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 th...