ME 349 Quizzes

Variable x=‐0.1<0.0<0.1 is equal to a. 1 b. 0

0

Variable x=‐1&‐2 is equal to a. 1 b. ‐3

1

Variablex=0 | 3 is equal to a. 1 b. 0

1

Variablex=~(‐1<‐2)isequalto a. 1 b. 0

1

If function follows to a power scaling law then the plot of this function is linear in a. Double logarithmic scale (log(x), log(y)). b. Semi‐logarithmic scale (x, log(y)) .

Double logarithmic scale (log(x), log(y)).

Euler method for the initial value problems is the method of: a. First order of approximation and the local truncation error Tau(i) ~∆x. b. First order of approximation and the local truncation error Tau(i) ~∆x^2. c. Second order of approximation and the local truncation error Tau(i) ~∆x^2.

First order of approximation and the local truncation error Tau(i) ~∆x^2.

Let's assume that the MATLAB function Fun (x) calculates value of an integrand in a definite integral for the interval from a to b. Then the value of the integral I can be determined as: a. I=quad(Fun,[a,b]). b. I=quad(@Fun,a,b).

I=quad(@Fun,a,b)

Theline[xy]=MyFun(z)is a. Correct function definition line. b. Incorrect function definition line.

Incorrect function definition line.

The expression [1:3] + [ 1 2 ] is a. Valid MATLAB expression b. Invalid MATLAB expression

Invalid MATLAB expression

The expressionx=sin[pi/3.0] a. Valid MATLAB arithmetic expression b. Invalid MATLAB expression

Invalid MATLAB expression

Thelineforx<3is a. Valid beginning of the for‐end loop. b. Invalid beginning of the for‐end loop.

Invalid beginning of the for‐end loop.

Matrix product [1 2][3 4] a. Is equal to 11. b. Is invalid and cannot be calculated

Is invalid and cannot be calculated

InMATLAB,matrixproduct[1 2]*[3 4] a. Is equal to 11. b. Is invalid and cannot be calculated.

Is invalid and cannot be calculated.

Matrix I [1 1] [1 1] a. Is the identity matrix. b. Is no the identity matrix.

Is no the identity matrix.

CHECK QUIZ #9 Q5

JJL;AF

Local truncation error of numerical solution of an IVP is a. Difference between numerical and accurate solutions of the problem appeared after one integration step due to finite‐difference approximation of derivatives in the ODE. b. Round‐off error appeared due to computer calculations with finite number of digits.

a. Difference between numerical and accurate solutions of the problem appeared after one integration step due to finite‐difference approximation of derivatives in the ODE.

Symmetric matrix is a matrix which elements satisfy to a. aij =1, all other elements are zeros. b. aij=aji

aij=aji

Variable c=(x<3)|(x>5) at x=4 is equal to a. 1 b. 0.

b. 0

Expression [ 1 2 ].^2 + 1 is equal to a. Value 10. b. Array[2 5]. c. Array[1 4 1].

b. Array[2 5].

The singular matrix is a matrix with a. det 0. b. det 1.

det 0.

Function C = polyfit ( X, Y, 2 ) returns array of coefficients of a fitting polynomial a. f(x)=C[1]x^2+C[2]x+C[3] b. f(x)=C[1]x+C[2] c. f(x)=C[3]x^2+C[2]x+C[1]

f(x)=C[1]x^2+C[2]x+C[3]

Assume that function Fun (x) calculates the LHS of the equation 0. Then the root of this equation in the interval , can be found with a. fzero ( Fun, a, b ). b. fzero(@Fun,[ab]).

fzero(@Fun,[ab]).

In order to plot two graphs, y1 vs. x1 and y2 vs. x2, one can use the command a. plot(x1,y1,x2,y2) b. plot(x1,x2,y1,y2)

plot(x1,y1,x2,y2)

Rectangle quadrature rule allows us to numerically calculate a. Definite integral. b. Antiderivative (indefinite integral).

Definite integral.

Theexpressionplot([1:3],sin([1:3]))is a. Valid MATLAB expression b. Invalid MATLAB expression

Valid MATLAB expression

The constant 1.275e‐3 represents the number a. 1.275 ∙ 10^-3 b. 1.275 ∙e^-3

1.275 ∙ 10^-3

Arithmetic expression 2*3^2 is equal to a. 36 b. 18

18

det([100;0 4 0;0 0 2])is equal to a. 0. b. 8.

8

Matrix product [1] [2 0] [2] [0 2] a. Is equal to row vector [2 4 ]. b. Cannot be evaluated.

Cannot be evaluated.

Command format allows us to: a. Change visual appearance of commands in the MATLAB command window b. Change representation of float‐point values in the MATLAB command window

Change representation of float‐point values in the MATLAB command window

Command clc a. Deletes all variables in the MATLAB workspace b. Clears the content of the MATLAB command window

Clears the content of the MATLAB command window

The MATLAB command OO = eye ( 4 ) a. Create zero matrix of size 4 x 4. b. Creates identity matrix of size 4 x 4.

Creates identity matrix of size 4 x 4.

If the data table contains N points, the degree of the interpolation polynomial is equal a. N+1 b. N‐1 c. N

N‐1

Initial value problem for the first‐order differential equation y'=f(x,y) includes a. One initial condition y(a)=K0 . b. Two initial conditions y(a)=K0,y'(a)=K1

One initial condition y(a)=K0

The for‐end loop is designed to a. Implement a loop with arbitrary condition. b. Perform calculations a pre‐determined number of times.

Perform calculations a pre‐determined number of times.

Command plot ( y, x ) a. Plots values of array y along horizontal axis b. Plots values of array y along vertical axis

Plots values of array y along horizontal axis

The function call polyval ( [ 2 1 ‐3 ], x ) returns the value of a. Polynomial 2x^2+x-3. b. Polynomial -3x^2+x+ 2.

Polynomial 2x^2+x-3.

Finite difference equation yi'= (yi+1-yi-1)/(2deltax) has a. First oder of approximation b. Second order of approximation

Second order of approximation

Finite difference equation (yi+1-2yi+yi-1)/deltax^2 a. Second-order derivative yi'' b. First-order derivative yi'

Second-order derivative yi''

The Newton‐Raphson method for the root finding requires a. Two initial values a and b defining the interval ,[a b] , where the root presents. b. Single initial approximation for the root.

Single initial approximation for the root.

The major advantage of the Newton‐Raphson method is a. Speed and ability to solve systems of equations. b. Robustness.

Speed and ability to solve systems of equations.

eye(2)\[1;2]is a. Valid MATLAB expression. b. Invalid MATLAB expression.

Valid MATLAB expression.

Let's assume that MATLAB function Fun ( x, y ) calculates the RHS of ODE y′=f(x,y). Then the initial value problem for this equation with initial condition y(a)=ya in the interval from a to b can be solved with a. [x,y]=ode45(@Fun,[a,b],ya). b. [x,y]=ode45(@Fun,ya,a,b).

[x,y]=ode45(@Fun,[a,b],ya).

After execution of the command x = [1:2:5].^2, variable x is a. The array[1 9 25] b. The array[1 5 9 13 17 21 25]

The array[1 9 25]

Coefficients of the fitting polynomial are calculated assuming that a. The mean square difference between the polynomial and data points is at minimum. b. The polynomial goes precisely through every point in the data table.

The mean square difference between the polynomial and data points is at minimum.

Coefficients of the interpolation polynomial are calculated assuming that a. The mean square difference between the polynomial and data points is at minimum. b. The polynomial goes precisely through every point in the data table.

The polynomial goes precisely through every point in the data table.

Solution of an interpolation problem allows us a. To find values of a function given in tabulated form for arbitrary . b. To find the root of the equation 0.

To find values of a function given in tabulated form for arbitrary .

One integration step with the Runge‐Kutta method of the 2nd order requires a. One calculation of the RHS function , f(x,y) . b. Two calculations of the RHS function , f(x,y).

Two calculations of the RHS function , f(x,y).

The Bisection method for the root finding requires a. Two initial values a and b defining the interval [a,b], where the root presents. b. Single initial approximation for the root.

Two initial values a and b defining the interval [a,b], where the root presents.

MATLABcommandA=[2 3;‐2 ‐3; 0 0]defines a. One dimensional array of size 6. b. Two‐dimensional array of size 2 x 3. c. Two‐dimensional array of size 3 x 2.

Two‐dimensional array of size 3 x 2.