Higher Engineering Mathematics - BS Grewal

184 Higher Engineering Mathematics and graphs of y = ln x and y = −ln x are shown in Problem 2. Sketch the following graphs, showing Fig. 18.19(b). relevant points: Problem 1. Sketch the following graphs, showing (a) y = 5 − (x + 2)3 (b) y = 1 + 3 sin 2x relevant points: (a) Figure 18.22(a) shows a graph of y = x3. (a) y = (x − 4)2 (b) y = x3 − 8 Figure 18.22(b) shows a graph of y = (x + 2)3 (see f (x + a), Section (iii) above). (a) In Fig. 18.20 a graph of y = x2 is shown by the bro- ken line. The graph of y = (x − 4)2 is of the form y y = f (x + a). Since a = −4, then y = (x − 4)2 is translated 4 units to the right of y = x2, parallel to 20 the x-axis. y ϭx 3 (See Section (iii) above). 10 y ϭx2 y y ϭ (x Ϫ 4)2 8 4 2x Ϫ4 Ϫ2 0 2 4 6x Ϫ2 0 Figure 18.20 –10 –20 (b) In Fig. 18.21 a graph of y = x3 is shown by the (a) broken line. The graph of y = x3 − 8 is of the y form y = f (x) + a. Since a = −8, then y = x3− 8 is translated 8 units down from y = x3, parallel to 20 the y-axis. y 5 (x 1 2)3 (See Section (ii) above). 10 y y ϭx 3 20 10 y ϭx 3 Ϫ 8 24 22 0 2x –10 Ϫ3 Ϫ2 Ϫ1 0 123 x –20 –10 –20 –30 (b) Figure 18.21 Figure 18.22

Functions and their curves 185 y y x 20 1 y 5 sin x y ϭ Ϫ(x ϩ 2)3 10 Ϫ4 Ϫ2 0 0 ␲ ␲ 3␲ –10 22 –20 21 (a) y 1 y 5 sin 2x 2x 0␲ ␲ 3␲ 2␲ x 2 2 21 (b) y y 5 3 sin 2x 3 (c) 2 y y ϭ 5 Ϫ (x ϩ 2)3 20 1 10 0␲ ␲ 3␲ 2␲ x 2 2 21 Ϫ4 Ϫ2 0 2x 22 –10 –20 23 (c) y y ϭ1 ϩ 3 sin 2x 4 (d) 3 2 Figure 18.22 (Continued) 1 0 Figure 18.22(c) shows a graph of y = − (x + 2)3 Ϫ1 ␲ ␲ 3␲ 2␲ x (see − f (x), Section (v) above). Figure 18.22(d) Ϫ2 2 2 shows the graph of y = 5 −(x + 2)3 (see f (x) + a, Section (ii) above). Figure 18.23 (d) (b) Figure 18.23(a) shows a graph of y = sin x. Figure 18.23(b) shows a graph of y = sin 2x (see f (ax), Section (iv) above). Figure 18.23(c) shows a graph of y = 3 sin 2x (see a f (x), Section (i) above). Figure 18.23(d) shows a graph of y = 1 + 3 sin 2x (see f (x) + a, Section (ii) above).

186 Higher Engineering Mathematics Now try the following exercise also periodic of period 2π and is defined by: −1, when −π ≤ x ≤ 0 Exercise 77 Further problems on simple transformations with curve sketching f (x) = 1, when 0 ≤ x ≤ π Sketch the following graphs, showing relevant points: 18.4 Continuous and discontinuous (Answers on page 200, Fig. 18.39) functions 1. y = 3x − 5 If a graph of a function has no sudden jumps or breaks it 2. y = − 3x + 4 is called a continuous function, examples being the 3. y = x2 + 3 graphs of sine and cosine functions. However, other 4. y = (x − 3)2 graphs make finite jumps at a point or points in the inter- 5. y = (x − 4)2 + 2 val. The square wave shown in Fig. 18.24 has finite 6. y = x − x2 discontinuities as x = π, 2π, 3π, and so on, and is 7. y = x3 + 2 therefore a discontinuous function. y = tan x is another 8. y = 1 +2 cos 3x example of a discontinuous function. 9. y = 3 −2 sin x + π 18.5 Even and odd functions 4 10. y = 2 ln x Even functions 18.3 Periodic functions A function y = f (x) is said to be even if f (−x) = f (x) for all values of x. Graphs of even functions are always A function f (x) is said to be periodic if f (x + T ) = symmetrical about the y-axis (i.e. is a mirror image). f (x) for all values of x, where T is some positive Two examples of even functions are y = x2 and y = cos x number. T is the interval between two successive repe- as shown in Fig. 18.25. titions and is called the period of the function f (x). For example, y = sin x is periodic in x with period 2π since y sin x = sin(x + 2π) = sin(x + 4π), and so on. Similarly, 8 y = cos x is a periodic function with period 2π since 6 cos x = cos(x + 2π) = cos(x + 4π), and so on. In gen- 4 y5x2 eral, if y = sin ωt or y = cos ωt then the period of the 2 waveform is 2π/ω. The function shown in Fig. 18.24 is 23 22 21 0 1 2 3 x f (x) 1 (a) y y 5cos x 2␲ 2␲ 0 ␲ ␲x 2 2 Ϫ2␲ Ϫ␲ 0␲ 2␲ x Ϫ1 ( b) Figure 18.24 Figure 18.25

Functions and their curves 187 Odd functions Problem 3. Sketch the following functions and state whether they are even or odd functions: A function y = f (x) is said to be odd if f (−x) = − f (x) (a) y = tan x for all values of x. Graphs of odd functions are always symmetrical about the origin. Two examples ⎧ when 0 ≤ x ≤ π of odd functions are y = x3 and y = sin x as shown in ⎪⎨⎪⎪⎪⎪ 2, 2 Fig. 18.26. (b) f (x) = ⎪⎪⎪⎪⎩⎪−22,, when π ≤ x ≤ 3π , Many functions are neither even nor odd, two such 22 examples being shown in Fig. 18.27. when 3π ≤ x ≤ 2π y y5x3 2 and is periodic of period 2π. 27 23 0 3x (a) A graph of y = tan x is shown in Fig. 18.28(a) and y 5 sinx is symmetrical about the origin and is thus an odd 227 function (i.e. tan(−x) = −tan x). (a) (b) A graph of f (x) is shown in Fig. 18.28(b) and y is symmetrical about the f (x) axis hence the 1 function is an even one, ( f (−x) = f (x)). y y ϭ tan x Ϫ␲ 0 ␲ 2␲ x 23␲ 2␲ 2␲2 0␲ ␲ 3␲ 2␲ x 2 2 2 21 ( b) Figure 18.26 (a) y f(x ) 2 20 y ϭe x 10 Ϫ2␲ Ϫ␲ 0 ␲ 2␲ x Ϫ2 Ϫ1 0 1 2 3 x ( b) (a) Figure 18.28 y 0 x Problem 4. Sketch the following graphs and state ( b) whether the functions are even, odd or neither even nor odd: Figure 18.27 (a) y = ln x (b) f (x) = x in the range −π to π and is periodic of period 2π.

188 Higher Engineering Mathematics (a) A graph of y = ln x is shown in Fig. 18.29(a) (a) odd (b) even and the curve is neither symmetrical about the (c) odd (d) neither y-axis nor symmetrical about the origin and is thus neither even nor odd. 3. State whether the following functions, which are periodic of period 2π, are even or odd: (b) A graph of y = x in the range −π to π is shown in Fig. 18.29(b) and is symmetrical about the origin θ, when −π ≤ θ ≤ 0 and is thus an odd function. (a) f (θ) = −θ, when 0 ≤ θ ≤ π y ⎧ when − π ≤ x ≤ π ⎨x, 2 3π2 1.0 y ϭ In x ≤ (b) f (x) = ⎩0, when π x ≤ 0.5 22 [(a) even (b) odd] 0 1234 x Ϫ0.5 18.6 Inverse functions (a) If y is a function of x, the graph of y against x can be y yϭx used to find x when any value of y is given. Thus the ␲ graph also expresses that x is a function of y. Two such functions are called inverse functions. Ϫ2␲ Ϫ␲ 0␲ 2␲ x In general, given a function y = f (x), its inverse may Ϫ␲ be obtained by interchanging the roles of x and y and then transposing for y. The inverse function is denoted ( b) by y = f −1(x). For example, if y = 2x + 1, the inverse is obtained by (i) transposing for x, i.e. x = y − 1 = y − 1 and 2 22 Figure 18.29 (ii) interchanging x and y, giving the inverse as y=x − 1 Now try the following exercise 22 Exercise 78 Further problems on even and Thus if f (x) = 2x + 1, then f −1(x) = x − 1 odd functions 22 In Problems 1 and 2 determine whether the given A graph of f (x) = 2x + 1 and its inverse f −1(x) = functions are even, odd or neither even nor odd. x − 1 is shown in Fig. 18.30 and f −1(x) is seen to be 22 1. (a) x4 (b) tan 3x (c) 2e3t (d) sin2 x a reflection of f (x) in the line y = x. (a) even (b) odd Similarly, if y = x2, the inverse is obtained by (c) neither (d) even (i) transposing for x, i.e. x = ±√y and (ii) interch√anging x and y, giving the inverse y=± x. Hence the inverse has two values for every value of x. Thus f (x) = x2 does not have a single inverse. In 2. (a) 5t 3 (b) ex + e−x cos θ (d) ex such a case the domain of the original function may (c) θ be restricte√d to y = x2 for x > 0. Thus the inverse is then y = + x. A graph of f (x) = x2 and its inverse

Functions and their curves 189 y Hence if f (x) = x − 1, then f−1(x) = x + 1 y5 2x11 (b) If y = f (x), then y = x2 − 4 √(x > 0) y5x Transposing for x gives x = y +√4 Interchanging x and y gives y = x + 4 4 Hence if √f (x) = x2 − 4 (x > 0) then f−1(x) = x+4 if x> −4 2 y5 x 2 1 (c) If y = f (x), then y = x2 + 1 √ 1 2 2 Transposing for x gives x = y√− 1 21 0 Interchanging x and y gives y = x − 1, which has 21 1 2 3 4x Figure 18.30 two values. y Hence there is no inverse of f(x) = x2 + 1, since the domain of f (x) is not restricted. 4 y5x2 Inverse trigonometric functions 2 y5x If y = sin x, then x is the angle whose sine is y. y 5 Œx„ Inverse trigonometrical functions are denoted by pre- fixing the function with ‘arc’ or, more commonly,−1. 0 1 2 3x Hence transposing y = sin x for x gives x = sin−1 y. Figure 18.31 Interchanging x and y gives the inverse y = sin−1 x. Similarly, y = cos−1 x, y = tan−1 x, y = sec−1 x, y =cosec−1x and y =cot−1 x are all inverse trigono- metric functions. The angle is always expressed in radians. Inverse trigonometric functions are periodic so it is necessary to specify the smallest or principal value of the angle. For sin−1 x, tan−1 x, cosec−1 x and cot−1 x, the principal value is in the range − π < y < π . For cos−1 x 22 and sec−1 x the principal value is in the range 0 < y < π. Graphs of the six inverse trigonometric functions are shown in Fig. 33.1, page 335. √ Problem 6. Determine the principal values of x f −1(x ) = for x > 0 is shown in Fig. 18.31 and, again, (a) arcsin 0.5 (b) arctan(−1) √ √ f −1(x) is seen to be a reflection of f (x) in the line y = x. (c) arccos − 3 (d) arccosec( 2) It is noted from the latter example, that not all func- 2 tions have an inverse. An inverse, however, can be determined if the range is restricted. Problem 5. Determine the inverse for each of the Using a calculator, following functions: (a) f (x) = x − 1 (b) f (x) = x2 − 4 (x > 0) (a) arcsin 0.5 ≡ sin−1 0.5 = 30◦ (c) f (x) = x2 + 1 = π rad or 0.5236 rad 6 (a) If y = f (x), then y = x − 1 (b) arctan(−1) ≡ tan−1(−1) = −45◦ Transposing for x gives x = y + 1 Interchanging x and y gives y = x + 1 = − π rad or −0.7854 rad 4

190 Higher Engineering Mathematics √√ 7. tan−1 1 π or 0.7854 rad (c) arccos − 3 ≡ cos−1 − 3 = 150◦ 4 22 = 5π rad or 2.6180 rad 8. cot−1 2 [0.4636 rad] 6 9. cosec−1 2.5 [0.4115 rad] √1 10. sec−1 1.5 [0.8411 rad] (d) arccosec( 2) = arcsin √ 11. sin−1 √1 π or 0.7854 rad 2 2 4 ≡ sin−1 12 = 45◦ 12. Evaluate x, correct to 3 decimal places: √ = π rad or 0.7854 rad x = sin−1 1 + cos−1 4 − tan−1 8 4 3 59 [0.257] Problem 7. Evaluate (in radians), correct to 13. Evaluate y, correct to 4 significant figures: 3 decimal places: sin−1 0.30 + cos−1 0.65. y = 3 sec−1 √ − 4 cosec−1 √ 2 2 sin−1 0.30 = 17.4576◦ = 0.3047 rad + 5 cot−1 2 cos−1 0.65 = 49.4584◦ = 0.8632 rad [1.533] Hence sin−1 0.30 + cos−1 0.65 18.7 Asymptotes = 0.3047 +0.8632 = 1.168, correct to 3 decimal places. If a table of values for the function y = x + 2 is drawn Now try the following exercise x +1 Exercise 79 Further problems on inverse up for various values of x and then y plotted against x, functions the graph would be as shown in Fig. 18.32. The straight lines AB, i.e. x = −1, and CD, i.e. y = 1, are known as Determine the inverse of the functions given in asymptotes. Problems 1 to 4. An asymptote to a curve is defined as a straight 1. f (x) = x + 1 [ f −1(x) = x − 1] line to which the curve approaches as the distance from the origin increases. Alternatively, an asymp- 2. f (x) = 5x − 1 f −1(x ) = 1 (x + 1) tote can be considered as a tangent to the curve at 3. f (x) = x3 + 1 5 infinity. [ f −1(x ) = √ − 1] Asymptotes parallel to the x- and y-axes 3x There is a simple rule which enables asymptotes paral- 4. f (x) = 1 + 2 f −1(x ) = x 1 2 lel to the x- and y-axis to be determined. For a curve x − y = f (x): Determine the principal value of the inverse func- (i) the asymptotes parallel to the x-axis are found by tions in Problems 5 to 11. equating the coefficient of the highest power of x to zero. 5. sin−1(−1) − π or −1.5708 rad 2 (ii) the asymptotes parallel to the y-axis are found by equating the coefficient of the highest power of y 6. cos−1 0.5 π or 1.0472 rad to zero. 3

Functions and their curves 191 Ay 5 4 3 x 12 y 5 x 11 2 C D 1 24 23 22 21 0 1 2 3 4x 21 22 x 12 23 y 5 x 11 24 25 B Figure 18.32 With the above example y = x + 2 , rearranging gives: Problem 8. Determine the asymptotes for the x + 1 x −3 y(x + 1) = x + 2 function y = 2x + 1 and hence sketch the curve. i.e. yx + y − x − 2 = 0 and x(y − 1) + y − 2 = 0 (1) Rearranging y = x −3 gives: y(2x + 1) = x −3 2x + 1 The coefficient of the highest power of x (in this case x1) i.e. 2x y + y = x − 3 is (y − 1). Equating to zero gives: y − 1 = 0 or 2x y + y − x + 3 = 0 and x(2y − 1) + y + 3 = 0 x +2 From which, y = 1, which is an asymptote of y = x + 1 Equating the coefficient of the highest power of x to as shown in Fig. 18.32. zero gives: 2y−1=0 from which, y = 1 which is an 2 asymptote. Returning to equation (1): yx + y − x − 2 = 0 Since y(2x + 1) = x − 3 then equating the coefficient of the highest power of y to zero gives: 2x + 1 = 0 from from which, y(x + 1) − x − 2 = 0. which, x = − 1 which is also an asymptote. 2 x − 3 −3 When x = 0, y = 2x + 1 = 1 = −3 and when y = 0, The coefficient of the highest power of y (in this case x −3 2x + 1 y1) is (x + 1). Equating to zero gives: x + 1 = 0 from 0= from which, x −3 = 0 and x = 3. + which, x = −1, which is another asymptote of y = x + 2 A sketch of y = x −3 is shown in Fig. 18.33. x 1 2x + 1 as shown in Fig. 18.32.

y5 x23 y 2x 11 6 4 2 28 26 24 22 21 0 2 2 Figure 18.33

192 Higher Engineering Mathematics 8x 1 46 2 y5 x 23 2x 11 5 y 1 2 2 1 x 52 24 26

Functions and their curves 193 Problem 9. Determine the asymptotes parallel to (iii) Equating the coefficient of the highest power the x- and y-axes for the function of x to zero gives m − 1 = 0 from which, x2 y2 = 9(x2 + y2). m = 1. Equating the coefficient of the next highest power Asymptotes parallel to the x-axis: of x to zero gives m + c + 1 =0. Rearranging x2 y2 = 9(x2 + y2) gives and since m = 1, 1 + c + 1 = 0 from which, c = −2. x2 y2 − 9x2 − 9y2 = 0 Hence y = mx + c = 1x − 2. hence x2(y2 − 9) − 9y2 = 0 i.e. y = x − 2 is an asymptote. Equating the coefficient of the highest power of x to zero gives y2 − 9 = 0 from which, y2 = 9 and y = ±3. To determine any asymptotes parallel to the x-axis: Asymptotes parallel to the y-axis: Rearranging y(x + 1) = (x − 3)(x + 2) Since x2 y2 − 9x2 − 9y2 = 0 then y2(x2 − 9) − 9x2 = 0 gives yx + y = x2 − x −6 Equating the coefficient of the highest power of y to zero The coefficient of the highest power of x (i.e. x2) is 1. gives x2 − 9 = 0 from which, x2 = 9 and x = ±3. Equating this to zero gives 1 =0 which is not an equation Hence asymptotes occur at y = ±3 and x = ±3. of a line. Hence there is no asymptote parallel to the Other asymptotes x -axis. To determine asymptotes other than those parallel to x- and y-axes a simple procedure is: To determine any asymptotes parallel to the y-axis: Since y(x + 1) = (x − 3)(x + 2) the coefficient of (i) substitute y = mx + c in the given equation the highest power of y is x + 1. Equating this to (ii) simplify the expression zero gives x + 1 = 0, from which, x = −1. Hence x = −1 (iii) equate the coefficients of the two highest powers is an asymptote. When x = 0, y(1) = (−3)(2), i.e. y = −6. of x to zero and determine the values of m and c. When y = 0, 0 =(x − 3)(x + 2), i.e. x = 3 and x = −2. y = mx + c gives the asymptote. A sketch of the function y(x + 1) = (x − 3)(x + 2) is Problem 10. Determine the asymptotes for the shown in Fig. 18.34. function: y(x + 1) = (x − 3)(x + 2) and sketch the curve. Problem 11. Determine the asymptotes for the function x3 − x y2 + 2x − 9 =0. Following the above procedure: Following the procedure: (i) Substituting y = mx + c into y(x + 1) = (x − 3) (x + 2) gives: (i) Substituting y = mx + c gives x3 − x(mx + c)2 + 2x − 9 =0. (mx + c)(x + 1) = (x − 3)(x + 2) (ii) Simplifying gives (ii) Simplifying gives x3 − x[m2 x2 + 2mcx + c2] + 2x − 9 = 0 mx2 + mx + cx + c = x2 − x − 6 i.e. x3 − m2 x3 − 2mcx2 − c2 x + 2x − 9 = 0 and x3(1 − m2) − 2mcx2 − c2 x + 2x − 9 = 0 and (m − 1)x2 + (m + c + 1)x + c + 6 =0 (iii) Equating the coefficient of the highest power of x (i.e. x3 in this case) to zero gives 1 −m2 = 0, from which, m = ±1. Equating the coefficient of the next highest power of x (i.e. x2 in this case) to zero gives −2mc = 0, from which, c = 0.

194 Higher Engineering Mathematics y y 5x 22 6 x 521 4 2 26 24 22 0 2 4 6x y (x 11) 5 (x 23)(x 12) 22 y (x 11) 5 (x 23)(x 12) 24 26 28 210 Figure 18.34

Functions and their curves 195 Hence y = mx + c = ±1x + 0, i.e. y = x and y = −x Hence 1= 1 and x2 = 1, from which, x = ±1. are asymptotes. x2 When x = 1, To determine any asymptotes parallel to the x- and y-axes for the function x3 − x y2 + 2x − 9 =0: x2 +1 1+1 y= = =2 Equating the coefficient of the highest power of x term to zero gives 1 = 0 which is not an equation of a line. x1 Hence there is no asymptote parallel with the x-axis. and when x = −1, Equating the coefficient of the highest power of y term to zero gives −x = 0 from which, x = 0. (−1)2 + 1 y = −1 = −2 Hence x = 0, y = x and y = − x are asymptotes for the function x3 − xy2 + 2x − 9 =0. i.e. (1, 2) and (−1, −2) are the co-ordinates of the turning points. d2 y = 2x−3 = 2 when x = 1, d2 y is positive, dx2 x3 ; dx2 which indicates a minimum point and when x = −1, Problem 12. Find the asymptotes for the function d2 y x2 +1 dx2 is negative, which indicates a maximum point, as shown in Fig. 18.35. y = and sketch a graph of the function. x Now try the following exercise Rearranging y = x2 + 1 gives yx = x2 + 1. Exercise 80 Further problems on x asymptotes Equating the coefficient of the highest power x term to In Problems 1 to 3, determine the asymptotes zero gives 1 =0, hence there is no asymptote parallel to parallel to the x- and y-axes. the x-axis. 1. y = x − 2 [y = 1, x = −1] Equating the coefficient of the highest power y term to x + 1 zero gives x = 0. 2. y2 = x [x = 3, y = 1 and y = −1] Hence there is an asymptote at x = 0 (i.e. the x −3 y-axis). To determine any other asymptotes we substitute y = mx + c into yx = x2 + 1 which gives (mx + c)x = x2 + 1 3. y = (x x(x + 3) 1) i.e. mx2 + cx = x2 + 1 + 2)(x + and (m − 1)x2 + cx − 1 = 0 [x = −1, x = −2 and y = 1] In Problems 4 and 5, determine all the asymptotes. Equating the coefficient of the highest power x term to 4. 8x − 10 + x3 − x y2 = 0 zero gives m − 1 = 0, from which m = 1. [x = 0, y = x and y = −x] Equating the coefficient of the next highest power x term to zero gives c = 0. Hence y = mx + c = 1x + 0, i.e. y = x 5. x2(y2 − 16) = y is an asymptote. [y = 4, y = −4 and x = 0] x2 +1 In Problems 6 and 7, determine the asymptotes and A sketch of y = x is shown in Fig. 18.35. sketch the curves. It is possible to determine maximum/minimum points x2 − x −4 6. y = on the graph (see Chapter 28). x +1 Since y = x2 + 1 = x2 + 1 = x + x−1 x = −1, y = x − 2, x xx see Fig 18.40, page 202 then dy = 1 − x−2 =1− 1 = 0 7. x y2 − x2 y + 2x − y = 5 dx x2 x = 0, y = 0, y = x, see Fig. 18.41, page 202 for a turning point.

196 Higher Engineering Mathematics y y 5x 6 y 5 x 211 4x 2 24 22 0 2 4 x 22 y 5 x 211 24 x 26 Figure 18.35 18.8 Brief guide to curve sketching (a) If the equation is unchanged when −x is substi- tuted for x, the graph will be symmetrical about The following steps will give information from which the y-axis (i.e. it is an even function). the graphs of many types of functions y = f (x) can be sketched. (b) If the equation is unchanged when −y is substi- tuted for y, the graph will be symmetrical about (i) Use calculus to determine the location and nature the x-axis. of maximum and minimum points (see Chap- ter 28) (c) If f (−x) = − f (x), the graph is symmet- rical about the origin (i.e. it is an odd (ii) Determine where the curve cuts the x- and y-axes function). (iii) Inspect the equation for symmetry. (iv) Check for any asymptotes.

Functions and their curves 197 18.9 Worked problems on curve y sketching 20 Problem 13. Sketch the graphs of 15 (a) y = 2x2 + 12x + 20 (b) y = −3x2 + 12x − 15 y 5 2x 21 12x 1 20 10 5 2 (a) y = 2x2 + 12x + 20 is a parabola since the equa- 24 23 22 21 0 1 2 3x tion is a quadratic. To determine the turning 23 point: 25 Gradient = d y = 4x + 12 = 0 for a turning point. 210 y 5 23x 21 12x 2 15 dx 215 Hence 4x = −12 and x = −3. 220 When x = −3, y = 2(−3)2 + 12(−3) + 20 =2. Hence (−3, 2) are the co-ordinates of the turning 225 point Figure 18.36 d2 y = 4, which is positive, hence (−3, 2) is a dx2 minimum point. When x = 0, y = 20, hence the curve cuts the Problem 14. Sketch the curves depicting the y-axis at y = 20. following equations: (a) x = 9 − y2 (b) y2 = 16x Thus knowing the curve passes through (−3, 2) and (0, 20) and appreciating the general shape (c) x y = 5 of a parabola results in the sketch given in Fig. 18.36. (a) Squaring both sides of the equation and trans- posing gives x2 + y2 = 9. Comparing this with (b) y = −3x2 + 12x − 15 is also a parabola (but ‘upside down’ due to the minus sign in front of the standard equation of a circle, centre ori- the x2 term). gin and radius a, i.e. x2 + y2 = a2, shows that x2 + y2 = 9 represents a circle, centre origin and Gradient = d y = −6x + 12 = 0 for a turning point. dx radius 3. A sketch of this circle is shown in Hence 6x = 12 and x = 2. Fig. 18.37(a). When x = 2, y = −3(2)2 + 12(2) − 15 =−3. (b) The equation y2 = 16x is symmetrical about the x-axis and having its vertex at the origin (0, 0). Hence (2, −3) are the co-ordinates of the turning Also, when x = 1, y = ±4. A sketch of this point parabola is shown in Fig. 18.37(b). d2 y dx2 = −6, which is negative, hence (2, −3) is a (c) The equation y = a represents a rectangular maximum point. x hyperbola lying entirely within the first and third When x = 0, y = −15, hence the curve cuts the axis quadrants. Transposing xy=5 gives y= 5 and at y = −15. , x The curve is shown sketched in Fig. 18.36. therefore represents the rectangular hyperbola shown in Fig. 18.37(c).

198 Higher Engineering Mathematics with the x- and y-axes of a rectangular co-ordinate system, the major axis being 2(3), i.e. 6 units long y and the minor axis 2(2), i.e. 4 units long, as shown in Fig. 18.38(a). 3 y x (a) x 5!(92y 2) 4 x y 14 6 (a) 4x 2 5 36 29y 2 y 1 x 24 x (b) y 2 516x 2Œ„3 y (b) 3y 2 11555x 2 Figure 18.38 (b) Dividing 3y2 + 15 = 5x2 throughout by 15 and x2 y2 x transposing gives − = 1. The equation 35 x2 y2 a2 − b2 = 1 represents a hyperbola which is sym- metrical about both the x- and y-axes, the distance (c) xy 5 5 between the vertices being given by 2a. x2 y2 Figure 18.37 Thus a sketch of − = 1 is as shown in 35 √ Fig. 18.38(b), having a distance of 2 3 between Problem 15. Sketch the curves depicting the its vertices. following equations: Problem 16. Describe the shape of the curves (a) 4x2 = 36 −9y2 (b) 3y2 + 15 =5x2 represented by the following equations: (a) By dividing throughout by 36 and transposing, y 2 y2 (a) x = 2 1 − (b) = 2x the equation 4x2 = 36 − 9y2 can be written as 28 x2 + y2 = 1. The equation of an ellipse is of the x 2 1/2 94 (c) y = 6 1 − x2 y2 a2 b2 16 form + = 1, where 2a and 2b represent the x2 y2 (a) Squaring the equation gives x2 = 4 1 − y 2 length of the axes of the ellipse. Thus 32 + 22 = 1 2 represents an ellipse, having its axes coinciding and transposing gives x2 = 4 − y2, i.e.

Functions and their curves 199 x2 + y2 = 4. Comparing this equation with Now try the following exercise x2 + y2 = a2 shows that x2 + y2 = 4 is the equa- Exercise 81 Further problems on curve tion of a circle having centre at the origin (0, 0) sketching 1. Sketch the graphs of (a) y = 3x2 + 9x + 7 and of radius 2 units. 4 (b) Transposing y2 =2x gives y = √ . Thus (b) y = −5x2 + 20x + 50. 8 4x ⎡⎤ y2 = 2x is the equation of a parabola having its (a) Parabola with minimum 8 axis of symmetry coinciding with the x-axis and ⎢⎢⎢⎢⎢⎢⎢⎣(b) pvPvaaaallsruuaseiebnoaaglttat(h−w2r,oi237tuh,0g−)mh5aan0xda,inm1pd43uasms.ing⎥⎦⎥⎥⎥⎥⎥⎥ through (0, 50). its vertex at the origin of a rectangular co-ordinate In Problems 2 to 8, sketch the curves depicting the system. equations given. (c) y=6 1− x2 1/2 y= y2 2. x = 4 1 − can be transposed to 16 6 4 1 − x2 1/2 and squaring both sides gives [circle, centre (0, 0), radius 4 units] 16 √y 3. x = y2 = 1− x2 i.e. x2 + y2 = 1. , 9 36 16 16 36 parabola, symmetrical about x-axis, vertex at (0, 0) This is the equation of an ellipse, centre at the ori- gin of a rectangular co-ordinate system, the m√ajor 4. y2 = x2 − 16 axis coinciding with the y-axis and being 2 36, 4 i.e. 12 units long. √The minor axis coincides with ⎡⎤ the x-axis and is 2 16, i.e. 8 units long. hyperbola, symmetrical about ⎢⎢⎣bxe-tawnedeny-vaexretsic,edsi8staunncites along⎦⎥⎥ Problem 17. Describe the shape of the curves x -axis represented by the following equations: y2 x2 x y 2 y 15 5. = 5 − (a) = 1 + (b) = 5 2 4 2x 52 ⎡⎤ x y2 ellipse, centre (0, 0), major axis (a) Since = 1 + ⎣1√0 units along y-axis, minor axis⎦ 2 10 units along x-axis 52 6. x = 3 1 + y2 x2 y 2 ⎡⎤ =1+ hyperbola, symmetrical about ⎣⎢⎢bxe-tawnedeny-vaexretsic,edsi6staunncites along⎦⎥⎥ 25 2 x -axis i.e. x2 − y2 = 1 7. x2 y2 = 9 25 4 rectangular hyperbola, lying in first and third quadrants only This is a hyperbola which is symmetrical a√bout both the x- and y-axes, the vertices being 2 25, i.e. 10 units apart. (With reference to Section 18.1 (vii), a is equal to ±5) (b) The equation y = 15 is of the form y= a a= , 4 2x x 60 = 30. 2 This represents a rectangular hyperbola, sym- metrical about both the x- and y-axis, and lying entirely in the first and third quadrants, similar in shape to the curves shown in Fig. 18.9.

200 Higher Engineering Mathematics 8. x = 1 (36 − 18y2) ⎡⎤ 3 hyperbola, symmetrical about x- ⎡ ⎤ ellipse, centre (0, 0), ⎣and y-axes, vertices 2 units ⎦ apart along x-axis ⎢⎢⎣mmianjoorr axis 4√units along x -axis,⎦⎥⎥ axis 2 2 units √ 12. y = 9 − x2 along y-axis [circle, centre (0, 0), radius 3 units] 9. Sketch the circle given by the equation 13. y = 7x−1 ⎡rectangular hyperbola, lying ⎤ x2 + y2 − 4x + 10y + 25 =0. ⎣⎢⎢sinymfirmstetarnicdatlhaibrdouqtuxad- raanndts, ⎥⎥⎦ [Centre at (2, −5), radius 2] In Problems 10 to 15 describe the shape of the y-axes curves represented by the equations given. 14. y = (3x)1/2 10. y = [3(1 − x2)] parabola, vertex at (0, 0), sym- ⎡⎤ metrical about the x-axis ⎣2el√lip3suen, ictsenatlroen(g0y, 0-a)x,ism,amjoirnaoxris⎦ axis 2 units along x-axis 15. y2 − 8 =−2x⎡2 ⎤ ⎢⎣⎢aeyxl-laiispxsi2se√,, m8ceuinnntoritresa(ax0li,os0n4)g,utmhneiatsjor⎥⎥⎦ 11. y = [3(x2 − 1)] along the x-axis Graphical solutions to Exercise 77, page 186 1. 2. y y 4 10 2 5 y 5 3x 25 0 123 x0 1 2 3x 25 22 y 5 23x 14 3. 4. y y 22 21 8 y 5 x 213 8 6 1 2x y 5(x 23)2 4 2 4 0 0 2 4 6x Figure 18.39

5. 6. Functions and their curves 201 y y y 5x 2x 2 15 0.50 1x 10 0.25 5 y 5(x24) 212 0 0 2 4 6 8x 7. y5x312 8. y 5 11 2 cos 3x y 12 x y 10 3 ␲ ␲ 3␲ 2␲ x 22 5 2 y 5 2 ln x 22 21 0 1 1 2 3 4x 25 0 210 21 9. 10. y y 6 4 3 2 y 5 3 2 2 sin(x 1 ␲ ) 2 4 1 0 p p 3p 2p x 0 22 21 22 Figure 18.39 (Continued)

202 Higher Engineering Mathematics y 5x 22 Graphical solutions to Problems 6 and 7, Exercise 80, page 195 y 6 x 521 4 2 26 24 22 0 2 4 6x 22 x 2 2x 2 4 x 2 2x 2 4 x 11 x 11 y 5 y 5 24 26 Figure 18.40 y y 5x Figure 18.41 xy 2 2 x 2y 1 2x 2y 5 5 6 4 2 26 24 22 0 2 4 6x xy 2 2 x 2y 1 2x 2y 5 5 22 xy 2 2 x 2y 1 2x 2y 5 5 24 26

Chapter 19 Irregular areas, volumes and mean values of waveforms 19.1 Areas of irregular figures (iii) Areas PQRS Areas of irregular plane surfaces may be approximately =d y1 + y7 + y2 + y3 + y4 + y5 + y6 determined by using (a) a planimeter, (b) the trapezoidal 2 rule, (c) the mid-ordinate rule, and (d) Simpson’s rule. Such methods may be used, for example, by engineers In general, the trapezoidal rule states: estimating areas of indicator diagrams of steam engines, surveyors estimating areas of plots of land or naval Area = ⎡ ⎛ + ⎞ ⎤ architects estimating areas of water planes or transverse first sum of sections of ships. width of ⎣ 1 ⎝last ⎠ + remaining⎦ interval (a) A planimeter is an instrument for directly mea- 2 ordinate ordinates suring small areas bounded by an irregular curve. (c) Mid-ordinate rule (b) Trapezoidal rule To determine the area ABCD of Fig. 19.2: To determine the areas PQRS in Fig. 19.1: QR BC y1 y2 y3 y4 y5 y6 y7 y1 y2 y3 y4 y5 y6 PS AD dddddd dddddd Figure 19.1 Figure 19.2 (i) Divide base PSinto any number of equal inte- (i) Divide base AD into any number of equal rvals, each of width d (the greater the number intervals, each of width d (the greater the of intervals, the greater the accuracy). number of intervals, the greater the accuracy). (ii) Accurately measure ordinates y1, y2, y3, etc. (ii) Erect ordinates in the middle of each interval (shown by broken lines in Fig. 19.2).

204 Higher Engineering Mathematics (iii) Accurately measure ordinates y1, y2, y3, etc. 30 Graph of speed/timeSpeed (m/s) (iv) Area ABCD = d(y1+ y2+ y3+ y4+ y5+ y6) 25 In general, the mid-ordinate rule states: 20 15 width of sum of 10 Area = 5 interval mid-ordinates 0 123456 (d) Simpson’s rule Time (seconds) Figure 19.3 To determine the area PQRS of Fig. 19.1: 1.25 2.5 (i) Divide base PS into an even number of inter- 4.0 vals, each of width d (the greater the number 5.5 of intervals, the greater the accuracy). 7.0 8.75 10.75 12.5 15.0 17.5 20.25 24.0 (ii) Accurately measure ordinates y1, y2, y3, etc. (iii) Area PQRS = d [(y1 + y7) + 4(y2 + y4 + Thus 3 y6) + 2(y3 + y5)] In general, Simpson’s rule states: area = (1) 0 + 24.0 + 2.5 + 5.5 2 1 width of first + last Area = ordinate + 8.75 + 12.5 + 17.5 3 interval = 58.75 m sum of even (b) Mid-ordinate rule (see para. (c) above) +4 The time base is divided into 6 strips each of width 1 second. ordinates Mid-ordinates are erected as shown in Fig. 19.3 by the broken lines. The length of each mid-ordinate sum of remaining is measured. Thus +2 odd ordinates Problem 1. A car starts from rest and its speed is area = (1)[1.25 + 4.0 + 7.0 + 10.75 measured every second for 6 s: + 15.0 + 20.25] Time t (s) 0 1 2 3 4 5 6 = 58.25 m Speed v (m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0 (c) Simpson’s rule (see para. (d) above) Determine the distance travelled in 6 seconds (i.e. The time base is divided into 6 strips each of the area under the v/t graph), by (a) the trapezoidal width 1 s, and the length of the ordinates measured. rule, (b) the mid-ordinate rule, and (c) Simpson’s Thus rule. area = 1 (1)[(0 + 24.0) + 4(2.5 + 8.75 A graph of speed/time is shown in Fig. 19.3. 3 (a) Trapezoidal rule (see para. (b) above) + 17.5) + 2(5.5 + 12.5)] The time base is divided into 6 strips each of width 1 s, and the length of the ordinates measured. = 58.33 m