MTH 102 Part Five PDF

Exercises 1. Find the values of the following definite integrals ( i) 5 (ii) (4 − ) (iii) 2 sin 2 (iv) 3 +4 (v) √ 2. The average value of a...

Exercises 1. Find the values of the following definite integrals ( i) 5 (ii) (4 − ) (iii) 2 sin 2 (iv) 3 +4 (v) √ 2. The average value of a complex voltage waveform is given by the formula 1 = (10 sin + 3 sin 3 + 2 sin 5 ) ( ) APPLICATIONS SOME APPLICATIONS OF DIFFERENTIATION Velocity and acceleration If a quantity depends on and varies with a quantity then the rate of change of with respect to is. Thus, for example, the rate of change of pressure with respect to height ℎ is. A rate of change with respect to time is usually just called ‘rate of change’ the ‘with respect to time’ being assumed. Thus, for example, a rate of change of current, , is. And a rate of change of temperature, , is , etc. Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 63 Example The length metres of a certain metal rod at temperature °C is given by = 1 + 0.00005. Determine the rate of change of a length. In mm/°C, when the temperature is (a) 100°C and (b) 400°C. Solution The rate of change of length is denoted. Since length = 1 + 0.00005 + 0.0000004 , then = 0.00005 + 0.0000008 (a) When = 100°C, = 0.00005 + (0.0000008)(100) = 0.00013m/°C = 0.13mm/°C (b) When = 400°C, = 0.00005 + (0.0000008)(400) = 0.00037m/°C = 0.37mm/°C Example The distance metres travelled by a vehicle in time seconds after the brakes are applied is given by = 20 −. Determine (a)the speed of the vehicle (in km/h) at the instant the brakes are applied, and (b) the distance the care travels before it stops. Solution (a) Distance, = 20 −. Hence, velocity = = 20 −. At the instant the brakes are applied, time = 0. Hence, velocity, = 20m/s 20 × 60 × 60 = km/h 1000 Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 64  = 72km/h (b) When the car finally stops, the velocity is zero, that is = 20 − = 0, from which , 20 =. This implies = 6s. Hence, the distance travelled before the care stops, is: = 20 − = 20(6) − (6) = 120 − 60 = 60m. Stationary Points In the figure below, the gradient (or rate of change) of the curve changes from positive, between O and P, to negative between P and Q. It then changes again to positive between Q and R. At point P, the gradient is zero and, as increases, the gradient of the curve changes from positive just before P to negative just after P. Such a point is called a maximum point and appears at the ‘crest of a wave’. At point Q, the gradient is also zero and, as increases, the gradient of the curve changes from negative just before Q to positive just after Q. Such a point is called a minimum point, and appears as the ‘bottom of a valley’. Points such as P and Q are called turning points. R Maximum point Negative P gradient Positive gradient Positive gradient O Q Minimum point Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 65 It is possible to have a turning point, the gradient on either side of which is the same. Such a point is called a point of inflexion. The following is a diagram showing points of inflexion Maximum point Points of inflexion Procedure for finding and distinguishing between stationary points (i) Given = ( ), determine. (ii) Put = 0 and solve for the values of. (iii) Substitute the values of into the original equation, = ( ) to find the corresponding value of. To determine the nature of the stationary points: Either (iv) Find and substitute into it the values of found in (2). If the result is: (a) positive – the point is a minimum point, (b) negative – the point is a maximum point, (c) zero – the point is a point of inflexion. Or (v) Determine the sign of the gradient of the curve just before and just after the stationary points. If the sign change for the gradient of the curve is: Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 66  (a) positive to negative – the point is a maximum point. (b) negative to positive – the point is a minimum one. (c) positive to positive or negative to negative – the point is a point of inflexion. Example Locate the turning point on the curve =3 − 6 and determine its nature by examining the sign of the gradient on either side. Solution Following the above procedure: (i) Since =3 −6 , = 6 − 6. (ii) At a turning point, = 0. Hence 6 − 6 = 0. It follows that = 1. (iii) When = 1, = 3(1) − 6(1) = −3. Hence the coordinates of the turning points are (1, −3). (iv) If is slightly less than 1, say, 0.9, then = 6(0.9) = −0.6, which is negative. If is slightly greater than 1, say, 1.1, then = 6(1.1) − 6 = 0.6, which is positive. It follows that the point (1, −3) is a minimum point. Example Locate the turning point on the following curve and determine whether it is a maximum or minimum point: = 4 + , then =4− = 0, for a maximum or minimum value. Hence 4 = , , giving = = −1.3863 (. ) When = −1.3863, = 4(−1.3863) + = 5.5452 + 4.0000 = −1.5452. Thus, (−.3863, −1.5452) are the co-ordinates of the turning point. =. When = −1.3863, = = 4.0, which is positive, hence, (−.3863, −1.5452) is a minimum point. Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 67 Exercise 1. The distance metres moved by a car in a time seconds is given by =3 − 2 + 4 − 1. Determine the velocity and acceleration when (a) = 0 and (b) = 1.5 s. 2. The displacement cm of the slide valve of an engine is given by = 2.2 cos 5 + 3.6 sin 5. Evaluate the velocity (in m/s) when time = 30ms. 3. Determine the coordinates of the maximum and minimum values of the graph = − 3 + 5 by (a) examining the gradient on either side of the turning points and (b) determining the sign of the second derivative SOME APPLICATIONS OF INTEGRATION 1. Application to length of arcs Using the line segment to approximate a small arc piece, the length of the small arc piece can be approximated by ≃ ( ) +( ) = 1+ = 1+. The total length of the whole arc can then be obtained by adding up all lengths of the small arc pieces in the Riemann sum sense under a limiting process, which leads to Arc length = 1+. Example Find the length of an arc which is given by = + from = 1 to =3 Solution First compute and. 1 1 = −. 2 2 Thus, Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 68  1 1 1 2 1 = − = − + , 2 2 4 4 4 ( + ) + = 1+ = =. 4 2 It follows that + 14 Arc length = = 2 3 Example Find the length of arc C which is given by = ( − 1) from = 1 to = 5. Solution First compute and. 2 3 = ∙ ( − 1) , 3 2 and so = − 1. Thus, 2 10√5 − 2 Arc length = 1 + ( − 1) = =. 3 3 Exercises 1. Find the length of an arc which is given by = +2 from = 2 to =4 2. Find the length of arc C which is given by = ( + 2) from = 1 to = 4. Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 69 2. Surface area computing. The surface generated by rotating a smooth arc C around an axis. Then area of the corresponding surface generated by small piece of arc with length equals =2 The area of the whole surface then is Area = 2 1+ If the axis of rotation is parallel to the -axis, then changes should be made accordingly. Example Find the area of the surface of revolution generated by revolving the curve which is given by = ,1 ≤ ≤ 2 about the -axis. Solution First compute and. =3 , and so = 1+9. For each with 1 ≤ ≤ 2, the distance from the corresponding arc piece to the axis of rotation is. Thus, Surface area = 2 1+9 = 145√145 − 10√10. 27 Exercise Find the area of the surface of revolution generated by revolving the curve which is given by =3 ,0 ≤ ≤ 3 about the -axis. Prepared by Chinedu Peter & supervised by Prof. J. O. Fatokun 70