Modelling a Suspension Bridge

KAPS1.A piece of line was tied to 2 chairs so that the stick to hangs down fairly loose. 2.Both chairs had the aforementioned(prenominal) superlative degree withdraw the show, and the rush was do sure that they were tied to the same position. (88cm)3.As it washstand be square upn on the remand below, the line of longitude from the floor to the situate was heedful from eleven impactly distant flowers along the floor. Points (cm)Height from the visor to the mountain range (cm)0884070.5805712048.516042.520040.524042.528047.532056.536069.5400884.A represent was drawn utilise the supports as the y-axis and the floor as the x-axis. (Attached on the end of assignment criteria sheet provided)5.From looking at the spread out biz, the function looks like a quadratic polynomial function. Therefore, I will subscribe to that y = ax2 + bx + c. ?The constant quantity c is the survey of y when x=0? [http://www.ucl.ac.uk/mathematics/geomath/rev/quadnb/ql11c.html]. Theref ore,c = y-intercept = 88Now we mustiness summon out a, being the dilation. To find this out, we moldiness add up by taking the midway baksheesh of the twist till the final transmit of the curve, indeed, ap flowers 6, 7, 8, 9, 10 and 11. We will make the sixth point starting from 0cm, as if we start get through from 200, the x pry will non be constant. To make it constant, we must make the dissimilarity between the y-intercept to the sixth point and the difference from 6th point to the 7th point to be scarcely the same. ( master table below)y 40.542.547.556.569.588x04080120160200Now, we must square the x value to see if it forms a bang-up linear graph. (see table andy40.542.547.556.569.588x2016006400144002560040000As it terminate be seen on the graph above, the graph is in truth close to a perfectly smashing line, therefore the supposition of a quadratic homunculus was correct. Now,m = △y/△xUsing points, (0,40) and (40000, 88)m = (88 ? 40) / (40000-0 )m = 48 / 40000m = 0.0012Since a is the grad! ient, a = m= 0.0012Now that we start out the value of a and c, we could now substitute these value into the quadratic formula along with the point from the original table. (NOTE: must be from the original table, and the point must be the point which is located where the curve is button upwards, as that is where we got the value of a before.)Therefore,y = ax2 + bx + cy = 0.0012x2 + bx + 88Substitute point (320,56.5)56.5= 0.0012 x 3202 + 320b + 8856.5= 122.88 + 320b + 88320b = -154.38Therefore,b = -0.4824375b = -0.48Therefore, the traffic pattern for the data is, y = 0.0012x2 ? 0.48x + 886.The points were entered into excel to produce a graph of the data. (See graph on beside page.)7.As it can be seen above, the mathematical sham for the data collect wasy = 0.0012x2 ? 0.4739x + 87.853, with an R2 value of 0.9992, which is extraordinarily close to 1, therefore proving us distinctly that the data forms a parabolic function. 8.Using the fine art computer, a gr aph, a model and a table were produced. See pictures below. Graph produced employ nontextual matter computing machine model produced victimization graphics computer #1 (R value and the model itself) object lesson produced employ graphics information processing system #2 (Used as line of best fit)Table produced using graphics calculating machine9.aslkdjfMAPS10.Calculate the tokenish vizor of the pull back from the undercoat algebraic aloney using:Thinking logic everyy, if some(a)thing straight (in this case, the sop up) is hanged down fairly loose from 2 supports, the point which is just now in the middle - the point where the distance from the two supports equal - must be the point which has the minimum lift arrive at the floor, assuming that the string itself is non out of proportion in tummy or in shape. Therefore, the point with the concluding height bump off the floor is the 6th point, which are is located on the nose 2 metres from the chair on both sides. a ) Original banquet plot:Referring to the data utili! se to sketch the scatter plot, the low point must be the 6th point, with a minimum height of exactly 40.5cm off the free-base. Therefore, the minimum height off the appreciation is 40.5cm. To find out the minimum height off the floor for the next 3 models, we must substitute the 6th point, which has a value of 200cm as the x value. b) homunculus produced in shade 5 of KAP:y = 0.0012x2 ? 0.48x + 88 (model produced in Q5)substitute, (200, y)y = 0.0012 x 2002 ? 0.48 x 200 + 88y = 40Therefore, the minimum height off the ground is 40cm. c) Model produced using excel:y = 0.0012x2 ? 0.4739x + 87.853(model produced in Q6)substitute, (200, y)y = 0.0012 x 2002 ? 0.4739 x 200 + 87.853y = 41.073Therefore, the minimum height off the ground is 41.073cm. d) Model produced using graphics calculator:y = 0.00118043414918x2? 0.47387820512823x + 87.853146853148(model produced in Q8)substitute, (200, y)y = 0.00118043414918 x 2002 ? 0.47387820512823 x 200 +87.853146853148y = 40.29487179Therefore, y = 40.295Therefore, the minimum height off the ground is 40.295cm. As it can be seen on the 4 diverse types of model, the minimum height off the ground varies, wherefore could this be possible, when the values we started off with were all the same? And this point leads me to the next scruple. 11.Differences in ResultsScatter plot: 40.5cmModel produced in wonder 5:40cmModel produced using excel:41.

073cmModel produced using GC:40.295cmAs it can be seen above, all the different types of model vary in the results. This was out-of-pocket to the different mathematical formula we used to calculate. If we assume that t he string was straight and the string had the same wi! dth and mass throughout the whole string, the scatter plot should be the about consummate model, as the height was physically measured at the lowest point. But if the string was old, not straight and some split were torn making the width and mass different throughout, it would in spades influence the results. And that?s when we have to use other(a) methods. For example, the graphics calculator model can be used to see the pattern, and find out the lowest point, which could possibly be situated not in the middle of the two supports totally when somewhere else. In this investigation, the string used was new, and the string was do sure that it was straight and had no sign of variable, which might have influenced the results. But, if the string was not straight, a statement could be do. ?That the middle point of the scatter plot was not the lowest point.? If that was the case, using the graphics calculator model would be the most exact, as it is the most on the nose equipment t o find the pattern/function of the curve. In any case, I prefer not to use the model produced in uncertainty 5, as it was basically made by using only few of the points instead of all. Also, the model produced in excel can not be preferred as well, as all the calculation in this model was made only up to 4 decimal places, where as, the decimals for the graphics calculator had oer 10 decimal places. Summarise Q11Scatter plot: approximately precise. It is the physical criterion interpreted. (NOTE: if the condition of the string is bad, this method is not precise.)Model produced in question 5: The least precise. Since this model was made by using few points, it does not take in consideration of all points. Model produced using excel: little precise. It was only taken to 4 decimal places which only gives a ?approximate? result. Model produced using GC: Precise. It was taken to over 10 decimal places, but it is still not a perfectly defined model of the result. Bibliography:Search Engines:www.google.comWebsites:http://www.ucl.ac.uk/M! athematics/geomath/rev/quadnb/ql11c.htmlAll accessed online on 01.09.06 If you penury to get a proficient essay, order it on our website: BestEssayCheap.com

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