|Strips||Strip Width||Area (LB)|
Click/tap in the right/left halves of the graph area to decrease/increase the number of strips
Suppose we wish to find the displacement of a body moving along a straight line with an increasing velocity. We can not use displacement = (velocity) (time) since the velocity is varying . That equation would have worked if the velocity was constant. But changes over time can be treated as a constant for a small duration. After all, in a small duration the quantity could not have varied much. ( Consider this: could you treat the temperature of an open room constant for a day, an hour, a minute, a sec, a millisec) We could always lead ourselves to a duration where the variation seems small enough to elicit a disdainful comment " neglect it!". We could then use displacement = velocity x time without feeling bad about it. We could then move on to the next small duration, find the velocity at the beginning of the duration and treat that as a constant for that small duration. This way we have beaten the variation.
Sounds cumbersome- but it is not. Not at least in case when velocity varies in an organized manner. Let us assume a very simple variation of velocity with time and work this out. The velocity of a body varies from 1 m/s to 2 m/s in 1 s linearly (The v-t relation here is v=1+1.0*t). What is its displacement?
We would go about it this way. Divide the duration of 1 s in to ten intervals of 0.1s each. Since the velocity varies linearly, the velocities at the beginning of the intervals are 1, 1.1, 1.2,.....,1.9 m/s. We will now use our old trick. Treat the velocity at the beginning of the duration as a constant for the entire duration, and while dealing with next duration switch to the value of the velocity at the beginning of that duration and so on. So the displacement in the first 0.1 s duration is 1 x 0.1 = 0.1 m, the second 0.1 s duration is 1.1 x 0.1=0.11 m, and the third 1.2 x 0.1=0.12 m, so on till for the tenth 0.1 s duration it is 1.9 x 0.1= 0.19 m. Adding up all these would give a decent approximation to the displacement. 0.1+0.11+0.12+.........+0.19 = 1.45 m. This is less than the needed displacement by a certain amount, because we took for each interval the value of the velocity at the beginning of the interval which was lower than the velocity at the end of the interval. But we at least have some value for the displacement.
If we use the velocity at the end of the duration as constant for the entire duration. we would get the velocity for the ten durations of 0.1 s each as 1.1,1.2,1.3,.....,2 m/s. Summing the corresponding displacements of 0.11+0.12+..........+0.2 would yield 1.55m, which is more than the needed displacement, because the velocity at the end of the interval is greater than the velocity during the interval.
If you are not happy with it - try splitting the duration of 1 s in to 100 intervals and use the same trick. The velocity at the beginning of the first 0.01 s duration is 1 m/s, for the second it is 1.01 m/s and so on. The corresponding displacements are 1 x 0.01= 0.01 m, 1.01 x 0.01 = 0.0101 m, ........... .......1.99 x 0.01 = 0.0199 m. Summing up 0.01+0.0101+............+0.0199 = 1.495 m. This is definitely a better approximation, but a little lower. Choosing the value of the velocity at the end of the durations as the velocity for the duration, the corresponding values are 0.0101+0.0102+............+0.02 = 1.505 m and and this is a little higher. You can see the gap between the higher and lower estimates for displacements getting smaller.
We could proceed this way, using smaller and smaller durations to make our approximation of what varies a little is almost a constant work better and better. We would get a value a little more than what we need or a little less than what we need. This gap can be reduced by increasing the number of intervals.
Suppose now we draw a graph of the velocity versus time for the body. Since the area under velocity time graph is the displacement our problem would be to find this area. The area under the curve between given limits gives us the displacement for the duration between the limits. To find this area we split up the area in to rectangular strips- we have no formula for the area of irregular shapes- and find the area of the rectangular strips instead. This would leave out some of the irregular portions of the area. But we will take care of that.
You should be able to see the connection - Seeing the area as the sum of areas of rectangles using a rectangle is same as assuming the velocity as constant for the duration represented by the width of the rectangle.
If we divided the area under the v-t diagram in to 10 strips, that would correspond to dividing the motion in to ten intervals. The height of the strip representing the velocity at the beginning of the duration and the width of the strip the duration of time. The area of the strip represents the displacement during that interval and we would get a lower estimate to the displacement. We could also treat the height at the end of the duration as constant for the entire duration. We would then get the upper estimate. Increasing the number of strips will give us better and better approximations to lower and upper estimates and the gap between them decreases.
In the applet we are seeking the area under the curve (y = x*x) or st. line (y = x) from x= 1 to x=3. We are dividing this interval between x=1 and x = 3 in to 2,4,8,16,..... parts. The orange part is either the area left out of the calculations or included in them and so represents the inaccuracy. By increasing the number of steps, the orange colored area becomes smaller and smaller. We can not here draw a rectangle of less than one pixel wide, so the maximum number of rectangles is limited to 128. But you could imagine increasing the number to get a more accurate value for the area. By increasing the number of rectangles indefinitely, we could reduce the inaccuracy to limits that are acceptable in evaluating a given quantity.
If this sounds unscientific- reflect on this. When you say the value a physical quantity is this, you are only making an approximation to the value. After all, when you say m = 1 kg, you are saying that a given body's mass can be taken as 1 kg within some limits of error. We do not really mean a mathematical equality of any kind. When we talk of a real physical body taking m as equal to 1 kg is meaningless- The accuracy of the value is limited by the fact that all measuring instruments can only give you approximations, the degree of accuracy of which is limited by the sensitivity of the instrument. In that sense all measurements are approximations. Also it should be very reassuring to know that we do not really need the value of a physical quantity to some infinite degree of accuracy. We can merrily go along with these approximations. None of the calculations we make, be it those involved in sending a probe to a distant planet, or those in sending a beam of radiation responsible for the illumination of a particular point on a TV screen - you would end up achieving what you desire, through the approximations made to sufficient degree of accuracy and the instruments designed to implement those approximations.
For the straight line the sequence of terms for 2,4,6,...,128 strips are
2 strips of width 1 unit lower bound 1*1+2*1=3.0 upper bound 2*1+3*1=5.0
4 strips of width 0.5 unit lower bound 1*0.5+1.5*0.5+2*0.5+2.5*0.5 = (1+1.5+2.0+2.5)*0.5=7*0.5 = 3.5 upper bound 1.5*0.5+2*0.5+2.5*0.5+3*0.5 = (1.5+2.0+2.5+3)*0.5=9*0.5 = 4.5
8 strips of width 0.25 unit lower bound 1*0.25+1.25*0.25+............+2.75*0.25 = (1+1.25+....+2.75)*0.25=3.75 upper bound 1.25*0.25+1.5*0.25...............+3*0.25 = (1.25+1.5+....+3)*0.25 = 4.25
16 strips of width 0.125 unit lower bound 1*0.125+1.125*0.125+..........+2.875*0.125 = (1+1.125+......+2.875)*0.125 = 3.875 upper bound 1.125*0.125+1.25*0.125+............+3*0.125 = (1.125+......+3)*0.125 = 4.125
64 strips of width 0.03125 unit lower bound (1+1.03125+............+2.96875)*0.03125 = 3.96875 upper bound (1.03125+1.0325+..........+3)*0.03125 = 4.03125
128 strips of width 0.015625 unit lower bound (1+1.015625+............ +2.984375)*0.015625 = 3.984375 upper bound (1.015625+............ +3)*0.015625 = 4.015625