Advanced Mathematics - Repetitorium PDF

Repetitorium Advanced Mathematics DLMDSAM01 Repetitorium advanced mathematics DLMDSAM01 Publisher: IU Internationale Hochschule GmbH IU International University of Applied Sciences Juri-Gagarin-Ring 152 D-99084 Erfurt Mailing address: Kaiserplatz 1 D-83435 Bad Reichenhall media-relations@iu....

Repetitorium Advanced Mathematics DLMDSAM01 Repetitorium advanced mathematics DLMDSAM01 Publisher: IU Internationale Hochschule GmbH IU International University of Applied Sciences Juri-Gagarin-Ring 152 D-99084 Erfurt Mailing address: Kaiserplatz 1 D-83435 Bad Reichenhall [email protected] https://www.iu.org DLMDSAM01 Version No.: xxx-xxxx-xxx (This draft is preliminary and may be subject to further changes prior to release) ©2021 IU Internationale Hochschule GmbH This course book is protected by copyright. All rights reserved. This course book may not be reproduced and/or electronically edited, duplicated, or distributed in any kind of form without written permission by the IU Internationale Hochschule GmbH. Contents 1 Unit 1 1 1.1 Unit 1.1.............................. 3 1.2 Unit 1.2.............................. 16 1.3 Unit 1.3.............................. 24 1.4 Unit 1.4.............................. 33 2 Unit 2 43 2.1 Unit 2.1.............................. 43 2.2 Unit 2.2.............................. 58 3 Unit 3 71 3.1 Unit 3.1.............................. 71 3.2 Unit 3.2.............................. 81 3.3 Unit 3.3.............................. 90 4 Unit 4 103 4.1 Unit 4.1.............................. 103 4.2 Unit 4.2.............................. 109 4.3 Unit 4.3.............................. 117 4.4 Unit 4.4.............................. 123 5 Unit 5 135 5.1 Unit 5.1.............................. 135 5.2 Unit 5.2.............................. 144 5.3 Unit 5.3.............................. 154 5.4 Unit 5.4.............................. 164 6 Unit 6 171 6.1 Unit 6.1.............................. 171 6.2 Unit 6.2.............................. 180 6.3 Unit 6.3.............................. 190 6.4 Unit 6.4.............................. 198 i 1 Unit 1 Formulas Definition of the first derivative of function f (x) with respect to the variable x: f (x + ∆x) − f (x) f 0 (x) = lim (1.1) ∆x→0 ∆x Definition of the n-th derivative of function f (x) with respect to the vari- able x: f (n−1) (x + ∆x) − f (n−1) (x) f (n) (x) = lim (1.2) ∆x→0 ∆x Differentiation of products of two functions: f (x) = v(x)u(x) f 0 (x) = v 0 (x)u(x) + v(x)u0 (x) (1.3) Chain Rule of differentiation: f (x) = f (u(x)) f (u) u(x) f 0 (x) = (1.4) u x Differentiation of quotients of two functions: u(x) f (x) = v(x) u0 (x)v(x) − u(x)v 0 (x) f 0 (x) = (1.5) v 2 (x) 1 Integration by parts: Z Z u(x)v (x)dx = u(x)v(x) − v(x)u0 (x)dx 0 (1.6) First derivatives of the fundamental functions: d n x = nxn−1 (1.7a) dx d sin(ax) = a cos(ax) (1.7b) dx d cos(ax) = −a sin(ax) (1.7c) dx d a tan(ax) = 2 (1.7d) dx cos (ax) d 1 ln(ax) = (1.7e) dx x d exp (ax) = a exp (ax) (1.7f) dx Integral of the fundamental functions: un+1 Z un du = + c, n 6= −1 (1.8a) n+1 Z du = ln |au| + c (1.8b) u au Z au du = +c (1.8c) ln a Z eu du = eu + c (1.8d) Z cos(u)du = sin(u) + c (1.8e) Z sin(u)du = − cos(u) + c (1.8f) Euler Equation for F = F (y, y 0 , x): ∂F d ∂F − =0 (1.9) ∂y dx ∂y 0 2 1.1 Unit 1.1 Questions 1. (L) Using the definition of differentiation in Eq. (1.1), compute the first derivative of the function f (x) = 2x + 1. 2. (L) Using the definition of differentiation in Eq. (1.1) and of n-th differentiation in Eq. (1.2), calculate the second derivative of the function f (x) = ax2 + b. 3 Solutions 1. Substitute x + ∆x for x in the expression of f (x) = 2x + 1 and calculate the limit of the difference quotient in Eq. (1.1). f 0 (x) = 2. Substitute x + ∆x for x in the expression of f (x) = 2x + 1 and then using the definition in Eq. (1.1) we find f (x + ∆x) − f (x) f 0 (x) = lim ∆x→0 ∆x [2(x + ∆x) + 1] − [2x + 1] = lim ∆x→0 ∆x 2x + 2∆x − 2x = lim ∆x→0 ∆x 2∆x = lim ∆x→0 ∆x = lim 2 ∆x→0 = 2. 2. Substitute x + ∆x for x in the expression of f (x) = ax2 + b and compute the limit of the difference quotient in Eq. (1.1) to find the first derivative. Repeat the same procedure for the first derivative to obtain the second derivative. Substitute x + ∆x for x in the expression of f (x) = ax2 + b and compute the limit of the difference quotient in Eq. (1.1) to find 4  the first derivative. f (x + ∆x) − f (x) f 0 (x) = lim ∆x→0 ∆x [a(x + ∆x)2 + b] − [ax2 + b] = lim ∆x→0 ∆x a(x2 + 2x∆x + (∆x)2 ) − ax2 = lim ∆x→0 ∆x 2ax∆x + a(∆x)2 = lim ∆x→0 ∆x = lim (2ax + a∆x) ∆x→0 = 2ax. Therefore, 2ax is the first derivative. Repeating the same pro- cedure for the first derivative i.e., 2ax, we will find the sec- ond derivative: substitute x + ∆x for x in the expression of f 0 (x) = 2ax and compute the limit of the difference quotient in Eq. (1.1) to find the second derivative. f 00 (x) = 2a f 0 (x + ∆x) − f 0 (x) f 00 (x) = lim ∆x→0 ∆x 2a(x + ∆x) − 2ax = lim ∆x→0 ∆x 2ax + 2a∆x − 2ax = lim ∆x→0 ∆x 2a∆x = lim ∆x→0 ∆x = lim 2a ∆x→0 = 2a. 5  3. (L) Find the stationary points of the function f (x) = 3x2 − 6x and determine if the stationary points are minimum, maximum or saddle points. 4. (M) We need to build a rectangular window, with sides x and y. For this we have a total of 6 meters of framing material. How do we need to choose x and y in order for the window area to me maximum? 6 3. Tip: The stationary points are defined by the first derivative and the type of the stationary points by the second derivative. x=1 is the only stationary point which is a minimum. The solution of f 0 (x) = 0 gives us the stationary points (let us call it point x = a). Then we find the sign of f 00 (x) at a to find if the stationary point is a minimum (f 00 (a) > 0), a maximum (f 00 (a) < 0), or a saddle point (f 00 (a) = 0). Using the formula in Eq. (1.7a) to find the first derivative f 0 (x) = 6x − 6 f 0 (x) = 0 → x = 1 Then x=1 is a stationary point. To see if it is minimum or maximum we should find the sign of f 00 (x) at x = 1. Using again the formula in Eq. (1.7a) on the first derivative f 0 (x) = 6x − 6 to find the second derivative f 00 (x) = 6 > 0 As the f 00 (x) at x = 1 is positive, then x = 1 is a minimum. 4. Hint: Express the area in terms of a function A(x) of which we can find the maximum. Answer: x = y = 1, 5m Detailed Solution: a) The restriction of having 6m of framing material means that the perimeter of our rectangle must be 6m: 2x + 2y = 6. From the condition 2x + 2y = 6 we can express y in terms of x as y = 3 − x. b) The area of the rectangle is given by xy. c) If we substitute the expression of y = 3 − x in the formula for the area xy we can express the area of the window as a function of x as A(x) = xy = x(3 − x) = 3x − x2 7  d) In order to find the maximum area, we should find the sta- tionary points of the function A(x) (A0 (x) = 0) and find the maximum stationary point (A00 (x) < 0). e) First, we calculate the first derivative of the function A(x) to find the stationary points: A0 (x) = 3 − 2x A0 (x) = 0 ⇒ x = 1.5 f) To find if it is a minimum or a maximum, we should cal- culate the second derivative of the function A(x) in point x = 1.5: A00 (x) = −2 < 0 g) Therefore, x = 1.5 is a maximum. As y = 3 − x, then we have y = 1.5 8 Questions 5. (S) Calculate the first derivative of the following function with respect to the x: (sin(3x) − 1)2 f (x) = x3 R1 6. (M) Evaluate the integral 0 x2 sin(x)dx. 9 Answers 5. Hint: Using the chain rule in Eq. (1.4) and differentiation of quotations Eq. (1.5). Answer: 3(sin(3x) − 1)(2x cos(3x) − (sin(3x) − 1)) x4 Detailed Solution: a) we can rewrite f (x) as u(x) f (x) = v(x) where u(x) = (sin(3x) − 1)2 and v(x) = x3. b) Then we use the differentiation of quotations in Eq. (1.5) to calculate the derivative of f (x) with respect to x: 0 u0 (x)v(x) − u(x)v 0 (x) f (x) = v 2 (x) c) To find the derivative of u(x) with respect to the x, we rewrite u(x) = w2 (x), where w(x) = sin(3x) − 1 and using the chain rule in Eq. (1.4) to find u0 (x) as u(x) du dw(x) dw(x) = = 2w(x) dx dw dx dx = 2(sin(3x) − 1)(3 cos(3x)) = 6 cos(3x)(sin(3x) − 1) d) Therefore, for f 0 (x) we have (using step b) u0 v − uv 0 f 0 (x) = v2 6 cos(3x)(sin(3x) − 1)x3 − (sin(3x) − 1)2 (3x2 ) = x6 3(sin(3x) − 1)(2x cos(3x) − (sin(3x) − 1)) = x4 10 6. Use integration by parts from Eq. 1.6 R1 0 x2 sin(x)dx = 2 sin(1) + cos(1) − 2 ≈ 0.22. We integrate by parts by setting u = x2 and v 0 = sin(x). Then we have u0 = 2x and v = − cos x. The formula for integration by parts yields Z 1 1 Z 1 2 x sin(x)dx = uv − u0 vdx 0 0 0 Z 1 2 = [−x cos(x)]10 +2 x cos(x)dx 0 Z 1 = − cos(1) + 2 x cos(x)dx 0 R1 In the last expression we need to compute the integral 0 x cos(x)dx, which we do again by parts, this time setting u = x and v 0 = cos(x). Then u0 = 1 and v = sin(x). So now the formula for integration by parts yields Z 1 Z 1 1 x cos(x)dx = [x sin(x)]0 − sin(x)dx 0 0 1 = sin(1) + cos(x) 0 = sin(1) + cos(1) − 1 Putting all together we find Z 1 Z 1 2 x sin(x)dx = − cos(1) + 2 x cos(x)dx 0 0 = − cos(1) + 2 (sin(1) + cos(1) − 1) = cos(1) + 2 sin(1) − 2 ≈ 0.22 11 Questions R 7. (S) Evaluate the integral e3x cos(x)dx. 8. (S) A car is moving with a speed v(t) that depends on time according to the function t v(t) = 2 t +1 What is the distance the car travels with this speed between times t0 = 0s and t1 = 5s? 12 Answers 7. Use the integration by parts method from Eq. (1.6) Answer: e3x (sin(x) + 3 cos(x)) 10 Solution: Consider the integration by parts formula Z Z u(x)v (x)dx = u(x)v(x) − v(x)u0 (x)dx 0 we define: u(x) = e3x → u0 (x) = 3e3x v 0 (x) = cos(x), → v(x) = sin(x) Therefore Z Z 3x e cos(x)dx ≡ u(x)v 0 (x)dx Z = e sin(x) − 3e3x sin(x)dx 3x = e3x sin(x) − 3I(x), where Z I(x) = e3x sin(x)dx p(x)q 0 (x)dx = R To calculateRI(x) we use again the integration by parts p(x)q(x) − q(x)p0 (x)dx: p(x) = e3x , → p0 (x) = 3e3x q 0 (x) = sin(x), → q(x) = − cos(x) Therefore Z I(x) = p(x)q 0 (x)dx Z 3x = e (− cos(x)) − 3e3x (− cos(x)) dx 13  Then we have Z e3x cos(x)dx = e3x sin(x) − 3I(x)  Z  3x 3x 3x = e sin(x) − 3 e (− cos(x)) − 3e (− cos(x))dx Z = e (sin(x) + 3 cos(x)) − 9 e3x cos(x)dx 3x R Moving the integral e3x cos(x)dx to the left side of the equation Z 10 e3x cos(x)dx = e3x (sin(x) + 3 cos(x)) ⇒ e3x Z e3x cos(x)dx = (sin(x) + 3 cos(x)) 10 8. Hint: By definition, the velocity v(t) is the derivative of the distance s(t) with respect to time. Remember, the integral is the inverse of differentiation. Answer: s = 21 ln(26) Detailed solution: By definition, the velocity v(t) is the deriva- tive of the distance s(t) with respect to time: ds(t) v(t) = dt Therefore, to obtain the distance s(t), we should calculate the inverse of the differentiation of the both side of the above equa- tion. In the course, we have learnt that the integral is the inverse of the differentiation. It results in: Z s(t) = v(t)dt Therefore, the distance covered between times t0 = 0 and t1 = 5 is given by the integral: Z 5 Z 5 t s(t) = v(t)dt = 2 dt 0 0 t +1 14 We notice that if we multiply the numerator by 2 we get exactly the derivative of the denominator, which makes it convenient to multiply and divide the integral by two and use the formula for the rational functions: Z 5 1 5 2t 1 5 f 0 (t) Z Z t 2 dt = dt = dt 0 t +1 2 0 t2 + 1 2 0 f (t) with f (t) = t2 + 1. Using the integral formula Eq. (1.8b), we obtain: Z 5 t 1 2+1 dt = [ln(f (t))]50 0 t 2 1 5 = ln(t2 + 1) 0 2 1 1 = ln(26) − ln(1) 2 2 1 = ln(26). 2 Remember ln(1) = 0 15 1.2 Unit 1.2 Questions ∂f ∂f 1. (L) Find the first order partial derivatives (fx = ∂x and fy = ∂y ) of f (x, y) = xy 2 + x. ∂f ∂f 2. (M) Find the first order partial derivatives (fx = ∂x , fy = ∂y and fz = ∂f ∂z ) of f (x, y, z) = xy 2 + xz 2 + xyz. 16 Solutions 1. Hint: First, calculate the partial derivative with respect to x, treating y as a constant to obtain the partial derivatives fx ≡ ∂f ∂x , repeat the same for the variable y. Answer: fx = y 2 + 1, fy = 2xy. Detailed Solution: to calculate fx ≡ ∂f ∂x we assume the variable y as a constant and then calculate the derivative of f with respect to the variable x ∂f ∂ fx = = (xy 2 + x) = y2 + 1 ∂x ∂x y=const. To calculate fy ≡ ∂f ∂y , this time we assume the variable x as a constant and then we calculate the derivative of f with respect to the variable y ∂f ∂ fy = = (xy 2 + x) = 2xy + 0 = 2xy ∂y ∂y x=const. 2. Hint: First, calculate the partial derivative with respect to x, treating y as a constant to obtain the partial derivatives fx ≡ ∂f ∂x , repeat the same for the variables y and z. Answer: fx = y 2 + z 2 + yz, fy = 2xy + xz, fz = 2xz + xy. Detailed Solution: to calculate fx ≡ ∂f ∂x we assume the variables y and z as a constant. Then we calculate the derivative of f with respect to the variable x ∂f fx = = y 2 + z 2 + yz ∂x To calculate fy ≡ ∂f ∂y , this time we assume the variables x and z as a constant and calculate the derivative of f with respect to the variable y ∂f fy = = 2xy + 0 + xz = 2xy + xz ∂y 17  Finally, to calculate fz ≡ ∂f ∂z , this time we assume the variables x and y as a constant and calculate the derivative of f with respect to the variable z ∂f fz = = 0 + 2xz + xy = 2xz + xy ∂y 18 Questions 3. (L) Find the total derivatives of the function f (x, y) = cos(xy). 4. (S) Find the total derivatives of the function w(x, y) = cos (x2 + 2y). 19 Solutions 3. Hint: to find the total derivative, we should calculate df = fx dx + fy dy. Answer: df = −y sin(xy)dx − x sin(xy)dy. Detailed Solution: to calculate the total derivative, we should calculate fx and fy where fx ≡ ∂f ∂x and fy ≡ ∂f ∂y. To calculate fx we assume the variable y as a constant and calculate the derivative of f with respect to the variable x ∂f ∂ fx = = cos(xy) = −y sin(xy) ∂x ∂x To calculate it, we assumed y as a constant like a coefficient a. To calculate fy , this time we assume the variable x as a constant and calculate the derivative of f with respect to the variable y ∂f ∂ fy = = cos(xy) = −x sin(xy) ∂y ∂x Now, we substitute fx and fy in df = fx dx + fy dy to obtain the total derivative of the function f (x, y) df = fx dx + fy dy = −y sin(xy)dx − x sin(xy)dy 4. Hint: Use the chain rule. Answer: dw = −2 sin(x2 + 2y)(xdx + dy). Detailed Solution: a) We should calculate the total derivative of w(x, y) = cos (x2 + 2y), i.e., dw: dw = wx dx + wy dy ∂w wx ≡ ∂x ∂w wy ≡ ∂y 20 b) However, it is easier to use the chain rule and rewrite w(x, y) as w(u) w(u) = cos(u) ∂w = − sin(u) ∂u c) where: u =u(x, y) = x2 + 2y ∂u = 2x ∂x ∂u =2 ∂y d) Then using the chain rule ∂w ∂u ∂w ∂u dw = dx + dy ∂u ∂x ∂u ∂y dw = (− sin(u))(2x)dx + (− sin(u))(2)dy = −2 sin(u)(xdx + dy) = −2 sin(x2 + 2y)(xdx + dy) 21 Questions ∂f 5. (M) If f (x, y) = xy ln(2x), then calculate ∂x. 22 Solutions 5. Hint: assume the variable y as a constant and then calculate the derivative of f with respect to the variable x. Answer: ∂f ∂x = y(1 + ln(2x)). As mentioned in the hint, to calculate ∂f ∂x , we assume the variable y as a constant. ∂f ∂ = [xy ln(2x)] ∂x ∂x To calculate the derivative, we assume f (x, y) = A(x, y)B(x) where A(x, y) = xy and B(x) = ln(2x). Then ∂f ∂A(x, y)B(x) = ∂x  ∂x    ∂A(x, y) ∂B(x) = B(x) + A(x, y) ∂x ∂x   2 = (y) ln(2x) + xy 2x = y ln(2x) + y = y(1 + ln(2x)) 23 1.3 Unit 1.3 Questions x RR 1. (L) Evaluate the indefinite integral I(x, y) = y dxdy. RR 2. (L) Evaluate the definite integral R xy dxdy for −1 < x < 1 and −1 < y < 1. 24 Solutions Hint: The solution I(x, y) = RR x 1. y dxdy must satisfy the condi- tion ∂2 x I(x, y) =. (1.10) ∂x∂y y First find a particular solution to this differential equation and then obtain the general solution by addition a function c(x, y) that satisfies ∂2 c(x, y) = 0. ∂x∂y Answer: x2 I(x, y) = ln(|y|) + f (x) + g(y), 2 where f (x) and g(y) are arbitrary differentiable functions Detailed Solution: The integrand xy can be written in product form x 1 =x· , y y where the first factor only depends on x and the second factor only depends on y. A particular solution of (1.10) can therefore be found by integrating each factor separately with respect to x and y, respectively. In other words, if Jp (x) and Kp (y) are particular solutions of d J(x) = x, dx d 1 K(y) = , dy y respectively, then Ip (x, y) = Jp (x)Kp (y) is a particular solution of (1.10). Disregarding integration constants for now, standard integration gives the following particular solutions: x2 Jp (x) = 2 Kp (y) = ln(|y|) 25  Hence x2 Ip (x, y) = ln(|y|) 2 is a particular solution of (1.10). The general solution of (1.10) is obtained from the particular solution by adding any function c(x, y) that satisfies ∂2 c(x, y) = 0. ∂x∂y This is equivalent to c(x, y) being of the form c(x, y) = f (x) + g(y) for differentiable functions f (x) and g(y). So the general solu- tion of (1.10) has the form x2 I(x, y) = ln(|y|) + f (x) + g(y). 2 2. Hint: Integrate separately with respect to the two variables as the integrand is separable. Answer: R xy dxdy = 0. RR Detailed Solution: In this case the two variables x and y are independent and it is possible to write the integrand xy as mul- tiplication of two functions of x and y: xy = x × y and the same for the integral ZZ I(x, y) = xydxdy R = K(x) × J(y) Z 1 K(x) = xdx −1 Z 1 J(y) = ydy −1 26 Then we can calculate each integral separately: 1 1 x2 Z    1 1 K(x) = xdx = + c1 = ( + c1 ) − ( + c1 ) = 0 −1 2 −1 2 2 Z 1  2 1   y 1 1 J(y) = ydy = + c2 = ( + c2 ) − ( + c2 ) = 0 −1 2 −1 2 2 Therefore, I(x, y) = 0 × 0 = 0 27 Questions RR 3. (M) Calculate the double integral A (x + cos(y))dxdy over the area A defined by 0 < x < 2 and −π/2 < y < π/2. RR 4. (S) Find the integral A xy 2 dxdy where the area A is a disk of radius 1, i.e. x2 + y 2 ≤ 1. 28 Solutions 3. Hint: First calculate the integral over one variables (say x) and treat the other variable (say y) as a constant. Then perform the calculation over y variable. Answer: 2π + 4 Detailed Solution: In this case, it is not possible to write the integrand (x+cos(y)) as multiplication of two independent func- tions of x and y. Then we use another approach: a) First we calculate the integral over the x variable and we consider y as a constant: ZZ (x + cos(y))dxdy A Z π Z 2 2 = dy (x + cos(y)) dx − π2 0 π 2 x2 Z   2 = dy + x cos(y) + c1 − π2 2 0 Z π 2 = dy ((2 + 2 cos(y) + c1 ) − (0 + 0 cos(y) + c1 )) − π2 Z π 2 = dy (2 + 2 cos(y)) − π2 b) we continue by performing the integral over the variable y Z π 2 dy (2 + 2 cos(y)) − π2 π 2 = (2y + 2 sin(y) + c2 ) − π2 considering Z cos(y)dy = sin(y) + c 29   π   −π  = π + 2 sin( ) + c2 − −π + 2 sin( ) + c2 2 2 We know π  sin =1  2 −π sin = −1 2 Therefore,  π   −π  π + 2 sin( ) + c2 − −π + 2 sin( ) + c2 = 2π + 4 2 2 4. Hint: First calculate the integral over the variable y, while treat- ing the variable x as a constant. Then evaluate the integral for the variable x. Answer: 0 Detailed Solution: a) First we calculate the integral over the variable y and we assume the variable x as a constant. To obtain the range of the integration over the variable y in terms of variable x, we write: √ √ x 2 + y 2 ≤ 1 ⇒ − 1 − x2 ≤ y ≤ 1 − x2 The range of the variable x is also between −1 and 1, as the area A is a disk of radius 1. b) Then we have for the integral: √ ZZ Z 1 Z 1−x2 2 xy dxdy = dx √ xy 2 dy A −1 − 1−x2 c) Let us first to calculate the integral over the variable y Z √1−x2  3 √1−x2 xy √ xy 2 dy = + c1 √ − 1−x 2 3 − 1−x2 2
3/2 = x 1 − x2 3 30  RR d) Substituting the above term in the integral A xy 2 dxdy, we obtain: ZZ Z 1 Z √1−x2 xy 2 dxdy = dx √ xy 2 dy A −1 − 1−x 2 Z 1 2
3/2 = dx x 1 − x2 −1 3 e) Substituting 1 − x2 = u and −2xdx = du in the above integral, we obtain Z Z 2 2 3/2 1 (−du)u3/2
dx x 1 − x = 3 3   1 2 5/2 =− u + c2 3 5 2 = − (1 − x2 )5/2 + c2 15 Replacing the range of the variable x, we obtain: Z 1  1 2 2 3/2
2 2 5/2 dx x 1 − x = − (1 − x ) + c2 =0 −1 3 15 −1 Questions RR 5. (M) Evaluate the integral T xy 3 dxdy where T is the triangle com- prised between the straight line y = 2x and the x-axis between x = 0 and x = 2. 31 Solutions 5. Hint: First calculate the integral over the variable y, while treat- ing the variable x as a constant. Then evaluate the integral for the variable x. Answer: T xy 3 dxdy = 128 RR 3 Detailed Solution: a) Our triangle T is given by x varying between 0 and 2 and, for each value of x, y varying between 0 and 2x. We can write the integral as ZZ Z 2 Z 2x  3 3 xy dxdy = xy dy dx T 0 0 b) Here we cannot separate the two integrals as the range of the variable y is a function of the variable x. Therefore, we have to solve the inner one first, as this will then affect the outer one, because the result will be a function of x. In the inner integral, we treat the x in the integrand as a constant: Z 2 Z 2x  Z 2 Z 2x  3 3 xy dy dx = x y dy dx 0 0 0 0 2 4 2x Z   y = x dx 0 4 0 2 16x5 Z = dx 0 4 Z 2 =4 x5 dx 0  6 2 x =4 6 0 6 2 128 =4 =. 6 3 32 1.4 Unit 1.4 Questions 1. (L) Given the function f (y(x), y 0 (x), x) = y(x)y 0 (x) + x, find the functional Z 1 F [y] = f (y(x), y 0 (x), x)dx −1 for y(x) = x2. R1 2. (L) Find the stationary path of the functional F [y] = 0 (xy + y 2 + y 0 y) dx. 33 Solutions 1. Hint: Substitute y(x) in f (y(x), y 0 (x), x) and perform the inte- gral. Answer: F [y(x) = x2 ] = 0 Detailed Solution: a) We start by substituting y(x) = x2 and y 0 (x) = 2x in f (y(x), y 0 (x), x) where f (y(x), y 0 (x), x) = y(x)y 0 (x) + x = (x2 )(2x) + x = 2x3 + x R1 b) Then we perform the integral −1 f (y(x), y 0 (x), x)dx to find the functional F Z 1 2 F [y(x) = x ] = f (y(x), y 0 (x), x)dx −1 1 1 2x4 x2 Z  3
= 2x + x dx = + +c −1 4 2 −1 =0 2. Hint: Use the Euler equation defined in Eq. (1.9). Answer: y = −x/2 Detailed Solution: R1 a) Remember the extrema of the functional I[y] = −1 F (y(x), y 0 (x), x)dx is determined by the Euler equation: ∂F d ∂F − =0 ∂y dx ∂y 0 b) In this exercise, we have F (y(x), y 0 (x), x) = xy + y 2 + y 0 y. 34  Therefore, ∂F = x + 2y + y 0 ∂y ∂F =y ∂y 0 d ∂F d 0 = y = y0 dx ∂y dx c) Then the Euler equation will be ∂F d ∂F − =0→ ∂y dx ∂y 0 = (x + y 0 + 2y) − y 0 = 0 −→ y = −x/2 R 1 y = −x/2 Therefore, is an stationary path for the functional F [y] = 0 (xy + y + y 0 y) dx. 2 35 Questions 3. (M) Find the differential equation that the stationary path of the Rb distance functional S[y] = a (y 02 + y 2 + 2xy)dx satisfies. 4. (M) Find the differential equation that the stationary path of the Rb distance functional S[y] = a (y 02 + y)dx satisfies. 36 Solutions 3. Hint: Use the Euler equation in Eq. (1.9) to find the differential equation. Answer: y 00 − y = x Detailed Solution: a) Here for the F (x, y, y 0 ) in the Euler equation in Eq. (1.9) we have F (x, y, y 0 ) = y 02 + y 2 + 2xy ∂F = 2y + 2x ∂y ∂F 0 = 2y 0 ∂y d ∂F d 0 0 = 2y = 2y 00 dx ∂y dx b) Substituting in the Euler equation in Eq. (1.9) ∂F d ∂F − =0→ ∂y dx ∂y 0 = 2y + 2x − 2y 00 = 0 → y 00 − y = x 4. Hint: Use the Euler equation in Eq. (1.9) to find the differential equation. Answer: 2y 00 − 1 = 0 Detailed Solution: a) Here for F (x, y, y 0 ) in the Euler equation in Eq. (1.9) we 37  have F (x, y, y 0 ) = y 02 + y ∂F =1 ∂y ∂F = 2y 0 ∂y 0 d ∂F d 0 0 = 2y = 2y 00 dx ∂y dx b) Substituting in the Euler equation in Eq. (1.9) ∂F d ∂F − =0→ ∂y dx ∂y 0 = 1 − 2y 00 = 0 38  R2 5. (S) Consider the functional F [y] = 1 xy 02 dx. a) Find the differential equation that the stationary path y(x) sat- isfies. b) Find the stationary function y(x). R1 6. (S) Find the curve y(x) on which the functional 0 (y 02 + 12xy)dx takes it extreme values (its extremas) when y(0) = 0 and y(1) = 1. 39 Solutions 5. Hint: Use the Euler equation to find differential equation for the stationary path. Then solve the differential equation to obtain y(x). Answer: (a) The stationary path satisfies the equation y 0 x = constant. (b) y = c ln(|x|) + b Detailed Solution: a) Using the Euler equation in Eq. (1.9) where F (y, y 0 , x) = xy 02 : F (x, y, y 0 ) = xy 02 ∂F =0 ∂y ∂F 0 = 2xy 0 ∂y Then the Euler equations gives: ∂F d ∂F − =0→ ∂y dx ∂y 0 d = (2xy 0 ) = 0 → xy 0 = constant. dx b) To find y(x) we should solve the differential equation: y0x = c dy c = dx x In the above equation c is a constant. To solve the above differential equation we first multiply both sides with dx and then integrate over the variable x: Z Z dy c dx = dx dx x 40  The left-hand side integral is simply: Z dy dx =y dx To calculate the right-hand side integral, we use the Eq. (1.8b) Z c dx = c ln(|x|) + b x Now, we can solve the differential equation and finding y: Z Z dy c dx = dx dx x y = c ln(|x|) + b 6. Hint: Use the Euler equation to find differential equation for the stationary path and then solve the differential equation for y 00. Answer: y = x3 Detailed Solution: a) First we try to find the differential equation that the sta- tionary path should satisfy. For this purpose, we use the Euler equation in Eq. (1.9) for F (y, y 0 , x) = y 02 + 12xy: F (x, y, y 0 ) = y 02 + 12xy ∂F = 12x ∂y ∂F = 2y 0 ∂y 0 Then the Euler equations gives: ∂F d ∂F − =0→ ∂y dx ∂y 0 d = 12x − (2y 0 ) = 0 → y 00 = 6x. dx 41  b) To find y(x) we should solve the differential equation: y 00 = 6x To solve the above equation we use the relation d2 3 2 (x + ax + b) ≡ (x3 + bx + c)00 = 6x → dx y = x3 + bx + c 2 d Where a and b are constants and dx 2 is the operator to obtain the second derivative with respect to the variable x. Again, it is also possible to solve this differential equation by integrating two times both sides of the equation y 00 = 6x To find constants b and c, we use the conditions y(0) = 0 and y(1) = 1: y(0) = 0 → (0)3 + b(0) + c = 0 → c = 0 y(1) = 1 → (1)3 + b(1) = 1 → b = 0 Then y = x3 42 2 Unit 2 2.1 Unit 2.1 Questions 1. (L) Consider the integral form of the convolution of two functions (e.g. two signals) f (τ ) and g(τ ) Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ −∞ Write down the convolution integral for signals f (t) = 1 and g(t) = e|t|. 2. (L) Consider the integral form of the convolution of two functions (e.g. two signals) f (τ ) and g(τ ) Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ −∞ Write down the convolution integral for two signals f (t) = sin(t) and g(t) = (1 − cos(t))/t2. 43 Solutions 1. Hint: Substitute the functions f and g in the convolution inte- gral. Also remember f ∗ g = g ∗ f. Answer: Z∞ (f ∗ g)(t) = e|t−τ | dτ −∞ Detailed Solution: Consider the convolution integral: Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ −∞ Then substitute the two function f (t) = 1 and g(t) = e|t| Z∞ (f ∗ g)(t) = 1 × e|t−τ | dτ −∞ Z∞ = e|t−τ | dτ −∞ The following solution is also valid as (f ∗ g)(t) = (g ∗ f )(t) Z∞ (g ∗ f )(t) = e|τ | × 1dτ −∞ Z∞ = e|τ | dτ −∞ 2. Hint: Substitute the functions f and g in the convolution inte- gral. Remember f ∗ g = g ∗ f. Answer: Z∞ (f ∗ g)(t) = sin(τ )(1 − cos(t − τ ))/(t − τ )2 dτ −∞ 44  Detailed Solution: Consider the convolution integral: Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ −∞ Then substitute the two function f (t) = sin(t) and g(t) = (1 − cos(t))/t2 Z∞ (f ∗ g)(t) = sin(τ )(1 − cos(t − τ ))/(t − τ )2 dτ −∞ Also the following solution is valid because f ∗ g = g ∗ f Z∞ (g ∗ f )(t) = (1 − cos(τ ))/τ 2 sin(t − τ )dτ −∞ 45 Questions 3. (M) Consider the integral form of the convolution of two functions (such as two signals) f (t) and g(t) Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ −∞ What would be the upper and lower limits of the integral, if both functions f (t) and g(t) are 0 for t < 0. 4. (M) Write the convolution integral (f ∗ h)(t) of two functions f (t) and h(t): f (t) = exp (−t) h(t) = sin(t) where both functions are zero for t < 0. Hint: When both functions f (t) and g(t) are zero for t < 0, the convolution integral’s limits change from (−∞, ∞) to: Zt (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 46 Solutions 3. Hint: Investigate on which regions one of the functions f or g would be zero. Answer: Zt (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 Detailed Solution: Consider the general form of convolution in- tegral in the −∞ to +∞ domain Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ −∞ For −∞ < τ < 0, we have f (τ ) = 0 as f is zero for the negative a argument and then (f ∗ g)(t) = 0. Then the nonzero integral limits reduces to Z∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 However, for τ > t, g(t−τ ) = 0 as g = 0 for negative arguments. Then we can ignore the other null part of the integration in the t < τ < +∞ interval and rewrite the integral again as Zt (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 4. Hint: Remember from the textbook that for the convolution we have f ∗ h = h ∗ f. τR=t Answer: (f ∗ h)(t) = exp (−τ ) sin(t − τ )dτ τ =0 47  Detailed Solution: The integral for the convolution could be written as: Z∞ (f ∗ h)(t) = f (τ )g(t − τ )dτ −∞ And when both functions are zero for t < 0 we can change the integral limits into Zt (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 As both functions f and h are zero for τ < 0, i.e., we have Zt (f ∗ h)(t) = f (τ )g(t − τ )dτ 0 Zt = exp (−τ ) sin(t − τ )dτ 0 48 Questions 5. (S) Consider the two following functions: f (t) = exp (2t) g(t) = exp (6t) Where both functions are zero for t < 0 interval. a) Write down the integral form for the convolution of the two functions f and g at time t, i.e., (f ∗ g)(t). b) Calculate the integral of the convolution (f ∗ g)(t). Hint: When both functions f (t) and g(t) are zero for t < 0, for the convolution integral we have: Zt (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 6. (S) Consider the two following signals f1 and f2 f1 (t) = 0 t < 0 f1 (t) = 1 0 ≤ t ≤ 3 f1 (t) = 0 t > 3 and f2 (t) f2 (t) = 0 t < 0 f2 (t) = 2 0 ≤ t ≤ 1 f2 (t) = 0 t > 1 49  Figure 2.1: Signals f1 and f2. a) Find the interval, on which the convolution is not zero. b) Calculate the convolution of two signals f1 and f2. 50 Solutions 5. Hint: Remember from the textbook that f ∗ g = g ∗ f. Also remember the parameter of one of the functions should be sub- stituted with for example t − τ. Also the lower integral limit should be 0 and the upper limit should be t. Answer: a) Z τ =t (f ∗ g)(t) = exp (6t) exp (−4τ )dτ τ =0 b) 1 (f ∗ g)(t) = (g(t) − f (t)) 4 Detailed Solution: a) The integral for the convolution could be written as: Z τ =∞ (f ∗ g)(t) = f (τ )g(t − τ )dτ τ =−∞ As both functions are zero for t < 0, we have: Z t (f ∗ g)(t) = f (τ )g(t − τ )dτ 0 Where f (τ ) = exp (2τ ) g(t − τ ) = exp (6(t − τ )) 51  Then Z τ =t (f ∗ g)(t) = exp (2τ ) exp (6(t − τ ))dτ Zτ =0 τ =t = exp (2τ ) exp (6t) exp (−6τ )dτ Zτ =0 τ =t = exp (−4τ ) exp (6t)dτ τ =0 As the variable t does not depend on the variable τ : Z τ =t (f ∗ g)(t) = exp (6t) exp (−4τ )dτ τ =0 b) To calculate the integral of f ∗ g, remember Z 1 exp (ax)dx = exp (ax) a Then Z τ =t (f ∗ g)(t) = exp (6t) exp (−4τ )dτ τ =0 ! τ =t −1 = exp (6t) exp (−4τ ) 4 τ =0 1 = (exp (6t) − exp (2t)) 4 1 = (g(t) − f (t)) 4 6. Hint: Remember the convolution is non-zero only on the interval where both signals are non-zero and also overlap. Answer: a) 0 < t < 4 b)    0 −∞ < t < 0  2t 0≤t≤1   (f1 ∗ f2 )(t) = 2 1