Project Help Notebook by Keith Stroyan

Help NoteBook Copyright 1998 by Academic Press, Inc. - All rights reserved.
Modify this Notebook to prepare your project report.

Project 7: The Expanding House

by A. Student and B. Pupil

with technical help from Da LabMonitor and Help Notebook

My Project Introduction

Write this last.  Begin with the basic question and state your main finding clearly, but with few technical details.  You might even modify the title of this section to make it more interesting.

Beginning Explorations or Computer Experiments

Coefficient of expansion

The coefficient of expansion of a substance is defined by the formula

length at higher temperature = c times (length at starting temperature) times (change in temperature)

Variables

T = temperature in degrees Celcius

l = l[T] = lenght at temperature  T  in feet

w = w[T] = width at temperature  T  in feet

h = h[T] = height at temperature  T  in feet

Linear expansion in general

Following are symbolic expressions for changes in length, width, and height in terms of length width and height at a starting temperature.  Notice that we do not use the usual Mathematica sytax  f[x_] = (expression in x), but leave the underscore out.    This makes a general experssion when we don't have a formula. You might prefer to simply type the algebra for this project and not use Mathematica to perform the symbolic computations for you.

\(Clear[v, l, w, h, T, ΔT, Δv, 
      Δl, Δw, Δh, c]; 

    Δl\  = \ c\ l[T]\ ΔT\)Δw  =  c w[T] ΔTΔh  =  c h[T] ΔTl[T + ΔT]  =  Factor[l[T]  +  Δl]w[T + ΔT]  =  Factor[w[T]  +  Δw]h[T + ΔT]  =  Factor[h[T]  +  Δh]

Volume in general

Given expressions for length, width and height, the volume is givne by their product as in the next Mathematica definition.

v[T_] := l[T] w[T] h[T]v[T]v[T + ΔT]

Sample linear expansions for wood

In the following computations we replace the general letters with specific values to find values.  Explain these in your project

\(c = 4.0\/10\^6; 
l[0] = 40\)Δl /. {ΔT -> 16.0, T -> 0}l[T + ΔT] /. {ΔT -> 16.0, T -> 0}
        l[T + ΔT] - l[T] /. {ΔT -> 16.0, 
        T -> 0}

w[0] = 30w[T + ΔT] /. {ΔT -> 16.0, T -> 0}
        w[T + ΔT] - w[T] /. {ΔT -> 16.0, 
        T -> 0}

h[0] = 8h[T + ΔT] /. {ΔT -> 16.0, T -> 0}
        h[T + ΔT] - h[T] /. {ΔT -> 16.0, 
        T -> 0}

Volume in particular

v[T] /. {T -> 0}v[T + ΔT] /. {ΔT -> 16.0, T -> 0}
        v[T + ΔT] - v[T] /. {ΔT -> 16.0, 
        T -> 0}

A Thimble, a Bucket, a Bathtub

Answer Hint 7.1 by making and explaining reasonable estimates for the volume of these everyday objects.

Hints 7.2 & 7.3

Read and answer Hint 7.2 from the print project.  You can copy, paste and modify computations above to do your computations.

Volume as a Composition

Using only the initial volume (CAPITAL)  V0, we can define another formula (CAPITAL) V (for volume) as a function of an unknown  x  by the following (standard Mathematica formula-function) command.

\(Clear[V, x, Δx, Subscript]; 
V[x_] := V\_0\ x\^3\)V[x]

Explain why this formula also gives the volume when we make  x  the multiplier for any of the linear quantities, lenght, width, or height,

l[T+ΔT] =  l[T] x
w[T+ΔT] =  w[T] x
h[T+ΔT] =  h[T] x

where  x = (1 + c ΔT)

Hint 7.4

Read the print hint and use the following to explain your answer.

The change in V for a given change in  x  is given by

V[x + Δx] - V[x]

Expand[V[x + Δx] - V[x]]

This makes the rate of change of ___

Simplify[((V[x + Δx] - V[x])/Δ)x]

Hint 7.5

Read the hint and explain the following computation

xVrate = 
      Simplify[((V[x + Δx] - V[x])/Δ)x]

When  T  goes from  0  to  ΔT,  the multiplier changes by  Δx = c ΔT, because ___

Using this, we see that

\(TraditionalForm
    \`ΔV\/Δx\ 
        Δx\/ΔT = \ ?? \)

\(x = 1; 
Δx\  = \ c\ ΔT; 

    Expand[xVrate\ Δx\/ΔT]\)

and since  Δ? is small, __ and __ are small since __, so

\(TraditionalForm
    \`ΔV\/Δx\ Δx\/ΔT 
      ≈ 3\ V\_0\ c\)

Limit[
      3 c (Subscript[V, 0]) + 3 (c^2) ΔT (Subscript[V, 0]) + 
        (c^3) Δ(T^2) (Subscript[V, 0]), ΔT -> 0]

Hint 7.7

Read the hint and type your solution here.

Hint 7.8

Read the hint and type your solution here.

Volume Expansion and The Chain Rule

Read the print project and explain the following computation.  Perhaps the  dV and dT notation will be clearer.

\(Clear[V, x, T]; 
V[x_] = V\_0\ x\^3\)D[V[x[T]], T]

Hint 7.9 (1/L) (dL/dT) =  ?, (1/V) (dV/dT) =  ?

Read the hint and type your solution here.

Exploration 7.10

Read the hint and type your solution here.

\(Clear[L, T, L0, Subscript, c]; 

    soln = DSolve[{\(L'\)[T] == c\ L0, L[0] == L0}, L[T], T]; 

    L\_1[T_] = L[T] /. soln[\([1]\)]\)

\(soln = DSolve[{\(L'\)[T] == c\ L[T], L[0] == L0}, L[T], T]; 

    L\_2[T_] = L[T] /. soln[\([1]\)]\)

\(c = 4.0*10\^\(-6\); 
L0 = 40; 
L\_1[T]\)(Subscript[L, 2])[T]

Plot[{(Subscript[L, 1])[T], (Subscript[L, 2])[T]}, {T, -50, 50}, 
      AxesLabel -> {"°C", "length"}, 
      PlotStyle -> {{AbsoluteThickness[3], RGBColor[1, 0, 0]}, 
          RGBColor[0, 0, 1]}];

Plot[(Subscript[L, 1])[T] - (Subscript[L, 2])[T], {T, -50, 50}, 
      AxesLabel -> {"°C", "difference"}, 
      PlotStyle -> {RGBColor[0, 1, 0]}];