Mathematical Review Problems

Transcription

1 Fall 6 Louis Scuiero Mathematical Review Problems I. Polynomial Equations an Graphs (Barrante--Chap. ). First egree equation an graph y f() x mx b where m is the slope of the line an b is the line's intercept example fx ():= x x range is - to slope m = fx () intercept b = - 4 x. Secon egree equation an graph y f() x ax bx c root b b 4ac a root b b 4ac a example g() z := z 3 z gz () 5. Fin the roots by using the quaratic formula ( 3) ( 3) 4 z := z =.68 z

2 . By using MathCAD biult in "root" function z := initial guess value relatively close to the roots ( ) z root z 3 z, = Using the MathCAD "Given" an " Fin" solving block Z := 6 Given Z Fin( Z) =.68 3 Z. Roots of Polynomial equations ( 3) ( 3) 4 z := z =.38 y h( v) av 4 bv 3 cv v e Fining the roots of h(v) coul be very ifficult. The stanar way is to gragh the function example 4 ( ) := z kz 3 z 3 z z kz ( ) 3 3 The roots are the value of v for which k(v) is zero z. Use MathCAD "Symbolics" keywor coeffs from the Symbolic toolbar an the built in"polyroots" function Q := k z ( ) coeffs, z 3 value of e value of q := polyroots( Q) q T = ( ) q =

3 j :=,.. 3 ( ) ( ) kz kq j z, q j Plot the following in plane polar corrinates from to π. Examples r ( θ) 8 ( ) := 3 tan( θ) r θ θ 6 3 θ :=, cos( θ) 3 33 ( ) ( ) := 3 cos( θ) r θ 5 3 cos θ θ 3 Problem (Barrante).(b,i): Determine the roots of the equations an state whether in each case the roots are the zeros of the function. ( y ) 6x (Go to "Symbolic" toolbar an select "solve") x y 5 x 6 3.(b,e): Plot the following functions in Cartesian coorinates y 3x 9 PV k k is a constant

4 Fin the roots by the graph metho an then by calculation y x 6x 3 Fin the roots usinf the graph an the "root" function Problem (- (i,j), Barrante) Determine the roots of the following equations, an then state whether in each case the roots are the zeros of the function -(i) x y 4 -(j) ( x ) ( y 4) 9 - (b,e), Barrante) Plot the following functions in plane polar coorinates from to π - (b) - (e) r r 4 sinθ 4 sinθcosθ II. Partial Derivatives (Barrante--Chap. 4 ) Problem (n.q) 4.(n, q) Differentiate the following functions. (All upper case letters are constants.) (n) e E A z 7 8 z <-- Use <cntrl>= in equation. Set cursor at z, pull own Symbolics menu, select ifferentiate on variable. Then solve for z using again the Symbolics menu; select variable an solve with eiting cursor on z the fz ( ) (q) E A z 7 8 z <-- Repeat using new function that explicitly epens on z. 3 cos( θ) Then solve for θ using again the Symbolics menu; select variable an solve with the eiting cursor on θ

5 III. Integral Calculus (Barrante--Chap. 5) Two major approaches to integral calculus.. Consier the integral as an antierivative (inverse of ifferentiation). y = f(x) therefore y/x = f(x)/x = f(x)' = y' y = f(x)' x (ifferential form) The question to pose is what function f(x) when ifferentiate gives f(x)' the first erivative? The function f(x) is calle the integral of the ifferential an is note fx () fx ( )' x. Consier the integral as the sum of many similar, infinitesimal elements (area uner the curve). This assigns a physical meaning to the integral. For a more complete treatement of Integral Calculus you are encourage to rea Barrante Chapter 5 Problem (a), 5- (a) an 5-3 (h) 5- (a) 5 x 3 x 5- (a) e 4x x 5-3 (h) for a equals, 5 x ( ax ) e x IV. Matrices an Determinants (Barrante--Chap 9) A matrix is efine as a -D array of numbers. It can containe an equal numbers of rows an columns (rectangle) or ifferent number of rows an columns an it is calle an mxn matrix. The simplest matirx is a column or a row matrix with one column or one row. example a := b := c := 3 := 4 a b M := eteminant is obtaine by a x - b x c where a, b, c an are c the elements of the matrix

6 eterminant : a b c = MathCAD built in function M = Matrix Algebra Interchanging rows an columns oes not change the eteminant of the matrix. Interchanging any rows or columns will change the sign of the eterminant. The eterminant of a square matrix is zero if any rows or columns are ientical. The eterminant is multiplye by k if any row or column is multiplie by a number k. Two matrices are ae by aing their elements. Two matrices are multiplie as shown below Example M a b a b M c c Aition Multiplicatio M M M M a c b a c b c a c b a b a c b c a b Solving linear equations Example x y z 3 x y z 5 x y 3 z 5 This set of equation can also be represente by the prouct of matrices x M 3 := 3 coefficient of x, y an z M 4 y prouct M 3 M 4 3 z 5 5 a) calculate the eterminant of the coefficient D := M 3 D = 5 b) calculate the eterminant of the M 3 after substituing the first column with, -5, -5 M' 3 := D := M' 3 D = 5 c) calculate the eterminant of the M 3 after substituing the secon column with, -5, -5

7 M'' 3 := 3 5 D := M'' D = ) calculate the eterminant of the M 3 after substituing the thir column with, -5, -5 Finally M''' 3 := 3 5 D3 := M''' 3 5 D3 = 5 D D3 D y := z := x := D y = D z = 3 D x = Problem 6 Matrix algebra. Given the following matrices: A :=, (a) Compute the eterminant A. Bx () x x := z x x r s t u (b) Fin the roots of the polynomial equation Bx (). (c) Solve for the components of z when Az 4. V. Differential Equations (Barrante--Chap 7) Format require to solve a ifferential equation or a system of ifferential equations using one of the comman-line ifferential equation solvers such as rkfixe, Rkaapt, Raau, Stiffb, Stiffr or Bulstoer. For a numerical routine to solve a ifferential equation (DE), we must somehow pass the ifferential equation as an argument to the solver routine. A stanar form for all DEs will allow us to o this.

8 Orinary ifferential equations (first orer).. to solve this type of DE the solving block shown below shoul be use (Given an Oesolve) Given y(x) = f(x) initial conition y(x) = for instance y:= Oesolve (x, x en ) plot the solution The iea is to change the n-th orer ODE into a system of n couple first-orer ifferential equations. Basic iea: get ri of any secon, thir, fourth, etc. erivatives that appear, leaving only first erivatives. Example: 3 y 3 x x y ( x ) x y e 5 x This is thir orer, has non constant coefficients an is neither linear nor homogeneous. To solve it we will introuce 3 new functions: Ax () y, Bx () y, ' Cx ( ) y''. Then the erivatives of A, B, C are simply given by x A B, x B C. An the erivative of C can be foun by solving the given ifferential equation: ( ) B x C x A x e 5 x Suppose the initial conitions are: y()=, y'()=, y''()=-. We will fin an plot the values of y(x) in the interval <x<4. y := r := D( xy, ) steps := y y x y ( x ) ( y ) e 5 x x := x := 4 the interval The integration is efine as follows: 5 arguments, y which is a vector (here a 3x matrix) with the BC of the solution an the first an secon erivative at the initial value, then x an x which represent the initial an final values of x, r the number of points between initial an final values an finally the vector D that contains the first an the secon erivative of the solution

9 i :=.. r inex for steps Q := rkfixe( y, x, x, r, D) Runge-Kutta integration Plot the solution its erivative its secon erivative Q.5 Q.5 4 Q 4 4 Q Q Q 3 Q = x y y' y'' Problem 5 (7- (a, c, e)) () a x y 3 y interval < u < initial conitions y() = nstep := y x x y y () c interval < u < r := initial conitions y() =, y'() = an y'' () = number of steps () e y 9 y x interval < u< 5 n := 5 initial conitions y() = -, y'() = an y''() =