(*^ ::[ Information = "This is a Mathematica Notebook file. It contains ASCII text, and can be transferred by email, ftp, or other text-file transfer utility. It should be read or edited using a copy of Mathematica or MathReader. If you received this as email, use your mail application or copy/paste to save everything from the line containing (*^ down to the line containing ^*) into a plain text file. On some systems you may have to give the file a name ending with ".ma" to allow Mathematica to recognize it as a Notebook. The line below identifies what version of Mathematica created this file, but it can be opened using any other version as well."; FrontEndVersion = "NeXT Mathematica Notebook Front End Version 2.2"; NeXTStandardFontEncoding; fontset = title, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e8, 24, "Times"; ; fontset = subtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e6, 18, "Times"; ; fontset = subsubtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, italic, L1, e6, 14, "Times"; ; fontset = section, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, grayBox, M22, bold, L1, a20, 18, "Times"; ; fontset = subsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, blackBox, M19, bold, L1, a15, 14, "Times"; ; fontset = subsubsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, whiteBox, M18, bold, L1, a12, 12, "Times"; ; fontset = text, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = smalltext, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 10, "Times"; ; fontset = input, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeInput, M42, N23, bold, L1, 12, "Times"; ; fontset = output, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L-5, 12, "Courier"; ; fontset = message, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = print, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = info, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = postscript, PostScript, formatAsPostScript, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeGraphics, M7, l34, w282, h287, L1, 12, "Courier"; ; fontset = name, inactive, noPageBreakInGroup, nohscroll, preserveAspect, M7, italic, B65535, L1, 10, "Times"; ; fontset = header, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, italic, L1, 12, "Times"; ; fontset = leftheader, L0, 12; fontset = footer, inactive, nohscroll, noKeepOnOnePage, preserveAspect, center, M7, italic, L1, 12, "Times"; ; fontset = leftfooter, L0, 12; fontset = help, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = clipboard, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = completions, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12, "Courier"; ; fontset = special1, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special2, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special3, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special4, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special5, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; currentKernel; ] :[font = title; inactive; preserveAspect; startGroup] AREAFIT :[font = section; inactive; Cclosed; preserveAspect; startGroup] BRIEF ABSTRACT :[font = subsection; inactive; preserveAspect; endGroup] We seek to determine parameters in functions so that integrated areas will meet required specifications. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect] FileName: AREAFIT :[font = subsection; inactive; preserveAspect] Full title: Area Fitting :[font = subsection; inactive; preserveAspect] Last Revision Date: 27 May 1996. :[font = subsection; inactive; preserveAspect] Developer: Robert Feller, Richmond High School, Richmond IN 47374 :[font = subsection; inactive; preserveAspect] Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. :[font = subsection; inactive; preserveAspect; endGroup] Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering. :[font = section; inactive; Cclosed; preserveAspect; startGroup] STATEMENT OF PROBLEM :[font = subsection; inactive; preserveAspect] Part A Find the necessary vertical shift, h, of f(x) = -x^2 to x^2 + h so that the area bounded above by the new function and below by the x-axis which lies between the points of intersection of the function with the x-axis is equal to 2. :[font = subsection; inactive; preserveAspect] Part B Same as part A, but make the area equal to 4. Can you see the pattern to set the area equal to any positive number n? :[font = subsection; inactive; preserveAspect; endGroup] Part C The area bounded above by y = f(x) = x^3 - 4x^2 - x + 4, bounded below by the x-axis, and between the two roots x = -1 and x = 1 of f(x) = 0 is 5.3333. If the value of the constant is increased by a small amount, the area enclosed (from x = -1 and x = 1 ) by that portion of the curve will increase. If the value of the constant is decreased, the area will become smaller. By what value must the constant be changed to enclose an area of 4 units by that portion of the curve? :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsection; inactive; preserveAspect; endGroup] Integration, solving equations. :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; preserveAspect] ISSUES RELATED TO THE PROBLEM :[font = subsection; inactive; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] Integration, solving equations. :[font = subsection; inactive; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect; endGroup] This can be a discovery exercise to confirm understanding of area and use of the CAS. :[font = subsection; inactive; preserveAspect; startGroup] Expectations :[font = subsubsection; inactive; preserveAspect; endGroup] Students will be able to integrate a function with a constant and use the resulting integrated function in problem formulation. :[font = subsection; inactive; preserveAspect; startGroup] Future payoffs :[font = subsubsection; inactive; preserveAspect; endGroup] Students will be able to deal with an additional layer of abstraction with regard to use of integrals. :[font = subsection; inactive; preserveAspect; startGroup] Extensions :[font = subsubsection; inactive; preserveAspect; endGroup] Confirm that the area under the sine curve from x=0 to x=pi is 2 units. This integer value is usually an unexpected answer. This leads to similar investigations of the same family such as y = 3 sin(x) and y = sin(2x). :[font = subsection; inactive; preserveAspect; endGroup] References and Sources :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION(S) :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Part A :[font = subsubsection; inactive; preserveAspect; startGroup] We know that the roots of the function f will be ħSqrt[h]. Therefore we want the definite integral of f between the roots to equal 2. :[font = input; preserveAspect; startGroup] area[h_] =Integrate[h-x^2,{x,-Sqrt[h],Sqrt[h]}] :[font = output; output; inactive; preserveAspect; endGroup] (4*h^(3/2))/3 ;[o] 3/2 4 h ------ 3 :[font = input; preserveAspect; startGroup] Solve[area[h]==2,h] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {{h -> (3/2)^(2/3)}} ;[o] 3 2/3 {{h -> (-) }} 2 :[font = subsubsection; inactive; preserveAspect; endGroup] thus the correct vertical shift would be the cube root of 1.5 squared :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Part B Same as part A, but make the area equal to 4. Can you see the pattern to set the area equal to any positive number n? :[font = input; preserveAspect; startGroup] Solve[area[h]==4,h] :[font = output; output; inactive; preserveAspect; endGroup] {{h -> 3^(2/3)}} ;[o] 2/3 {{h -> 3 }} :[font = input; preserveAspect; startGroup] Solve[area[h]==n,h] :[font = output; output; inactive; preserveAspect; endGroup] {{h -> ((9/2)^(1/3)*n^(2/3))/2}, {h -> -((-1)^(1/3)*(9/2)^(1/3)*n^(2/3))/2}, {h -> ((-1)^(2/3)*(9/2)^(1/3)*n^(2/3))/2}} ;[o] 9 1/3 2/3 1/3 9 1/3 2/3 (-) n -((-1) (-) n ) 2 2 {{h -> -----------}, {h -> ----------------------}, 2 2 2/3 9 1/3 2/3 (-1) (-) n 2 {h -> -------------------}} 2 :[font = subsubsection; inactive; preserveAspect; endGroup] We can use the first root in the list above and confirm that this pattern works for our two examples. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Part C :[font = subsubsection; inactive; preserveAspect; startGroup] We start by looking at a plot of the original function. :[font = input; preserveAspect] h =4; :[font = input; preserveAspect] f[x_] =x^3-4*x^2-x+h; :[font = input; preserveAspect; startGroup] Plot[f[x],{x,-2,3}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] Confirm the area :[font = input; preserveAspect; startGroup] Integrate[f[x],{x,-1,1}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 16/3 ;[o] 16 -- 3 :[font = subsubsection; inactive; preserveAspect; startGroup] We compute the area with the function shifted up by k. :[font = input; preserveAspect; startGroup] area[k_] = Integrate[f[x] + k,{x,-1,1}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 16/3 + 2*k ;[o] 16 -- + 2 k 3 :[font = subsubsection; inactive; preserveAspect; startGroup] We find that the following value of k is appropriate. :[font = input; preserveAspect; startGroup] sol = Solve[area[k]==4,k] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {{k -> 3^(2/3)}} ;[o] 2/3 {{k -> 3 }} :[font = subsubsection; inactive; preserveAspect; startGroup] And we confirm this. :[font = input; preserveAspect; startGroup] Integrate[f[x] + k/.sol[[1]],{x,-1,1}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup; endGroup] 16/3 + 2*3^(2/3) ;[o] 16 2/3 -- + 2 3 3 :[font = section; inactive; Cclosed; preserveAspect; startGroup] ISSUES IN SOLUTION :[font = subsection; inactive; preserveAspect; endGroup; endGroup] Again in this problem it is the notion of a parameter being used and held in solving a multi-step problem involving integration. ^*)