# 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 (1,m+b), or sqrt(1+m^2). The arclength formula
# gives  
> Int(sqrt(1+diff(m*x+b,x)^2),x=0..1)
> =int(sqrt(1+diff(m*x+b,x)^2),x=0..1);

                     1
                    /
                   |         2 1/2            2 1/2
                   |   (1 + m )    dx = (1 + m )
                   |
                  /
                    0

# # 6  Find the length of the curve y = x^2 fro 1 to 2 to
# within .005
# 
# 
# Solution:  The length is the integral, which can
# be evaluated easily by Maple.
# 
> Int(sqrt(1+diff(x^2,x)^2),x=1..2)=int(sqrt(1+diff(x^2,x)^2),x=1..2);
> evalf(");

   2
  /
 |           2 1/2
 |   (1 + 4 x )    dx =
 |
/
  1

      1/2                 1/2         1/2               1/2
    17    - 1/4 ln(-4 + 17   ) - 1/2 5    - 1/4 ln(2 + 5   )


                            3.168 = 3.168

# We want to see how many partitions we would need
# to approximate this number to within .005 using
# the right rule Rn.
# 
# We can use theorem 2 to bound the error,
# taking k1 to be the maximum  of the absolute value
# of the derivative, which we can get 
# 
# 
> restart;
> f := x->sqrt(1+4*x^2);

                                             2
                       f := x -> sqrt(1 + 4 x )

> fp :=  D(f);

                                        x
                     fp := x -> 4 --------------
                                              2
                                  sqrt(1 + 4 x )

> plot(fp, 1..2);

>  
# So a suitable value for k1 is 2
# We want n large enough that  k1(b-a)^2/(2n) = 1/n < .005
# 
# solving the inequality for n, we see that any n > 200
# will work by theorem 2.  Using Maple,
>  bound := 2*(2-1)^2/(2*n); 

                             bound := 1/n

> solve(bound = .005,n);

                                 200.

# checking this 
> R200 := (2-1) /(200)* sum(f(1+i/200.) ,i=1..200) ;

                         R200 := 3.172558813

> evalf(R200-int(f(x),x=1..2) );

                             .0047179088

# The actual error using R200 is just slightly under the maximum error
# predicted by theorem 2.
# 
# 
# 8.4  Work
# 
# #1  Suppose that the spring on  a bathroom scale is stretched by .06
# inches when a 115 lb person stands
# on the scale.
# 
# a) how much work is done on the spring when this person steps on the
# scale?
# 
# Solution:  Get the spring constant k using Hooke's law
# 
# 115 = k * .06, so 
> k := 115/.06;

                              k := 1917.

# The work is the integral of the force  k*x with respect to distance dx
# as x goes from 0 to .06  
> work := int(k*x,x=0..(.06));

                            work := 3.451

# So around 3 and 1/2 inch-pounds of work is done on the 
# spring.
#
# Note: The statement of the problem is subject to interpretation. We have 
# calculated the amount of work needed to stretch the spring out .06 
# inches.  This means that as the person steps on the scale, the spring
# begins to stretch out  before his full weight gets on the scale.
#
# b) How much work is done when a 175 lb person steps on
# the scale?
# 
# Solution:  How far will the spring stretch?
# Then integrate the force from 0 to that distance
> eqn := 175  = k*x0;

                        eqn := 175 = 1917. x0

> x0 := solve(eqn,x0);

                             x0 := .09129

> work := int(k*x,x=0..x0);

                            work := 7.988

# Note that increasing the weight by 50 lbs (less than  50
# percent) increases the work by more than 100 percent.
# 
# We can see that more generally by letting the weight
# increase by p percent and asking to calculate the percentage increase
# in work
> eqn := 115 + p*115 = k*xp;

                    eqn := 115 + 115 p = 1917. xp

> xp := solve(eqn,xp);

                       xp := .05999 + .05999 p

> work := int(k*x,x=0..xp);

                                                   2
                  work := 3.449 + 6.899 p + 3.449 p

> wp := (work - 3.45)/3.45*100; 

                                                 2
                   wp := -.01 + 200.0 p + 99.99 p

> plot(wp,p=0..20);

# We can conclude that the percentage increase 
# in work wp is a quadratic  function of p, a percentage increase  in
# the weight on the spring. 
#     
# #4  A cylindrical gasoline tank with radius 4 and
# length 15 is buried on its side under a service 
# station. The top of the tank is 10 feet below 
# ground. Find the work needed to empty the tank
# through a nozzle 3 feet above ground.
# 
# Solution:
# 
# We think of a thin sheet of gas located at
# height x in the tank, with x between -4 and 4
# The thickness is dx, the length is 15, and
# the width is 2sqrt(16-x^2).  That sheet needs to
# be lifted 13 + 4-x  feet.  The work needed to do
# that is  42*dx * 15 *2sqrt(16-x^2) * (17-x), so
# the total work is
> Int(15*42*2*sqrt(16-x^2)*(17-x),x=-4..4);

                     4
                    /
                   |               2 1/2
                   |   1260 (16 - x )    (17 - x) dx
                   |
                  /
                    -4

> value("); evalf(");

                              171360 Pi


                             538343.3172

# So, we get about a half-million ft pounds of work
# or 2500 ton feet of work to empty this gas tank
# which contains
> 42*Pi*16*15; evalf(");

                               10080 Pi


                             31667.25395

# lbs of gas.  The work is the same as moving this weight from
# center of mass of the tank 17 feet up to the nozzle.