Suppose we are asked to compute the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is constant and is the only force on the projectile after the launch.
Eliminating the parameter t, we find the object's trajectory will be given by:
y=tan(θ)x−gsec2(θ)2v20x2
dydx=tan(θ)−gsec2(θ)v20x
s=∫v20gsin(2θ)0√1+(dydx)2dx
At this point I wanted to be slick and use the derivative form of the FTOC to differentiate directly, but I couldn't make it work. :lol:
So, here goes...
Let:
u=gv20sin(2θ)x∴du=gv20sin(2θ)dx
and we have:
1+(tan(θ)−gsec2(θ)v20x)2=1+(tan(θ)−2tan(θ)u)2=cos2(θ)+sin2(θ)(1−2u)2cos2(θ)=
cos2(θ)+sin2(θ)(1−4u+4u2)cos2(θ)=cos2(θ)+sin2(θ)−4usin2(θ)+4u2sin2(θ)cos2(θ)=
1−4usin2(θ)(1−u)cos2(θ)
And now we may write:
s=∫10√1−4usin2(θ)(1−u)cos2(θ)v20sin(2θ)gdu
s=2v20sin(θ)g∫10√1−4sin2(θ)u(1−u)du
If we observe that this integral is symmetric about the line u=12, we may write:
s=4v20sin(θ)g∫120√1−4sin2(θ)u(1−u)du
Let:
v=4u(1−u)∴dv=4(1−2u)du=4√1−vdu and we have:
s=v20sin(θ)g∫10√1−sin2(θ)v1−vdv
Let:
w=√1−sin2(θ)v1−v∴dw=(w2−sin2(θ))22cos2(θ)wdv
and we have:
s=2v20sin(θ)cos2(θ)g∫∞1w2(w2−sin2(θ))2dw
Let:
w=sin(θ)csc(α)∴dw=−sin(θ)csc(α)cot(α)dα
and we have:
s=2v20cos2(θ)g∫θ0sec3(α)dα
With a bit of algebra, we may write this in a form we may integrate directly:
sec3(α)=12(sec3(α)+sec3(α))=12(sec3(α)+sec(α)(tan2(α)+1))=
12(sec3(α)+sec(α)tan2(α)+sec(α)tan(α)+sec(α)tan(α)+sec(α))=
12(sec(α)sec2(α)+sec(α)tan2(α)+sec(α)tan(α)+sec2(α)sec(α)+tan(α))=
12ddθ(sec(α)tan(α)+ln|sec(α)+tan(α)|)
and so we have (given 0<θ<π2):
s=v20cos2(θ)g∫θ0d(sec(α)tan(α)+ln(sec(α)+tan(α)))
s(θ)=v20cos2(θ)g(sec(θ)tan(θ)+ln(sec(θ)+tan(θ)))
s(θ)=v20g(sin(θ)+cos2(θ)ln(sec(θ)+tan(θ)))
Now, differentiating and equating to zero, we find:
s′(θ)=v20g(cos(θ)+cos2(θ)sec(θ)tan(θ)+sec2(θ)sec(θ)+tan(θ)−2cos(θ)sin(θ)ln(sec(θ)+tan(θ)))=0
s′(θ)=2v20cos(θ)g(1−sin(θ)ln(sec(θ)+tan(θ)))=0
And so we find we want to satisfy:
f(θ)=sin(θ)ln(sec(θ)+tan(θ))−1=0
Using Newton's method, we will need:
f′(θ)=tan(θ)+cos(θ)ln(sec(θ)+tan(θ))
And so Newton's method gives us the recursion:
θn+1=θn−sin(θn)ln(sec(θn)+tan(θn))−1tan(θn)+cos(θn)ln(sec(θn)+tan(θn))
Using θ0=π4 as the first guess, we find:
θ1≈1.01751306067
θ2≈0.986290147513
θ3≈0.985515191499
θ4≈0.985514737863
θ5≈0.985514737862
θ6≈0.985514737862
Converting from radians to degrees, we find the launch angle which maximizes the arc length is:
θ≈56.4658351275∘
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