’=-(1/RC)vc • With RC=10^-3, write a function like function vcprime=rc(t,vc) vcprime=-(10^3)*vc

33. SOLVING FOR vc • Use ode23 as follows [t,vc]=ode23(’rc’,0,1,2)

34. L ~ R CIRCUIT WITH FORCED RESPONSE • Take the following RL circuit(p.497) i

35. L ~ R WHAT IS i ? • From circuits analysis we know

36. SOME NUMBERS • Let R=2 ohm, L=1 H and vs=10cos3t • Here is what we have Z=sqrt(9+4)=sqrt(13) Vm=10 =56.3o

37. EXERCISE- VERIFY WITH MATLAB • Use ode23 to solve for current. Match the two results, graphically, to see if the amplitude, phase and frequencies match the theoretical result