Question

Questions 1-2: With the length L adjusted so that Q of water in Vessel #1 is: 1.yi(t), the differential equation describing the height The height of water in the collection chamber, which also has cross-sectional area A 1, is given by the integral Height in the collection chamber 0 Since A 1 for both vessel #1 and the collection chamber, the above equation just says the water is conserved The water clock would be ideal if and only if the height of the accumulated water in the collection vessel satisfies Hideal(t) t Using Laplace transforms, solve for Y1(s) in the transform domain, then use ilapace to find the solution y1(t) in the time domain. Here's a simultaneous plot showing Hideal(t) t, H (t) and y1(t) Water Clock with One Vessel An ideal water clock (which records 3 perfect time) is shown by the black dottedWaewile Vessel line Hideal ....Ideal Water Ciock Height in First Vessel 2.5 The blue line, shows H(t), which is the time predicted by our one-vessel water clock 2 You should produce a similar graph, but that will not be graded 1.5 For smal times, the water clock is quite accurate. Inside the yellow oval, the blue line and black dotted line are tangent at1 time t0 0.5 1.5 2.5 3 Question 1: Record your answer for H(t) 1-y,(t) here. Answer: H(t) Comment: You should check that the tangent line to your answer at the origin is indeed the line Hideal (t)t The code below assumes you have already defined the solution H(t) as a function. % Equation for the tangent line at the origin is: tangent-line = taylor (H (t), t, 0, 'Order', 2) Question 2: How well does the single-vessel water clock perform at time t1 hour? See the green dot! Hint: The error is quite large! Answer: H(1) Question 2: Record your answer for H(1) here. Give answer to at least three decimals. Comment: After one hour, our water clock is off by about one third. That's not so good! However, it works quite well for smaller units of time, for which y,() is approximately still For example, H(0.1) = 0.095 is only off by about 5%.

Answer 1

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