( x) Solutions 9(c) 1. Complete solutions to Exercise 9(c) 1. We first make a sketch of y = sin x ( ): Area A= Area B. By (9.6) = cos cos 0 = 2 = 2

Size: px

Start display at page:

Download "( x) Solutions 9(c) 1. Complete solutions to Exercise 9(c) 1. We first make a sketch of y = sin x ( ): Area A= Area B. By (9.6) = cos cos 0 = 2 = 2"

Maximillian Butler
5 years ago
Views:

1 Solutions 9(c) Complete solutions to Exercise 9(c). We first make a sketch of y = sin x : y Area A y =sin(x) x -.5 Area B By (9.6) Similarly Thus - Area A= sin x dx ( x) ( ) [ ] = cos = cos cos = = Area B = total area = + = 4( units). We only need to consider one cycle, say between t = to t =.: area The mean value is the. The interval is. and the area of the interval triangle = ( 4. ): So ( 4. ) mean value of v= = V. How do we find the R.M.S. value of v? We need to find an equation for v. Since v is a straight line it is of the form v= mt+ c where m is the gradient and c is the v intercept. What is the value of c? (9.6) Area = ydx a b (*)

2 Solutions 9(c) From graph c = 4 What is the value of the gradient, m? 4 m = =. Hence substituting m = and c = 4 into (*) gives v= 4 t To find the R.M.S. value we use (9.8):. ( R. MS. ) = ( 4 t) dt. ( RM S). = ( t) taking out the common factor ( t) dt. expanding. dt = 8 t+ 5t dt t 5t = 8 t + Integrating 5. = 8. ( 5. ) + [ ].. = 5. = How do we find the R.M.S. value from this? RM.. S= 5. =. V ( d.p.). Using (9.7) with a =, b =, y = i = I sin( t) and dx = dt we have: Mean value of i = Isin () t dt I = sin () t dt ( Taking Out I) I = cos() t ( Integrating) I I = cos( ) cos = ( ) = = Mean value of i = I A Using this result, the mean value of sin()= t A (substituting I = into the above). R. MS.. = b ydx a (9.8)

3 Solutions 9(c) 4. (i) If we sketch v =sin() t over the period of to then we have: v 5 v =sin(t) 4 5 t -5 - It can be clearly seen that the mean (average) value of v =sin() t over to is V. (ii) Similarly for v = cos() t we have: v 5 v =cos(t) 4 5 t -5 - So the mean value is V. 5. Using (9.7) with a =, b =, y = v and dx = d( ωt) gives Mean value of v = V sin( ωt)d( ωt) V = sin ωt d ωt Taking Out V V ωt= = cos( ωt) ( Integrating) ωt = V V = cos( ) cos = [ ] = V = = b (9.7) Mean value of y = y dx a

4 Solutions 9(c) 4 6. Using (9.8) with a =, b =, y = i = I cos( t) gives ( irm.. S. ) = ( Icos() t ) dt I = cos tdt * How do we integrate cos () t with respect to t? Need to use (4.67) cos () t = + cos( t) () () The remaining evaluation is similar to EXAMPLE. cos () tdt= cos( t) dt + = cos( t) dt Taking Out + ( t) sin = t + sin ( ) = + = = ( ) = Substituting cos () t dt = into (*) gives ( i ) ( ) i RM.. S. RM.. S. I I = = I I I = = = ( Integrating) b (9.7) Mean value of y = y dx a (9.8) b R. MS.. = ydx a

5 Solutions 9(c) 5 7. Using (9.7) we have Mean value of v = ( e.t )dt.t = ( e ) dt.t e = t. by (8.4) = t+ (.) = e.t = + e + e = + e = e = Mean value of v =e =.68V ( d.p.) To find the R.M.S. value we use (9.8): (. t RM.. S. = ) e dt.t.t = e e dt Expanding +.t.t = e + e dt..t.t e e = t +.. by (8.4) (. ) (. ) { e e e e } = = { + e 5e [ 5] } ( R.M.S. ) = 6.89 So the R.M.S. value R. MS.. = 6.89 = 4.V ( d.p.) area 8. The mean value is defined as the. The interval is 8 s and the interval area consists of a rectangle ( t = to t = ), a trapezium ( t = to t = 5) and a triangle ( t = 5 to t = 8). Rectangle area = 6 =8 ma s (8.4) = kt+ m kt+ m e dt e k b (9.7) Mean value of y = y b a dx a (9.8) b R. MS. = ydx a

6 Solutions 9(c) 6 Trapezium area = () ( 6 +)= 6 ma s Triangle area = ( ) = 5 ma s Total area = =9 ma s Mean value = 9 =.8 ma ( s.f.) 8 9. The mean value of v is evaluated by (9.7): Mean value of v = ( ωt)sin ( ωt)d( ωt) () How do we integrate this function? Use integration by parts formula (8.45): u = ωt v = sin ωt u = v= sin( ωt) d( ωt) = cos( ωt) ( ωt) sin ( ωt) d( ωt) = ωtcos( ωt) + cos ωt d ωt = cos + sin ( ωt) = [ ] + sin sin = Substituting this into ( ) gives: = Mean value of v = ( )= V To find the R.M.S. value we first obtain ( R.M.S. ) : R. MS.. = ωt sin ωt d ωt = ( ωt) sin ( ωt) d( ωt) By using (4.68) we can rewrite sin ( ωt) as: ( ωt) = ( ωt) sin cos So we have ( RM.. S. ) = ( ωt) d( ωt) ( ωt) cos( ωt) d( ωt) (*) First integral on the right of (*) is straightforward but how do we find ( ωt) cos( ωt) d( ωt)? Use integration by parts, (8.45): u = ωt v = cos ωt du d ( ω t) sin = ( ωt) v= cos( ωt) d( ωt) = (8.45) u v dt = uv u v dt (9.7) Mean value of b y = y dx a ( ω t)

7 Solutions 9(c) 7 Hence cos( ) ( ωt) sin ( ωt) ( ωt) sin ( ωt) ωt ωt d ωt = d ωt ( ωt) ( ωt) d( ωt) = sin ( ω t) cos( ω t) d( ω t) = ( ω t) sin ( ω t) d( ω t) (**) Use integration by parts again: u = ωt v = sin ωt cos u = v= sin( ωt) d( ωt) = Substituting into (**): cos( ) ( ω t) ωtcos ωt cos ωt ωt ωt d ωt = d ωt ( ωt) sin = 4 = Evaluating the first integral of (*): t ω ( ωt) d( ωt) = = Substituting these evaluations into (*) gives: ( RM.. S. ) = R. MS.. =.8V = The form factor:.8 f = =.8. The mean value, M, of i between to is given by: M = cos ( ωt) d( ωt) (* ) To find the integral we use Simpson's rule with 4 equal integrals: h = = 4 8 We establish a table of values: ωt cos ω t Applying Simpson's rule

8 Solutions 9(c) 8 8 cos( ωtd ) ( ωt) 4(.96.69) (.84) = [ 9. ] Substituting into (*): 8 9. M = [ 9.] = =.75V ( d.p.). Very similar to EXAMPLE 4. Mean force is. 86kN.. (i) Shaded Area = [Area of rectangle]-[area under y = x between and ](*) Area under y = x between and = x dx Replacing this in (*) gives (ii) (iii) x dx by x = = by (8.) Shaded Area = Area of rectangle = x = (8.) = x [ ] = = y y = x y = x x (iv) Results are the same,, because x is the inverse function of x, that is x reflects x in the line y = x. Hence the above shaded area and the shaded area of part (i) are equal.

Solutionbank Edexcel AS and A Level Modular Mathematics

Solutionbank Edexcel AS and A Level Modular Mathematics Page of Exercise A, Question Use the binomial theorem to expand, x