Math 2280-2, Project #1, Problem 5
We first solve the crossbow example 2.3.3 using improved Euler's method with n=100
with(LinearAlgebra): with(plots): Digits:=16:
t0:=0.; tn:=10.; # first and last points in the interval
v0:=49.; # initial velocity
n:=100; # number of steps
h:=(tn-t0)/n; # step size
f:=(t,v)-> -(0.0011)*v*abs(v) -9.8; # slope function (rhs in DE dy/dx = f(x,y))
v:=v0: t:=t0:
ts1:=Vector(n+1): vs1:=Vector(n+1): ts1[1]:=t0: vs1[1]:=v0:
printf("%15s, %15s\134n","t","v"):
for i from 1 to n do
k1:= f(t,v): # left hand slope
k2:= f(t+h,v+h*k1): # approximation to right hand slope
k:=(k1+k2)/2: # averaged slope
v:= v + h*k: # improved Euler update
t:= t+h: # increase x
ts1[i+1]:=t: vs1[i+1]:=v:
if frac(i/10)=0 then
printf("%15.1f, %15.5f\134n",t,v):
end if:
od: # end for i loop
JCIiIUYj
JCIjNSIiIQ==
JCIjXCIiIQ==
IiQrIg==
JCIxKysrKysrKzUhIzs=
Zio2JEkidEc2IkkidkdGJUYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCYqKCQiIzYhIiUiIiI5JUYvLUkkYWJzRyUqcHJvdGVjdGVkRzYjRjBGLyEiIiQiIykqRjVGNUYlRiVGJQ==
t, v
1.0, 37.15469
2.0, 26.24277
3.0, 15.94529
4.0, 6.00406
5.0, -3.80200
6.0, -13.51045
7.0, -22.93557
8.0, -31.89841
9.0, -40.25568
10.0, -47.90661
And now with 200 points
t0:=0.; tn:=10.; # first and last points in the interval
v0:=49.; # initial velocity
n:=200; # number of steps
h:=(tn-t0)/n; # step size
f:=(t,v)-> -(0.0011)*v*abs(v) -9.8; # slope function (rhs in DE dy/dx = f(x,y))
v:=v0: t:=t0:
ts2:=Vector(n+1): vs2:=Vector(n+1): ts2[1]:=t0: vs2[1]:=v0:
printf("%15s, %15s\134n","t","v"):
for i from 1 to n do
k1:= f(t,v): # left hand slope
k2:= f(t+h,v+h*k1): # approximation to right hand slope
k:=(k1+k2)/2: # averaged slope
v:= v + h*k: # improved Euler update
t:= t+h: # increase x
ts2[i+1]:=t: vs2[i+1]:=v:
if frac(i/20)=0 then
printf("%15.1f, %15.5f\134n",t,v):
end if:
od: # end for i loop
JCIiIUYj
JCIjNSIiIQ==
JCIjXCIiIQ==
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JCIxKysrKysrK10hIzw=
Zio2JEkidEc2IkkidkdGJUYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCYqKCQiIzYhIiUiIiI5JUYvLUkkYWJzRyUqcHJvdGVjdGVkRzYjRjBGLyEiIiQiIykqRjVGNUYlRiVGJQ==
t, v
1.0, 37.15474
2.0, 26.24292
3.0, 15.94555
4.0, 6.00444
5.0, -3.80159
6.0, -13.51019
7.0, -22.93545
8.0, -31.89845
9.0, -40.25588
10.0, -47.90696
We see that the two approximations agree with each other to two digits.
Here is a plot of both along with the true solution (from the file crossbow.mw that we used in class). All are indistinguishable!
rho2:=0.0011: g:=9.8:
C1:=arctan(v0*sqrt(rho2/g)): C2:=arctanh(v0*sqrt(rho2/g)):
tmax1:=C1/sqrt(rho2*g): tmax2:=C2/sqrt(rho2*g):
v3:=t->piecewise(t<tmax1,sqrt(g/rho2)*tan(C1-t*sqrt(rho2*g)),t>=tmax1,sqrt(g/rho2)*tanh(C2-(tmax2+t-tmax1)*sqrt(rho2*g)));
Zio2I0kidEc2IkYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLUkqcGllY2V3aXNlRyUqcHJvdGVjdGVkRzYmMjkkSSZ0bWF4MUdGJSomLUklc3FydEc2JEYrSShfc3lzbGliR0YlNiMqJkkiZ0dGJSIiIkklcmhvMkdGJSEiIkY4LUkkdGFuR0YzNiMsJkkjQzFHRiVGOComRi5GOC1GMjYjKiZGOUY4RjdGOEY4RjpGODFGL0YuKiZGMUY4LUkldGFuaEdGMzYjLCZJI0MyR0YlRjgqJiwoSSZ0bWF4MkdGJUY4Ri5GOEYvRjpGOEZBRjhGOkY4RiVGJUYl
unassign('t'):
p1:=pointplot(Matrix([ts1,vs1]),symbol=diamond):
p2:=pointplot(Matrix([ts2,vs2]),symbol=cross):
p3:=plot(v3(t),t=0..10):
display({p1,p2,p3});
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(a) to obtain the value where v=0, we run improved Euler's method and select the time for which v is closest to 0. Since all computations are already stored in a vector here, we can compute the minimum in Maple (as done below). Normally (in higher dimensional problems) you do not want to store the value of your solution at all points, so the numerical method's loop could be modified to find the time for which v is minimum (with no additional memory cost).
# find the index of the minimum value in the vector vs2
res:=rtable_scanblock( abs(vs2), [rtable_dims(vs2)],(val,ind,res) -> `if`(val<res[2],[ind,val],res),[[1],abs(vs2[1])]);
printf("approximation = %4.2f seconds\134n",ts2[res[1][1]]);
NyQ3IyIjJCokIjFpKj0uQEtMOyIhIzs=
approximation = 4.60 seconds
(b) To estimate the impact velocity after 9.41 seconds, we could use improved Euler with tn=9.41, to get:
t0:=0.; tn:=9.41; # first and last points in the interval
v0:=49.; # initial velocity
n:=200; # number of steps
h:=(tn-t0)/n; # step size
f:=(t,v)-> -(0.0011)*v*abs(v) -9.8; # slope function (rhs in DE dy/dx = f(x,y))
v:=v0: t:=t0:
for i from 1 to n do
k1:= f(t,v): # left hand slope
k2:= f(t+h,v+h*k1): # approximation to right hand slope
k:=(k1+k2)/2: # averaged slope
v:= v + h*k: # improved Euler update
t:= t+h: # increase x
od: # end for i loop
printf("v( %4.2f ) approx. equals to %4.2f m/s\134n",t,v);
JCIiIUYj
JCIkVCohIiM=
JCIjXCIiIQ==
IiQrIw==
JCIxKysrKysrMFohIzw=
Zio2JEkidEc2IkkidkdGJUYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCYqKCQiIzYhIiUiIiI5JUYvLUkkYWJzRyUqcHJvdGVjdGVkRzYjRjBGLyEiIiQiIykqRjVGNUYlRiVGJQ==
v( 9.41 ) approx. equals to -43.48 m/s