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\centerline{Mathematics 132 -- Calculus for Physical and Life Sciences 2}
\centerline{Lab Day 2 -- Applications of Integration}
\centerline{March 15, 2005}
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\noindent
{\it Background}
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Recall that in class yesterday, we derived an integral formula
for the arc length $L$ of the portion of the graph $y = f(x)$ from
$x = a$ to $x = b$:
$$L = \int_a^b \sqrt{1 + (f'(x))^2}\ dx.$$
Today, we want to apply this to a ``real-world'' problem.
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\noindent
{\it Maple Background}
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The integrals you need to compute today are ones that
can be expressed in elementary form using the functions
we know.  If you want a symbolic formula for an 
indefinite integral, you can either use our hand computation
methods, the table of integrals, or Maple's ${\tt int}$
command.  For instance, 
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\centerline{\tt int(sqrt(1+x\^{}2),x);}
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asks Maple to find $\int \sqrt{1 + x^2}\ dx$
in symbolic form.

As in Lab Day 1, we will also want to compute numerical approximations
to integrals for today's lab.  Instead of comparing how
accurate different integration methods are, though, we simply
want to let Maple determine an accurate value.  So for
any integrals to be evaluated today, just use the method
you used for the ``exact'' values in Lab Day 1.  For 
instance, if you wanted to compute $\int_0^2 e^{\sin(x)}\ dx$,
you would use:
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\centerline{\tt evalf(Int(exp(sin(x)),x=0..2));}
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\noindent
If you want to graph a function like $y = e^{\sin(x)}$, use 
the basic {\tt plot} command:
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\centerline{\tt plot(exp(sin(x)),x=0..2);}
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\noindent
Maple also has numerical routines for finding approximate
solutions of various types of equations.  The basic 
command here is called {\tt fsolve}.  For instance, suppose
we wanted to solve the equation $x - e^{\sin(x)} = 4$.  We could do
this as follows.  First we set up the equation to be
solved:
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\centerline{\tt eq:=x-5*sin(x) = 4;}
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Then the {\tt fsolve} command will determine a solution:
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\centerline{\tt fsolve(eq,x);}
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If you have reason to believe that your equation has more than
one solution (the equation $x - 5\sin(x) = 4$ has three!)
 and you want to look for one {\it near a given
$x$-value}, for instance near $x=4$, use a command like this:
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\centerline{\tt fsolve(eq,x=4);}
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\noindent
{\it Lab Problems}
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The distance between the towers of the main span of the Golden
Gate Bridge in San Francisco is about 1280 m.  The ``sag'' of
the cables supporting the roadway (the vertical distance from
the top of the towers down to the lowest point of the cable
halfway between the towers)
is about 143 m. on a cold winter day.  
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\noindent
A) If we ignore the fact that the roadway is pulling
the cables downward (in addition to the weight of the cables
themselves), then the shape of the cables would be that of 
a curve called a {\it catenary}.  The general equation of 
a catenary is 
$$y = c(x) = {e^{ax} + e^{-ax}\over 2}\leqno(1)$$
for some constant $a$.  
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\item{1)} First, let's determine a catenary that matches
the dimensions of the Golden Gate Bridge.  We will choose
coordinates so the lowest point on the cable is at $y = 0$.
The general equation (1) always gives $c(0) = 1$, so 
we need to shift down and consider:
$$y = c(x) = {e^{ax} + e^{-ax}\over 2} - 1$$
To determine the correct value for $a$, we want to solve
the equation $c(640) = 143$.  We can do this in Maple
as follows:
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\centerline{\tt cat:=x->(exp(a*x)+exp(-a*x))/2 - 1;}
\centerline{\tt eq:=cat(640) = 143;}
\centerline{\tt avalue:=fsolve(eq,a);}
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\item{} Note: {\tt cat} is a function whose graph is the shifted 
catenary. The output from the last command will be the appropriate
value of $a$.  
\item{2)}  Plot the catenary with the value of $a$ you 
found in part 1, and verify that you have a plausible
curve for the shape of a hanging cable.  You can substitute
the value for $a$ into the expression using a command:
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\centerline{\tt cats:=subs(a=avalue,cat(x));}
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\item{3)}  Determine the arclength of the catenary cable, using
the following:
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\centerline{\tt length:=evalf(Int(sqrt(1+diff(cats,x)\^{}2),x=-640..640));}
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(The first line substitutes the value for $a$ and calls
the resulting expression {\tt cats}.  The second computes
the arclength integral for the function using 
$$L = \int_{-640}^{640} \sqrt{1 + \left({d cats\over dx}\right)^2}\ dx$$
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\noindent
B) The shape of the ``loaded'' cables of a suspension
bridge is actually approximately {\it parabolic}.  
Find the equation of an appropriate {\it parabola} that approximates
the shape of the cables of the Golden Gate Bridge.  Note:  You'll
need to decide how to introduce coordinates in the diagram above.
The origin of the $xy$-plane can be anywhere in the picture
that is convenient.  Explain your choice in a text region.
When you have your equation, plot
your function using Maple to check that its graph matches
the diagram above(!)
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\noindent
C)  Using your parabola's equation, determine the {\it length of
the cables} (that is the length from one tower to the other along
the parabola with the shape of the hanging cable).  Note:  You'll
need to set up the function to integrate.  That can be done 
``by hand'' or by using some of Maple's built-in commands.
Look at the Help page for the {\tt diff} command, for instance. 
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\noindent
D)  As a check on your answer in part C, also compute the
length of the straight line joining the top of the left tower to 
the lowest point of the cable, then add the length of the
line back up to the top of the right tower.  Is your answer
in B consistent with this?  Explain.
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\noindent
E)  On a hot summer day, because of thermal expansion of the
metal, the cable is about $0.05\%$ longer (but the shape of the 
cable is still parabolic).  The effect of the heat on the 
concrete of the towers is negligible.  {\it How much does the 
``sag'' of the cable increase?}  Would you be likely to notice this
if you drove over the bridge?
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\noindent
{\it Assignment}
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Writeups due in class on Monday, March 21.
\bye

