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Notebook[{
Cell[CellGroupData[{
Cell["Calculus: The Language of Change", "Title"],
Cell[TextData[{
StyleBox["Exercise Set 7.5(CD7.1)\n",
FontVariations->{"Underline"->True}],
"Implicitly Linked Variables"
}], "Subtitle"],
Cell["by John Robeson", "Subsubtitle"],
Cell[CellGroupData[{
Cell[TextData[StyleBox[
"Exercise 1) Blowing Up a Balloon", "Subtitle"]], "Subtitle"],
Cell[TextData[{
"A child is blowing up a balloon by adding air at the rate of 2 cubic \
inches per second. Well before it bursts, its radius is 6 inches. Assume \
that the balloon is a perfect sphere and find the surface area S as an \
explicit function of volume V. Use your explicit function to say how fast the \
surface is stretching at this point. \n\nHINTS: Give variables for surface \
area, volume, time, and whatever else you need. Solve for S in terms of V. \
The \"chain\" in this exercise is that volume depends on time (through the \
child's efforts), V=V[t] and surface area depends on volume, S=S[V]. We want \
to know how area changes with time,",
Cell[BoxData[
\(dS\/dt\)]],
" , where S=S[V[t]]. (See review Exercise CD-28.3.2 to express surface area \
as an explicit function of volume.) "
}], "Text"],
Cell[CellGroupData[{
Cell["Solution to Exercise 1)", "Section"],
Cell[TextData[{
"We assume the following and will not consider the effect of pressure.\n\n\t\
\tV =",
Cell[BoxData[
\(\(4\ \[Pi]\ r\^3\)\/3\)]],
"\n\t\t\n\t\tS =",
Cell[BoxData[
\(4\ \[Pi]\ r\^2\)]],
"\n\t\t\nWe know that ",
Cell[BoxData[
\(dV\/dt\)]],
" is 2",
Cell[BoxData[
\(\(\ in\^3\)\/sec\)]],
"and that we need a relationship between V and S (which are both dependent \
on the radius r). So we first solve both of the above equations for r.\n\n\t\
\tr = ",
Cell[BoxData[
\(\((\(3\ V\)\/\(4\ \[Pi]\))\)\^\(1/3\)\)]],
"\n\t\t\n\t\tr = ",
Cell[BoxData[
\(\@\(S\/\(4\ \[Pi]\)\)\)]]
}], "Text"],
Cell[BoxData[{
FormBox[
RowBox[{
FormBox[
\(We\ can\ now\ set\ these\ equal\ to\ each\ other\ and\ then\ solve
\ for\ S . \n\n\t\t\((\(3\ V\)\/\(4\ \[Pi]\))\)\^\(1/3\)\ = \
\@\(S\/\(4\ \[Pi]\)\)\),
"TextForm"], "\t\t"}], TextForm],
FormBox[
RowBox[{
FormBox[
\(Raising\ both\ sides\ to\ the\ 6 th\ power\ gives\
\(us : \t\t\n\t\t\n\t\t\((\(3\ V\)\/\(4\ \[Pi]\))\)\^2\) =
\((S\/\(4\ \[Pi]\))\)\^3\),
"TextForm"], "\t\t"}], TextForm],
FormBox[
FormBox[\(\t\t\(9\ V\^2\)\/\(16\ \[Pi]\^2\) = S\^3\/\(64\ \[Pi]\^3\)\),
"TextForm"], TextForm]}], "Text"],
Cell[BoxData[{
FormBox[
RowBox[{
FormBox[\(We\ now\ solve\ this\ for\ S\),
"TextForm"], "\n", "\t\t"}], TextForm],
FormBox[
RowBox[{
FormBox[\(\t\tS\^3\ = \ 36\ \[Pi]\ V\^2\),
"TextForm"], "\n", "\t\t"}], TextForm],
FormBox[
RowBox[{
FormBox[\(so\t\tS\ = \((36\ \[Pi]\ V\^2)\)\^\(1/3\)\),
"TextForm"], "\n"}], TextForm],
FormBox[
FormBox[\(Since\ V\ is\ a\ function\ of\ time, \
we\ can\ now\ find\ \(dS\/dt\) . \n\n\n\t\t\(dS\/dt\) = \
\(\(dS\/dV\) dV\/dt\ = \ \
\(\(\((36\ \[Pi]\ V\^2)\)\^\(\(-2\)/3\)\ \((72\ \[Pi]\ V)\)\)\/3\
dV\/dt\n\t\t\n\t\t\n\t\t = \
\(24\ \[Pi]\ V\)\/\((36\ \[Pi]\ V\^2)\)\^\(2/3\)\ dV\/dt\)\)\),
"TextForm"], TextForm]}], "Text"],
Cell[BoxData[
\(TextForm
\`We\ know\ that\ dV\/dt\ is\ 2 \(\(\ in\^3\)\/sec\) and\ that\ when\ r\
is\ 6\ inches, \
V\ = \ \(\(4\ \[Pi]\ 6\^3\)\/3\ = \
\(288\ \ \[Pi]\ \ cubic\ inches . \n\nSo\ \ \ \ \tdS\/dt\ = \
\(24\ \[Pi]\
\((\ 288\ \ \[Pi])\)\)\/\((36\ \[Pi]\ \((288\ \[Pi])\)\^2)
\)\^\(2/3\)\ \((2)\)\)\)\)], "Text"],
Cell[CellGroupData[{
Cell[BoxData[
\(\(\n\n\
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Cell[BoxData[
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Cell[BoxData[
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" = ",
Cell[BoxData[
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Cell[BoxData[
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" which is the same answer we obtained implicitly in Example CD-7.4."
}], "Text"]
}, Closed]]
}, Closed]],
Cell[CellGroupData[{
Cell["Exercise 2) The Explicit Fast Ladder", "Subtitle"],
Cell[TextData[{
"\nSolve the equation ",
Cell[BoxData[
\(x\^2 + y\^2 = L\^2\)]],
" for y = ",
Cell[BoxData[
\(\@\(L\^2 - x\^2\)\)]],
". Use the fact that x = x[t] is a function of time together with the \
Chain Rule to show that\n\n\ty'[t] = ",
Cell[BoxData[
\(dy\/dt\)]],
" = ",
Cell[BoxData[
\(dy\/dx\)]],
" . ",
Cell[BoxData[
\(dx\/dt\)]],
" = ",
Cell[BoxData[
\(\(-\(x[t]\/\@\(L\^2 - \((x[t])\)\^2\)\)\)\)]],
" . x'[t]\n\t\nVerify that this model predicts that as x approaches L, the \
speed of the tip resting on the wall tends to infinity. \n\nThe result of the \
previous exercise is wrong for a real ladder. The tip of such a ladder cannot \
go faster than the speed of light. There is a physical condition that this \
simple mathematical model does not take into account. The real falling ladder \
is explored in the Scientific Projects Chapter on Mechanics. It uses \
Galileo's Law of Gravity from Chapter 9."
}], "Text"],
Cell[CellGroupData[{
Cell["Solution to Exercise 2)", "Section"],
Cell[TextData[{
"See Example CD-7.5, The Fast Ladder, for the implicit solution.\n\nHere we \
will solve for y explicitly.\n\n\t\t",
Cell[BoxData[
\(x\^2 + y\^2 = L\^2\)]],
"\n\t\t\n\t\t",
Cell[BoxData[
\(y\^2 = L\^2\)]],
" - ",
Cell[BoxData[
\(x\^2\)]],
"\n\t\t\n\t\ty = ",
Cell[BoxData[
\(\@\(L\^2 - x\^2\)\)]],
"\n\t\t\nThe length of the ladder L is constant, but x and y vary with \
time. So\n\n\t\t",
Cell[BoxData[
\(dy\/dt\)]],
" = ",
Cell[BoxData[
\(dy\/dx\)]],
". ",
Cell[BoxData[
\(dx\/dt\)]],
" = ",
Cell[BoxData[
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" ",
Cell[BoxData[
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Cell[BoxData[
\(\(-x\)\/\@\(L\^2 - x\^2\)\)]],
" ",
Cell[BoxData[
\(dx\/dt\)]],
" = ",
Cell[BoxData[
\(\(-x\)\/y\)]],
" ",
Cell[BoxData[
\(dx\/dt\)]],
" \n\t\t\nIn this problem we consider ",
Cell[BoxData[
\(dx\/dt\)]],
" as a given constant. So we \nnow find the limit of ",
Cell[BoxData[
\(dy\/dt\)]],
" (which is the speed of the tip resting on the wall) as x goes to L (which \
is as y goes to zero.)"
}], "Text"],
Cell[CellGroupData[{
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Cell[CellGroupData[{
Cell["Exercise 3) Related Rates Drill ", "Subtitle"],
Cell[TextData[{
"\n1. Each edge of a cube is expanding at the rate of 3 inches per second. \
How fast is the volume expanding at the point where the volume equals 64 \
cubic inches? \n2. Each edge of a cube is expanding at the rate of 3 inches \
per second. How fast is the surface area expanding at the point where the \
volume equals 64 cubic inches? \n3. A 6-foot man walks away from an 8-foot \
lamp at the rate of 5 feet per second. How fast is his shadow growing at the \
point where he is 7 feet from the lamp? \n4. A snowball melts at a rate \
proportional to its surface area, that is, it loses volume per unit time in \
this proportion. Say the constant of proportionality is k. What is ",
Cell[BoxData[
\(dr\/dt\)]],
" , the rate of change of radius with respect to time? \n5. Your snow-cone \
has melted filling the conical holder with sticky liquid. The liquid is \
dripping from the point at the bottom at the rate of 1 liter per hour. If the \
cone is 7 centimeters high and 5 centimeters in diameter at the top, what is \
the rate of change of the height of the liquid when half of it has leaked \
out? (Note: The volume of a cone of height h and base radius r is V = ",
Cell[BoxData[
\(1\/3\ \[Pi]\ r\^2\ h\)]],
".) "
}], "Text"],
Cell[CellGroupData[{
Cell["Solution to Exercise 3)", "Section"],
Cell[CellGroupData[{
Cell["Solution to 3.1)", "Subsection"],
Cell[TextData[{
"The volume of the cube is\n\t\tV = ",
Cell[BoxData[
\(x\^3\)]],
"\nwhere x is the length of a side. We know x is a function of t and that\n\
\t\t",
Cell[BoxData[
\(dV\/dt = dV\/dx\)]],
". ",
Cell[BoxData[
\(dx\/dt\)]],
"\n\n\t\t\n\t\t",
Cell[BoxData[
\(dV\/dx = 3\ x\^2\)]],
" and ",
Cell[BoxData[
\(dx\/dt\)]],
" = 3 in/sec\n\t\t\nso\t\t",
Cell[BoxData[
\(\(dV\/dt = \)\)]],
" (",
Cell[BoxData[
\(3\ x\^2\)]],
" ) 3 = ",
Cell[BoxData[
\(9\ x\^2\)]],
"\n\nWhen V = 64, x = 4 so at this point,\n\n\t\t",
Cell[BoxData[
\(dV\/dt\)]],
" = ",
Cell[BoxData[
\(\((9)\)\ 4\^2\)]],
"= 144 ",
Cell[BoxData[
\(in\^3\/sec\)]],
"\n\t\t"
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["Solution to 3.2)", "Subsection"],
Cell[TextData[{
"The surface area of the cube is\n\t\tS = 6 ",
Cell[BoxData[
\(x\^2\)]],
"\nwhere x is the length of a side. We know x is a function of t and that\n\
\t\t",
Cell[BoxData[
\(dS\/dt = dS\/dx\)]],
". ",
Cell[BoxData[
\(dx\/dt\)]],
"\n\n\t\t\n\t\t",
Cell[BoxData[
\(dS\/dx = 12\ x\)]],
" and ",
Cell[BoxData[
\(dx\/dt\)]],
" = 3 in/sec\n\t\t\nso\t\t",
Cell[BoxData[
\(\(dS\/dt = \)\)]],
" (",
Cell[BoxData[
\(12\ x\)]],
" ) 3 = ",
Cell[BoxData[
\(36\ x\)]],
"\n\nWhen V = 64, x = 4 so at this point,\n\n\t\t",
Cell[BoxData[
\(dS\/dt\)]],
" = ",
Cell[BoxData[
\(\((36)\)\ 4\)]],
"= 144 ",
Cell[BoxData[
\(in\^2\/sec\)]],
"\n\t\t"
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["Solution to 3.3)", "Subsection"],
Cell[TextData[{
"To solve this problem we use two similar right triangles. The larger one \
has a vertical side of 8 feet and a horizontal side of x + s, where x is the \
man's horizontal distance from the lamp and s is his shadow's length. The \
smaller right triangle has a vertical side of 6 feet and a horizontal side of \
s. For these two similar triangles we know that the ratio of their sides is \
equal.\n\n\t\t",
Cell[BoxData[
\(8\/\(x + s\) = 6\/s\)]],
"\n\t\t\nSolving for s we get\n\n\t\t8s = 6x + 6s\n\t\t2s = 6x\n\t\ts = 3x\n\
We also know that\n\n\t\t",
Cell[BoxData[
\(ds\/dt = ds\/dx\)]],
" . ",
Cell[BoxData[
\(dx\/dt\)]],
"\n\t\t\n\t\t",
Cell[BoxData[
\(ds\/dx = 3\)]],
" and ",
Cell[BoxData[
\(dx\/dt\)]],
" = 5\n\t\t\n\t\t\nso\t\t",
Cell[BoxData[
\(ds\/dt\)]],
" = (3) (5) = 15 ",
Cell[BoxData[
\(ft\/sec\)]],
"\n\n\nNote that both ",
Cell[BoxData[
\(ds\/dx\)]],
" and ",
Cell[BoxData[
\(dx\/dt\)]],
" are constant, so ",
Cell[BoxData[
\(ds\/dt\)]],
" is 15 ",
Cell[BoxData[
\(ft\/sec\)]],
" no matter what the man's distance is from the lamp. "
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["Solution to 3.4)", "Subsection"],
Cell[TextData[{
"We assume our snowball is a perfect sphere so\n\n\t\tV =",
Cell[BoxData[
\(\(4\ \[Pi]\ r\^3\)\/3\)]],
"\n\t\t\nand\t\tS =",
Cell[BoxData[
\(4\ \[Pi]\ r\^2\)]],
"\n\nThe information given tells us that\n\n\t\t",
Cell[BoxData[
\(dV\/dt = \(-k\)\ S\)]],
"\n\t\t\n\t\t\nWe know\t",
Cell[BoxData[
\(dr\/dt = dr\/dV\)]],
" . ",
Cell[BoxData[
\(dV\/dt\)]]
}], "Text"],
Cell[TextData[{
"Since\t\tV =",
Cell[BoxData[
\(\(4\ \[Pi]\ r\^3\)\/3\)]],
"\n\nImplicitly\n\t\tdV = ",
Cell[BoxData[
\(\(4\ \[Pi]\)\/3\)]],
"(",
Cell[BoxData[
\(3\ r\^2\)]],
") dr\n\t\t\n\t\t\nSo\t\t",
Cell[BoxData[
\(dr\/dV\)]],
" = ",
Cell[BoxData[
\(1\/\(4\ \[Pi]\ r\^2\)\)]],
"\n\nTherefore\t",
Cell[BoxData[
\(dr\/dt = dr\/dV\)]],
" . ",
Cell[BoxData[
\(dV\/dt\)]],
"\n\n\t\t= (",
Cell[BoxData[
\(1\/\(4\ \[Pi]\ r\^2\)\)]],
") ( -k S)\n\t\t\nBut\t\tS = ",
Cell[BoxData[
\(4\ \[Pi]\ r\^2\)]],
"\n\nSo\t\t",
Cell[BoxData[
\(dr\/dt\)]],
" = -k"
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["Solution to 3.5)", "Subsection"],
Cell[TextData[{
"We again use similar triangles in solving this problem. Our cone is 7 cm \
high and has a radius of 2.5 cm (half of the 5 cm diameter). As the liquid \
leaks out, the ratio of liquid height to liquid radius remains",
Cell[BoxData[
\(7\/2.5\)]],
"= 2.8\n\nSo\t\th = 2.8 r\nor\t\tr =",
Cell[BoxData[
\(1\/2.8\)]],
" h\n\nImplicitly\tdr = ",
Cell[BoxData[
\(1\/2.8\)]],
" dh"
}], "Text"],
Cell[TextData[{
"We also know that the volume of the cone is\n\n\t\tV = ",
Cell[BoxData[
\(1\/3\ \[Pi]\ r\^2\ h\)]],
"\n\t\t\nImplicitly\tdV = ",
Cell[BoxData[
\(\(1\/3\ \[Pi]\ \)\)]],
" ( ",
Cell[BoxData[
\(r\^2\ dh + h\ 2\ r\ dr\)]],
")\n\nSince\t\tdr = ",
Cell[BoxData[
\(1\/2.8\)]],
" dh\n\n\nwe have\tdV = ",
Cell[BoxData[
\(\(1\/3\ \[Pi]\ \)\)]],
" ( ",
Cell[BoxData[
\(r\^2\ dh + h\ 2\ r\ \ 1\/2.8\ dh\)]],
")\n\n\nor\t\t",
Cell[BoxData[
\(dV\/dt\)]],
" = ",
Cell[BoxData[
\(\(1\/3\ \[Pi]\ \)\)]],
" ( ",
Cell[BoxData[
\(\(r\^2\ + h\ 2\ r\ \ 1\/2.8\ \)\)]],
") ",
Cell[BoxData[
\(dh\/dt\)]],
"\n\n\n\t\t= ",
Cell[BoxData[
\(1\/3\ \[Pi] (\)]],
" ",
Cell[BoxData[
\(\((h\/2.8)\)\^2 + \(2\ h\^2\)\/2.8\^2\)]],
" )",
Cell[BoxData[
\(dh\/dt\)]],
"\n\n\n\t\t= \[Pi] ",
Cell[BoxData[
\(\((h\/2.8)\)\^2\)]],
" ",
Cell[BoxData[
\(dh\/dt\)]],
"\n\n\nso\t\t ",
Cell[BoxData[
\(dh\/dt\)]],
" = ",
Cell[BoxData[
\(2.8\^2\/\(\[Pi]\ h\^2\)\)]],
" ",
Cell[BoxData[
\(dV\/dt\)]],
"\n\nWe are given that dV/dt = 1 ",
Cell[BoxData[
\(liter\/hour\)]],
" = 1000 ",
Cell[BoxData[
\(cm\^3\/hour\)]],
"\n\n\nTherefore\t",
Cell[BoxData[
\(dh\/dt\)]],
" = ",
Cell[BoxData[
\(2.8\^2\/\(\[Pi]\ h\^2\)\)]],
" 1000\n\n\n\t\t",
Cell[BoxData[
\(dh\/dt\)]],
" = ",
Cell[BoxData[
\(7840\/\(\[Pi]\ h\^2\)\)]],
"\n\nNow we need to determine what h is when half of the liquid has leaked \
out. The total volume of the cone is\n\n\t\t",
Cell[BoxData[
\(TraditionalForm\`\(V\_c\ \)\)]],
" = ",
Cell[BoxData[
\(1\/3\ \[Pi]\ r\^2\ h\)]],
" = ",
Cell[BoxData[
\(1\/3\ \[Pi]\ 2.5\^2\ 7\)]],
" = ",
Cell[BoxData[
\(\(43.75\ \[Pi]\)\/3\)]]
}], "Text"],
Cell[TextData[{
"Half of that is ",
Cell[BoxData[
\(\(43.75\ \[Pi]\)\/6\)]],
Cell[BoxData[
\(cm\^3\)]],
"\n\nFor this volume of liquid in our cone\n\n\t\tV = ",
Cell[BoxData[
\(\(43.75\ \[Pi]\)\/6\)]],
" = ",
Cell[BoxData[
\(1\/3\ \[Pi]\ r\^2\ h\)]],
"\n\t\t\n\t\t\nRecall that \tr =",
Cell[BoxData[
\(1\/2.8\)]],
" h\n\n\nSo\t\tV = ",
Cell[BoxData[
\(1\/3\ \[Pi]\ \((1\/2.8\ h)\)\^2\ h\)]],
"\n\n\n\t\tV = ",
Cell[BoxData[
\(\[Pi]\/\(\((3)\)\ 2.8\^2\)\)]],
Cell[BoxData[
\(h\^3\)]],
" =",
Cell[BoxData[
\(\(\[Pi]\ h\^3\)\/23.52\)]],
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Cell[BoxData[
\(\((\(23.52\ V\)\/\[Pi])\)\^\(1/3\)\)]]
}], "Text"],
Cell[TextData[{
"So when V is ",
Cell[BoxData[
\(\(43.75\ \[Pi]\)\/6\)]],
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Cell[BoxData[
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" \n\t\t\n\t\t"
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Cell[CellGroupData[{
Cell[BoxData[
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Cell[TextData[{
"Finally we have\n\n\t\t",
Cell[BoxData[
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" = ",
Cell[BoxData[
\(7840\/\(\[Pi]\ h\^2\)\)]],
" =",
Cell[BoxData[
\(7840\/\(\[Pi]\ 5.5559\^2\)\)]],
"\n"
}], "Text"],
Cell[CellGroupData[{
Cell[BoxData[
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Cell[BoxData[
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Cell[TextData[{
"Therefore the rate of change of the height of the liquid when half of the \
liquid has leaked out 80.8458 ",
Cell[BoxData[
\(cm\/hr\)]],
". "
}], "Text"]
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End of Mathematica Notebook file.
***********************************************************************)