t hours This is the distance in miles travelled in 2 hours when the speed is 70mph. = 22 yards per second. = 110 yards.


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1 The area under a graph often gives useful information. Velocittime graphs Constant velocit The sketch shows the velocittime graph for a car that is travelling along a motorwa at a stead 7 mph. 7 The area under this graph is rectangular in shape. The shaded area = 7 = t hours This is the distance in miles travelled in hours when the speed is 7mph. Area under a velocittime graph = distance travelled. Constant acceleration Now consider the case when a car is accelerating steadil from to mph in seconds. ( d/s) The graph shows this situation, but note that the velocit is in miles per hour whilst the time is in seconds. The units need to be converted in order to find the distance travelled. mph = = ards per second. The distance travelled is given b the area under the graph: Distance travelled = = ards. ( mile = 76 ards) Note that this is equivalent to the car travelling for seconds at the average speed of ards per second. In each case sketch a velocittime graph and find the distance travelled in miles or ards: Car travels at a stead velocit of 6 mph for hours. Car decelerates steadil from a velocit of 6 mph until it stops seconds later. Car accelerates steadil from mph to 6 mph over a period of 6 seconds. Car decelerates steadil from 7 mph to mph over a period of seconds. The Nuffield Foundation
2 Variable velocit and acceleration The table and graph give the velocit of a car as it travels between sets of traffic lights. An estimate for the area under this graph can be found b splitting it into strips as shown. The strips at each end are approimatel triangular in shape and each strip between them is approimatel in the shape of a trapezium. t (s) 6 8 v (ms  ) 8 8 v ms  Area of A = Area of B = Area of C = ( + 8) ( 8 + ) 7 Total area = 7 This is an estimate of the distance travelled b the car (in metres) between the sets of traffic lights. A B C C B A 6 8 This is a simplified model of this situation. In practice the change in velocit is unlikel to be so smooth and smmetrical. Note that better estimates can be found b using more data and narrower strips. In each of the following: draw a velocittime graph on a graphic calculator or spreadsheet describe what happens during the given time interval estimate the distance travelled t (s) v (ms  ) t (s) v (ms  ) t (s) 6 8 v (ms  ) 6 t (s) v (ms  ) The Nuffield Foundation
3 Integration The area under a graph between = a and = b is: = b A = lim = a δ = b a d This means the limit of the sum of rectangles of area δ as δ tends to zero. The increment of area δa = δ δ A This can be written as =. δ δa da The limit of as δ tends to zero is δ d i.e. the derivative of A with respect to. So A = b a δ da d is equivalent to = d If the area under a graph is divided into ver narrow strips, each strip is approimatel rectangular in shape. a δ b δa i.e. the gradient of a graph of A against. Integration is the inverse of differentiation. Rules of Integration Reversing differentiation gives the rules shown: eg = gives d = d so d = In general, for an constant k, Also d = gives = d so d = k d = k d and = gives = d so d = In general, n d = + n n + The derivative of a constant is zero. This means that when a constant is differentiated it disappears. When integrating a constant appears this is called the constant of integration. Epression Integral k (constant) k + c + c + c + n + c + c n + c n + k n + k n + c n c + = + + c where c is the constant of integration. Because the value of the constant is not known (unless more information is given), these are known as indefinite integrals. The Nuffield Foundation
4 Finding areas b integration The area under a graph of a function is found b subtracting the value of the integral of the function at the lower limit from its value at the upper limit. The eample below shows the method and notation used. Eample Find the area under the curve = between = and = A = ( + 6) + d 6 = lower limit upper limit The sketch shows the area required. = = = The constant of integration has been omitted because it would be in both parts it alwas disappears when the parts are subtracted, so is normall left out of the working. Because integration with limits does not give a result involving an unknown constant, it is known as definite integration = [ ] [ ] A = The shaded area is square units. Integration can be used to find the area of an shape as long as its boundaries can be written as functions of. In each case: use a graphic calculator or spreadsheet to draw a graph of the function integrate to find the area under the graph between the given values of. = between = and = = between = and = = + between = and = = 6 between = and = 6 = + between = and = 6 = + between = and = 7 = + between = and = 8 = between = and = = between = and = = between = and = The Nuffield Foundation
5 More about velocittime graphs In velocittime graphs the area gives the distance travelled. The following eamples rework some of the previous eamples using integration. Eample Find the distance travelled when a car travels at a constant velocit of 7 mph for hours. s = 7 dt = [ 7t ] = [ 7 ] [ 7 ] = The car travels miles. Eample Find the distance travelled when a car accelerates steadil to a velocit of ards per second in seconds. In this case v is used instead of and t instead of. The distance is s. 7 s (intercept =, gradient =.) v ards/s v = 7 t hours s =.t.t dt =.t.. = = [ ] [ ] = [ ] s v =.t The car travels ards. Eample The velocit of a car as it travels between two sets of traffic lights is modelled b v =.t( t) where v is the velocit in metres per second and t is the time in seconds. Find the distance travelled. v ms  v =.t( t) s = ( t.t ) dt t =.t. = The car travels 7 metres. [ ] = 7 Compare the answers from these eamples with those given earlier. Use integration to check our answers to the questions at the bottom of page. 6 The Nuffield Foundation
6 The functions given below can be used to model the velocit v metres per second at time t seconds for the eamples given in questions to at the bottom of page. In each case: use a graphic calculator or spreadsheet to draw the graph of the model and compare the result with the graph ou drew earlier use integration to estimate the distance travelled over the given time interval and compare the result with our earlier answer. v = t between t = and t = 8 v = + t.t between t = and t = v = t t + between t = and t = v = 6 t between t = and t = Each of the following functions models the velocit v metres per second in terms of the time. In each case: use a graphic calculator or spreadsheet to draw the graph of the model briefl describe how the velocit varies over the given time interval integrate to find the distance travelled between the given values of t. v =.t between t = and t = 8 6 v = 8 t between t = and t = 7 v = t t + between t = and t = 8 v =. t between t = and t = 6 The Nuffield Foundation 6
7 Teacher Notes Unit Advanced Level, Modelling with calculus Skills used in this activit: finding areas under graphs estimating areas using triangles and trapezia integrating polnomial functions Preparation Students need to know how to use a graphc calculator or spreadsheet to draw graphs from a set of data values or given functions and how to find the area of rectangles, triangles and trapezia. Notes on Activit This activit uses areas under velocittime graphs to introduce integration. The main points are also included in the Powerpoint presentation of the same name. If our students have calculators that can integrate, this method could be used to check their answers. The answers to the questions, including sketches of the graphs, are given on the following two pages. The Nuffield Foundation 7
8 Answers Page miles ards 6 6 t hours ards ards Page v ms . 8 Accelerates from ms  to. ms  in 8 seconds 7. m v ms  Decelerates from ms  to ms  in seconds 8 m v ms  v ms  Accelerates from ms  to ms  in seconds 8. m Decelerates from ms  to ms  in seconds 7. m Page = = = The Nuffield Foundation 8
9 = + = + = = + 8 = = 8 =. Page 6 7. m (sf) 8 m 8 m 7 m (Graphs ver similar to sketches given for Page.) v ms  v =.t 6.6 Accelerates from ms  to.6 ms  in 8 seconds 8 v ms  v = 8 t Decelerates from 8 ms  to ms  in seconds.6 m m 7 v ms  Decelerates from 8 ms  to ms  v = t t + in seconds, then accelerates to ms  in the net seconds m v ms  v =. t Decelerates from ms  to ms  in 6 seconds m 6 The Nuffield Foundation
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