Working with calculus Assignment Essay Example for Free

Working with calculus Assignment EssayThe nightmare has make to pass. All of Kelleys extensive surgeries and nasal passage scrapings have (unfortunately) gone awry, and he waits in the Ear, Nose, and Throat doctors function waiting area spewing bloody snot into a conical paper cup at the tempo of 4 in3/min. The cup is being held with the vertex down (all the better to pool the snot in, my dear). The booger catcher has a cover of 5 inches and a base of 3 inches. How fast is the mucous level emanation in the cup when the snot is three inches deep? Investigating the problemThe volume of a chamfer V = where r is the gas constant of the cone and he is its height For the full cone or some(prenominal) part of it, the ratio of rh remains fixed, so As we are only interest in the rate of swap of the height we need to eliminate r so use r = 3h/10 for all levels So the new V = so to find h3 = and h = So devising a table to find for t= 0 to 25 and hence work out virtually how long t he cone takes to fill up, and the height value at each stage and also radius each time. As can be seen, the full height and radius is reached at about t 15 minutes. Lets hope the doctor is on time todayHere are the formulae used to hark back the tableHere is the interpret of h and r against time Both h and r increase rapidly in the 1st 5 minutes before the rate of increase slows as t increases. Using quantitative methods Various rates of smorgasbord could be investigated, including the rate of swap of h with appraise to V, the rate of change of r with respect to t and so on. However, the question asks about the rate of change of h with respect to t, so this will be investigated using the Leibnitz formula to estimate gradients using a spreadsheet.The following graph was obtained As can be seen this graph of the rate of change of height (the speed at which height changes) is not actually helpful, as there is a lot of change for t = 0 to t = 2 but after that the rate of c hange is much less. Some investigation shows that most of the change takes place between t = 0 and t = 1. So tracing the rate of change of the 2 sections on different graphs, with the one involving the first section in much more detail, will prepare a better picture. The table And the graph The reduction in speed of the heights rise is very markedThe table for t = 1 to t = 14 and the graph The question requires the rate of change at h = 3. From the table this can be see between t = 2 and t = 4, where the gradient is between 0. 46 and 0. 29 inches per minute Using differentiation V = so and we were also told So using the Chain Rule = Filling what is known 4 = so So when h = 3 = 0. 393 inches per minute Conclusion The numerical method does not give a very undefiled result and provided the Chain rule is used, the calculus method is much better