MS327. Diagnostic Quiz. Are you ready for MS327? Copyright c 2017 The Open University 1.4

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1 Faculty of Science, Technology, Engineering and Mathematics MS327 Deterministic and stochastic dynamics MS327 Diagnostic Quiz Are you ready for MS327? This diagnostic quiz is designed to help you decide if you are ready to study Deterministic and stochastic dynamics (MS327). This document also contains some advice on preparatory work that you may find useful before starting MS327 (see below and page 7). The better prepared you are for MS327 the more time you will have to enjoy the mathematics, and the greater your chance of success. The topics which are included in this quiz are those that we expect you to be familiar with before you start the module. If you have previously studied Mathematical methods, models and modelling (MST20) or Mathematical methods (MST224) then you should be familiar with most of the topics covered in the quiz. We suggest that you try this quiz first without looking at any books, and only look at a book when you are stuck. You will not necessarily remember everything; for instance, for the calculus questions you may need to look up a set of rules for differentiation or integration or use the table of standard derivatives or integrals provided on page 2. This is perfectly all right, as such tables are provided in the Handbook for MS327. You need to check that you are able to use them though. If you can complete the quiz with only the occasional need to look at other material, then you should be well prepared for MS327. If you find some topics that you remember having met before but need to do some more work on, then you should consider the suggestions for refreshing your knowledge on page 7. Try the questions now, and then see the notes on page 7 of this document to see if you are ready for MS327. (The answers to the questions begin on page 8.) Do contact your Student Support Team via StudentHome if you have any queries about MS327, or your readiness to study it. Copyright c 207 The Open University.4

2 Tables of standard derivatives and integrals Function Derivative Function Integral a 0 x a e ax ln(ax) sin(ax) cos(ax) tan(ax) cot(ax) sec(ax) cosec(ax) arcsin(ax) arccos(ax) arctan(ax) arccot(ax) arcsec(ax) arccosec(ax) ax a ae ax x a cos(ax) a sin(ax) asec 2 (ax) acosec 2 (ax) a sec(ax) tan(ax) acosec(ax) cot(ax) a a 2 x 2 a a 2 x 2 a +a 2 x 2 a +a 2 x 2 a ax a 2 x 2 a ax a 2 x 2 a x a (a ) ax+b ax e ax a eax x a+ a+ a ln ax+b ln(ax) x(ln(ax) ) sin(ax) cos(ax) tan(ax) cot(ax) sec(ax) cosec(ax) sec 2 (ax) x 2 +a 2 x 2 a 2 a cos(ax) a sin(ax) a ln cos(ax) a ln sin(ax) a ln sec(ax)+tan(ax) a ln cosec(ax) cot(ax) a tan(ax) cosec 2 (ax) a cot(ax) ( x ) x 2 +a 2 a arctan a (x a)(x b) a b ln a x x b ln(x+ ( x x 2 +a 2 ) or arcsinh a) a 2 x 2 ln x+ ( x x 2 a 2 or arccosh a) ( x arcsin a) page 2 of 6

3 Diagnostic quiz questions Differentiation Question Differentiate the following functions with respect to x. (a) f(x) = x2 +2x 3x+4 Question 2 (b) f(x) = sin(3x 2 +2) Find the second derivative with respect to x of f(x) = 3x 4 +2sin3x. Question 3 Find and classify the stationary points of y = x 3 +3x 2 24x+0. 2 Integration Question 4 Calculate the following integrals. 2 ( (a) x ) dx (b) x 2 (x 2 +5) 0 dx (c) x xcos3xdx 3 Ordinary differential equations Question 5 Find the particular solution of the differential equation y > 0, which satisfies y() = 5 2. dy dx = xy x 2 +, where Question 6 Find the general solution of the differential equation dy dx + 2y x = x. Question 7 Find the particular solution of the differential equation d 2 y dx 2 2dy +5y = 5x+8 dx that satisfies y(0) = 4 and dy (0) = 9. dx page 3 of 6

4 4 Partial differentiation Question 8 Find the second-order partial derivatives of the function u(x,y) = cos(4x+3y), and confirm that Question 9 2 u x y = 2 u y x. Use the Chain Rule to find du dt, where u(x,y) = ln( x 2 y 4) and x = cos2t, y = sint. 5 Matrices Question Let A = 2 0, B = 3 ( ) 3 and C = 4 2 ( ) Calculate each of the following, or state why the calculation is not possible. (a) AB (b) BA (c) 2A+3B (d) 2B+C Question Calculate the determinants of the following matrices. ( ) (a) (b) Question 2 Find the eigenvalues and eigenvectors of the matrix ( ) 8. 5 Question Find the eigenvalue of the matrix that corresponds to the eigenvector. 3 page 4 of 6

5 6 Vector algebra and calculus Question 4 If a = 3i+2j 5k and b = 2i j k, find (a) a+b, a+b, a b, and a b, (b) the angle between a and b, to the nearest degree. Question 5 If φ(x,y,z) = x 2 y +2xz, find φ (also known as gradφ) at the point (2, 2,3). Question 6 If F = x 2 yi 2xzj+2yzk, find F (also known as divf) and F (also known as curlf) at the point (,2, ). 7 Multiple integration Question 7 Evaluate the surface integral S xyda, where S is the region of the plane z = 0 bounded by the curve y = x 2 and by the line y = x. Question 8 Evaluate the volume integral B ( x 2 y +2xz 2) dv, where B is the interior of the cube with faces in the planes x = 0, y = 0, z = 0, x =, y = and z =. 8 Mechanics Question 9 The acceleration of particle of mass 5kg is given by a = 3sin6ti+3cos6tj+4k. At t = 0 the velocity, v(0) = i and the position, r(0) = j+k. (a) Find the magnitude of the force acting on the particle. (b) Find the position of the particle at any time t. page 5 of 6

6 Question 20 A particle connected to a spring moves in a straight line. The position of the particle along the line, x, is given by d 2 x 4x = 0. dt2 Initially, x(0) = 3 and x (0) = 0. (a) What is the speed of the particle as it passes the point x = 0 for the first time? (b) What is the period of the motion? 9 Complex numbers Question 2 If w = 2+i and z = 3 4i find, in Cartesian form (a) wz (b) w/z Question 22 Express w = +i and z = 3 i in complex exponential form (re iθ ), and hence express the following also in complex exponential form. (a) wz (b) w/z 0 Fourier series Question 23 Find a Fourier series for the function f(x) = +x for < x <. page 6 of 6

7 What can I do to prepare for MS327? Now that you have finished the quiz, how did you get on? The information below should help you to decide what, if anything, you should do next. Most of this quiz was unfamiliar to me. I don t think I have met any of the topics before. You should consider taking MST20 or MST224 before MS327. Contact your Student Support Team to discuss this. Some of these topics are familiar to me, but not all of them. If you have met less than 75% of these topics then you should consider taking MST20 or MST224 before MS327. Contact your Student Support Team to discuss this. If you have taken MST224, then the material in Section 8 might be new to you. If so, you might like to review some of the resources suggested below. I have met all these topics before, but I am a bit rusty on some, and had to look up how to do a few of them. You will benefit from doing some revision before starting MS327 see below for possible sources of help. I can do all these questions. You are probably well prepared for MS327, but do brush up on any topics you need to further enhance your confidence. If you have any queries contact your Student Support Team via StudentHome. What resources are there to help me prepare for MS327? If you have studied MST20, its predecessor MST209, or MST224, then you could use parts of these to revise for MS327. If you need to brush up some of the more basic topics like algebra, trigonometry and calculus, then you may find revising material from Essential mathematics (MST24) and Essential mathematics 2 (MST25) (or their predecessors MST2 and MS22) helpful. Alternatively, such material is often covered by standard A-level textbooks. If you have studied MST224 rather than MST20 then you might not have met mechanics before (Section 8 of the diagnostic quiz). You might like to review some material on this, such as the MST20 Bridging material available at mcs-notes2.open.ac.uk/webresources/maths.nsf/a/odm8/$file/mst20bridging.pdf, or an A-level mechanics textbook. The mathcentre web-site ( includes several teach-yourself books, summary sheets, revision booklets, online exercises and video tutorials on a range of mathematical skills. page 7 of 6

8 Diagnostic quiz solutions Solution (a) f (x) = (3x+4)(2x+2) 3(x2 +2x ) (3x+4) 2 (b) f (x) = 6xcos(3x 2 +2) = 3x2 +8x+ (3x+4) 2 Solution 2 f (x) = 2x 3 +6cos3x, so, f (x) = 36x 2 8sin3x. Solution 3 The stationary points occur when dy dx = 0, that is, when 3x2 +6x 24 = 0. This can be written 3(x 2)(x+4) = 0, so the stationary points are x = 2 and x = 4. At x = 2, y = 8 and at x = 4, y = 90. The second derivative, d2 y dx 2 = 6x+6. At x = 2, d2 y > 0, so (2, 8) is a minimum. dx2 At x = 4, d2 y < 0, so ( 4,90) is a maximum. dx2 Solution 4 2 ( (a) x ) dx = x (b) [ ] 2 3 x3 +2lnx = ln2 3 = ln2. x 2 (x 3 +5) 0 dx = 33 (x3 +5) +c, where c is an arbitrary constant. (Using the substitution u = x 3 +5.) (c) Integrating by parts gives xcos3xdx = x 3 sin3x 3 sin3xdx = 3 xsin3x+ 9 cos3x+c where c is an arbitrary constant. page 8 of 6

9 Solution 5 The differential equation is of first order, and the right-hand side can be written as f(x)g(y), for some functions f(x) and g(y). So the equation can be solved using separation of variables. Rearrange the equation so that all the terms involving y are on the left-hand side, and all those involving x are on the right: Integrating both sides gives hence, dy y dx = x x 2 +. y dy = x x 2 + dx lny = 2 ln(x2 +)+c, where c is an arbitrary constant. So y = exp ( 2 ln(x2 +)+c ) = A x 2 +,where A = e c. Since y = 5 2 when x =, we have 5 2 = A 2, so A = 5 and hence y = 5 x 2 +. Solution 6 The equation is of the form dy +g(x)y = f(x) so can be solved using the dx integrating factor method. The integrating factor is p(x) = exp ( g(x)dx ), which in this case is ( ) 2 p(x) = exp x dx = exp(2lnx) = x 2. (x So, x 2 y = 2 x ) dx = x 3 dx = x4 +c, where c is an arbitrary 4 constant. Hence y = x2 4 + c x 2. Solution 7 First solve the associated homogeneous equation The auxiliary equation is λ 2 2λ+5 = 0, d 2 y dx 2 2dy +5y = 0. dx The solutions are, λ = 2± 4 20 = 2±4i = ±2i. 2 2 Hence the complementary function is y c = e x (Acos(2x)+Bsin(2x)), where A and B are arbitrary constants. Returning to the inhomogeneous equation, d2 y dx 2 2dy +5y = 5x+8, dx the form of the right-hand side suggests a trial solution of the form y p = ax+b. So, dy p dx = a and d2 y p dx 2 = 0. page 9 of 6

10 Substituting into the inhomogeneous equation gives 0 2a+5(ax+b) = 5x+8 which can be written 5ax+(5b 2a) = 5x+8. Comparing coefficients gives 5a = 5 and 5b 2a = 8. So, a = and b = 2. The particular integral is y p = x+2. The general solution is y = y c +y p, that is, y = e x (Acos(2x)+Bsin(2x))+x+2. When x = 0, y = 4, so 4 = A+2, and A = 2. The derivative of the general solution is (using the Product Rule) dy dx = ex ((A+2B)cos(2x)+(B 2A)sin(2x))+. When x = 0, dy/dx = 9, so 9 = A+2B + = 2B +3 and so B = 3. The particular solution required is y = e x (2cos(2x)+3sin(2x))+x+2. Solution 8 u = sin(4x+3y) 4 = 4sin(4x+3y) x u y = sin(4x+3y) 3 = 3sin(4x+3y) 2 u x 2 = ( 4sin(4x+3y)) = 4cos(4x+3y) 4 = 6cos(4x+3y) x 2 u x y = ( 3sin(4x+3y)) = 3cos(4x+3y) 4 = 2cos(4x+3y) x 2 u y 2 = ( 3sin(4x+3y)) = 3cos(4x+3y) 3 = 9cos(4x+3y) y 2 u y x = ( 4sin(4x+3y)) = 4cos(4x+3y) 3 = 2cos(4x+3y) y Comparison of Solution 9 The Chain Rule gives We know, So, 2 u y x with 2 u x y confirms that 2 u x y = 2 u y x. u x = x 2 2x = 2 x, du dt = u dx x dt + u dy y dt. dx dt = 2sin2t, du dt = 2 x ( 2sin2t)+ 4 4cost cost = y y Now substitute for x and y in terms of t: 4sin2t. x u y = y 4 4y3 = 4 y, du dt = 4cost sint 4sin2t cos2t = 4cott 4tan2t. dy dt = cost. page 0 of 6

11 Solution (a) AB = (b) BA cannot be calculated, since B is 2 2 and A is 3 2. (c) 2A+3B cannot be calculated, since A is 3 2 and B is 2 2. ( ) 7 7 (d) 2B+C = 0 4 Solution (a) = = (b) = = 4(4 0) (3 2) = 25 Solution 2 The characteristic equation is given by λ 8 5 λ = 0, that is ( λ)( 5 λ)+8 = 0 or λ 2 +4λ+3 = 0. This can be factorised as (λ+)(λ+3) = 0, hence the eigenvalues are and 3. ( ) x The eigenvectors x = associated with the eigenvalue λ are found by y solving ( )( ) λ 8 x = 0. 5 λ y For λ = this gives 2x+8y = 0 x 4y = 0. ( )( ) 2 8 x = 0 which can be written as 4 y ( ) 4 These are satisfied when x = 4y, so an eigenvector is. ( )( ) 4 8 x For λ = 3 the equation gives = 0 which can be 2 y written as 4x+8y = 0 x 2y = 0. These are satisfied when x = 2y, so an eigenvector is ( ) 2. page of 6

12 Solution 3 Multiplying the matrix by the given eigenvector gives = 2 = The product is 2 times the eigenvector, so the corresponding eigenvalue is 2. Solution 4 (a) a+b = (3i+2j 5k)+(2i j k) = 5i+j 6k. a+b = 5i+j 6k = ( 6) 2 = = 62. a b = ( )+( 5) ( ) = = 9. i j k a b = = (2 ( ) ( 5) ( ))i (3 ( ) ( 5) 2)j+(3 ( ) 2 2)k = 7i 7j 7k = 7(i+j+k). (b) The cosine of the angle θ between a and b can be calculated using cosθ = a b a b. a = ( 5) 2 = 38 and b = 2 2 +( ) 2 +( ) 2 = 6. 9 So, cosθ = = ( ) 3 3 Hence, θ = arccos = 53 (to the nearest degree). 2 9 Solution 5 The scalar field is φ(x,y,z) = x 2 y +2xz. The partial derivatives are φ φ = 2xy +2z, x y = x2, and φ z = 2x. So, φ = gradφ = φ φ φ i+ j+ x y z k = (2xy +2z)i+x2 j+2xk. At the point (2, 2,3), φ = gradφ = (2 2 ( 2)+2 3)i+2 2 j+2 2k = 2i+4j+4k. page 2 of 6

13 Solution 6 If F = F i+f 2 j+f 3 k, then F = divf = F x + F 2 y + F 3 z i j k and F = curlf = / x / y / z = F F 2 F 3 ) i ( F3 y F 2 z ( F3 x F z ) ( F2 j + x F ) k. y Here, F = x 2 yi 2xzj+2yzk, so, F = divf = 2xy 0+2y = 2y(x+) i j k and F = curlf = / x / y / z x 2 y 2xz 2yz = (2z +2x)i 0j+( 2z x 2 )k = 2(x+z)i (x 2 +2z)k. At (,2, ), F = divf = 2 2(+) = 8 and F = curlf = 2( )i ( 2)k = k. Solution 7 y 2 y = x First sketch the region of integration, S: 0 S 0 x Note that y = x and y = x 2 intersect where x = 0 and x =, giving the minimum and maximum values of x for the region S. In Cartesian coordinates we can write xyda = xydydx, in which we integrate with respect to y first, and then respect to x, and hence divide the region of integration into strips as shown in the diagram above. Using the diagram to identify the limits of x and y we obtain xyda = y=x Now, xydy = x [ ( y=x 2 2 y2] y=x = x y=x 2 2 x2 ) 2 x4 = 2 (x3 x 5 ), x= and hence xyda = 2 (x3 x 5 )dx = [ S x=0 2 4 x4 ] 6 x6 = 0 2 S S S x= ( y=x x=0 ( 4 ) = ) xydy dx. y=x 2 page 3 of 6

14 Solution 8 z First sketch the region of integration, B: In Cartesian coordinates, B x ( x 2 y +2xz 2) dv = B y ( x 2 y +2xz 2) dxdydz. Since the boundaries of the volume of integration are coordinate planes, we have ( x 2 y +2xz 2) ( )dv = x 2 y +2xz 2) dxdydz B = = = = = [ 3 x3 y +x 2 z 2 ( ) 0 3 y +z2 dydz [ ] y= 6 y2 +z 2 y dz y=0 ( ) 6 +z2 dz [ 6 z + 3 z3 ] z= z=0 ] x= x=0 dydz dz = = 2. page 4 of 6

15 Solution 9 (a) The force, F, is given by Newton s second law: F = ma, where m is the mass of the particle. So, F = 5(3sin6ti+3cos6tj+4k) = 5sin6ti+5cos6tj+20k.. Hence, the magnitude of the force, F = 225sin 2 6t+225cos 2 6t+400 = = 625 = 25. (b) The velocity, v, is given by v = adt = (3sin6ti+3cos6tj+4k) dt = 2 cos6ti+ 2 sin6tj+4tk+c, where c is an arbitrary vector constant. Since v(0) = i, we have 2 i+c = i, so c = 3 2 i and v = ( 3 2 ) 2 cos6t i+ 2 sin6tj+4tk. The position, r, is given by (( 3 r = vdt = 2 ) 2 cos6t i+ ) 2 sin6tj+4tk dt ( 3 = 2 t ) 2 sin6t i 2 cos6tj+2t2 k+d, where d is an arbitrary vector constant. Since r(0) = j+k, we have 3 j+d = j+k, so d = j+k and 2 2 ( 3 r = 2 t ) ( 3 2 sin6t i+ 2 ) 2 cos6t j+ ( +2t 2) k. Solution 20 (a) The solution of the differential equation is for arbitrary constants A and B. x = Acos2t+Bsin2t, When t = 0, x = 3, so 3 = A and x = 3cos2t+Bsin2t.. Hence x (t) = 6sin2t+2Bcos2t, and since x (0) = 0, then 2B = 0 and so B = 0. So the motion is given by x = 3cos2t. The particle is at x = 0 when 3cos2t = 0, that is cos2t = 0. The first solution for t > 0 is 2t = π/2, that is t = π/4. We know x (t) = 6sin2t, so at t = π/4: x (t) = 6sin π 2 = 6. So the speed of the particle as it crosses x = 0 for the first time is 6. (b) The angular frequency, ω = 4 = 2. So the period is given by T = 2π ω = 2π 2 = π. page 5 of 6

16 Solution 2 (a) w +2z (2+i)(3 4i) = 6 8i+3i 4i 2 = 6 5i+4 = 0 5i. (b) w z = 2+i. This can be expressed in Cartesian form by rationalising 3 4i the denominator. 2+i 3 4i = 2+i 3 4i 3+4i 3+4i = 6+8i+3i+4i2 9 6i 2 = 2+i = i. Solution 22 Since w = +i then w = = 2 and arg(w) = arctan(/) = π/4, so w has complex exponential form w = 2e iπ/4. Likewise z = 3 i so z = = 2 and arg(z) = arctan( / 3) = π/6, so z has complex exponential form z = 2e iπ/6. (a) wz = 2e iπ/4 2e iπ/6 = 2 3/2 e iπ/2 2e iπ/4 (b) w/z = 2e iπ/6 = e 5iπ/2. 2 Solution 23 The Fourier series for a function f(x) over L < x < L is given by where a n = L F(x) = a ( a n cos nπx L +b nsin nπx ) L L L n= f(x)cos nπx L dx, and b n = L f(x)sin nπx L L L dx. Here, f(x) = +x and L =, so a 0 = a n = b n = (+x)dx = ] [x+ x2 = + ( ) = 2 2 [ (+x)cosnπxdx = (+x) ] ( nπ sinnπx ) nπ sinnπx dx = 0+ n 2 π 2 [cosnπx] = n 2 (cosnπ cos( nπ)) = 0 π2 [ (+x)sinnπxdx = (+x) ] ( nπ cosnπx + ) nπ cosnπx dx = 2 nπ cosnπ + n 2 π 2 [sinnπx] = 2 nπ cosnπ + 2 n 2 (sinnπ sin( nπ)) = π2 nπ cosnπ = 2 nπ ( )n Hence, the Fourier series is F(x) = n= 2 nπ ( )n sin nπx L. page 6 of 6