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4 Pages

### proj2

Course: MA 114, Fall 2008
School: Kentucky
Rating:

Word Count: 1086

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Lab # Project II # Two identical jars of boiling water, labeled A and B, stand beside # two identical jars of ice water , labeled C and D, in a room whose # temperature remains constant. The contents of jars A and C are # immediately mixed. Jars B and D are kept separate for a few minutes # and then mixed. How will the temperatures of the two mixtures # compare? One school of thought says that the AC mixture...

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Lab # Project II # Two identical jars of boiling water, labeled A and B, stand beside # two identical jars of ice water , labeled C and D, in a room whose # temperature remains constant. The contents of jars A and C are # immediately mixed. Jars B and D are kept separate for a few minutes # and then mixed. How will the temperatures of the two mixtures # compare? One school of thought says that the AC mixture will be # hotter because jar D, being very hot will lose its heat much more # rapidly than the cooler AC mixture and the result will be that more # heat is lost by B alone than by the AC misture. Another says that the # hot water from A and the cold water from C will act independently, # just like the separate portions in B and D and therefore the mistures # will be the same temperature. # # Run a number of (theoretical) experiments, varying the temperatures # of the room and water, the quantities of hot and cold water, and the # times. Prepare a report on the experiments, complete with an # explanation of the calculations (the theory) , a separate section of # results and a third of interpretation of the results relative to the # original question. In a final section provide a theoretical # explanation of the results of the experiments. # # Alternative Formulation # # Ms. Karen Heavin suggests the following alternative statement for the # lab. You may report on the original, this, or your own version. # # There were once three friends, Carl, Paul, and John. In an attempt # to survive the last two weeks of the semester they had begun a nightly # ritual of studying into the wee hours of the morning, replacing sleep # with coffee. During one of their sessions, after consuming about 15 # cups of coffee each, Carl decided to go run around the block to # control the shaking of his limbs from too much coffee. Just as he # was about to get up, John inadvertently poured Carl a full cup of # fresh coffee. Carl said not to worry about it he would drink it when # he got back. John then offered to put cream in the coffee so it would # be "just right" when Carl got back from his run. Carl objected # strenuously to this stating that the creme would cause his coffee to # be too cold when he got back. Upon hearing this Paul assured Carl his # coffee would not be any colder than if he left it alone and poured the # cream in when he got back. John felt Paul was in error and told him # so. # # Anyway, Carl started to prepare for his run but fell asleep as he # sat on his bed to put on his running shoes. John and Paul argued all # night over the timing the cream pouring (rather than studying # calculus). # # The next morning Carl found his friends still arguing. However, # after a nights sleep he could think straight. He drank the cold cup # of coffee, then quickly put the argument to rest by modeling the # problem with Newton's law of cooling (or whatever), doing some # exprimental calculations, looking at some graphs and then explaining # the results. Then Carl went off to school while John and Paul fell # asleep. # # # # GETTING STARTED: One will need to assume that the constant, k , in # Newton's law of cooling is the same for all aspects of the problem. In # addition one will need to assume that the temperature of a mixture of # two containers of water will be the weighted average of the # temperatures of the two components. # # The following takes one through a related problem and contains all of # the Maple commands and ideas one might need. # # Problem : Two of cups soup, the first at 90 C and the second at 100 C # are put in a room where the temp is maintained at 20 C. The first cup # cooled from 90 C to 60 C after 10 minutes, at which time the second # cup is put in a freezer at -5 C. Assume Newton's law of cooling and # answer the following questions. # # (a) Express the temperature of each cup as a function of time, and # draw their graphs. # # (b) How much longer will it take the two cups to reach the same # temperature? # # Solution: Assuming Newton's Law of cooling, the rate at which each # cup cools is proportional to the difference between the ambient # temperature and the temperature of the soup in the cup. The # differential equation for each cup looks like this: > restart; > Diff(f(t),t) = k*(f(t)-20); d -- f(t) = k (f(t) - 20) dt # This is a separable differential equation, whose solution can be # written in the # form > T := 20 + (T0-20)*exp(k*t); # T := 20 + (T0 - 20) exp(k t) # where T = T0 at t=0 and k is determined by the information given # that the first cup cools to 60 C in 10 minutes (i.e. T=60 at t=10). # # So if we denote the temperature of the first cup after t minutes by f # then > f := subs(T0=90,T); f := 20 + 70 exp(k t) > k :=solve(subs(t=10,f) =60,k); k := 1/10 ln(4/7) > f := unapply(f,t); f := t -> 20 + 70 exp(1/10 ln(4/7) t) # Assume that this value of k is also valid for the second cup. (Why # shouldn't it be? They are both soup.) So the second cup's temperature # is given by # > g := subs(T0=100,T); g := 20 + 80 exp(1/10 ln(4/7) t) > g := unapply(g,t); g := t -> 20 + 80 exp(1/10 ln(4/7) t) # # for the first 10 minutes. From that time on, the second cup's # temperature is given by # > h := t -> -5 + (itemp - (-5))*exp(k*t); h := t -> -5 + (itemp + 5) exp(k t) > # # where itemp is the temperature that the 2nd cup would start at in # ord...

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Kentucky - MA - 114
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Kentucky - MA - 114
# # Center of mass of a solid of revolution# # If y *`=` *f(x) &gt;= 0 for a &lt;= x *`&lt;=` b, then let S be the solid of# revolution obtained by rotating# the region under the graph of f around the x axis. We know how to# express the volume of S a
Kentucky - MA - 114
# Arclength # # Thm. If f' is continuous on [a,b], then the graph# of f over the interval has length &gt; len = Int(sqrt(1+(diff(f(x),x)^2),x=a.b); b /
Kentucky - MA - 114
# Homework from Ostobee Zorn# # 8.3 Arclength# # #2 Verify that the arclength formula gives the correct# answer for the length of the graph of y = mx +b from# x =0 to x = 1.# # Solution: The correct answer is the distance from (0,b)# to
Kentucky - MA - 114
# Present Value # # In the continuous compound interest formula, # A = P*e^(rt),# P is referred to as the 'present value' of A t years from# now if money grows at 100r percent yearly with continuous# compounding of interest. So for exampl
Kentucky - MA - 114
# Ostobee Zorn Chapter 9# # 9.1 #7 The hint is not needed.# &gt; with(student);Warning, new definition for D[D, Diff, Doubleint, Int, Limit, Lineint, Product, Sum, Tripleint, changevar, combine, completesquare, distance, equate, extrema
Kentucky - MA - 114
-# Notes on project 2:# # This problem involves ideas like those needed in project 2.# # Given a right triangle with legs a and b. Inscribe a circle c1, then# heading off towards one of the acute angles, inscribe an infinite # chain of ci
Kentucky - MA - 114
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Kentucky - MA - 114
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Kentucky - MA - 114
%!PS-Adobe-2.0 %Creator: dvipsk 5.58f Copyright 1986, 1994 Radical Eye Software %Title: ws3.dvi %Pages: 2 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %EndComments %DVIPSCommandLine: dvips ws3 %DVIPSParameters: dpi=300, comments removed %DVIPSSource:
Kentucky - MA - 114
# Ostobee Zorn vol 2&gt; # P. 46, no 3. # # This problem asks for the left, tight, and midpint regular estimates of an intgral, which the book abbreviates# as # &gt; L[n] = Sum(f(x[i])*Delta[n],i=0.n-1);M[n] = Sum(f(x[i]+x[i+1])/2)*Delta[n],i
Kentucky - MA - 114
&gt; # NOTES ON MA114 HOMEWORK 5.1-5.3# # # # p.12, no 7# # The car's velocity at time t is 60-20*t. # &gt; Int(60-20*t,t=0.4);\&gt; int(60-20*t,t=0.4);\# Since this number is positive and east is the positive direction the# # net effect of th
Kentucky - MA - 114
# Calculus II - notes 9/11# # Ostobee Zorn# # 11. page 47 # Evaluate limit 2/n *sum(2*j/n)^3,j=1.n) as n goes to infinity.&gt; s := n -&gt; 2/n *sum(2*j/n)^3,j=1.n); &gt; n - /
Kentucky - MA - 114
%!PS-Adobe-2.0 %Creator: dvipsk 5.58f Copyright 1986, 1994 Radical Eye Software %Title: hw3.dvi %Pages: 6 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %EndComments %DVIPSCommandLine: dvips hw3 %DVIPSParameters: dpi=300, comments removed %DVIPSSource:
Kentucky - MA - 114
# 7.1 &gt; with(student);[D, Diff, Doubleint, Int, Limit, Lineint, Product, Sum, Tripleint, changevar, combine, completesquare, distance, equate, extrema, integrand, intercept, intparts, isolate, leftbox, leftsum, makeproc, maximiz
Kentucky - MA - 114
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Kentucky - MA - 114
# Practice Skills Test 1# Instructions: This is a practice exam for the first 'skills' test# to be given on Tuesday, September# 23. You can use any calculator up to, but not including, a TI-92. # You must show your# work. # \medskip#
Kentucky - MA - 114
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Kentucky - MA - 114
%!PS-Adobe-2.0 %Creator: dvipsk 5.58f Copyright 1986, 1994 Radical Eye Software %Title: pt2.dvi %Pages: 1 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %EndComments %DVIPSCommandLine: dvips pt2 %DVIPSParameters: dpi=300, comments removed %DVIPSSource:
Kentucky - MA - 114
%!PS-Adobe-2.0 %Creator: dvipsk 5.58f Copyright 1986, 1994 Radical Eye Software %Title: pt3.dvi %Pages: 2 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %EndComments %DVIPSCommandLine: dvips pt3 %DVIPSParameters: dpi=300, comments removed %DVIPSSource:
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