Assignment, September 19: Consider (cos(t), sin(t)) 0≤t≤π/4, and (1/√(1+s2), s/√(1+s2)) 0≤s≤1. 1. Prove that those are two parametrizations of the same curve.
Aye aye! (cos(t), sin(t)) be describin’ a shape round as a doubloon, an the restriction on yon parameter be making it naught but a piece. Piece of eight, to talk all mathematical precise-like.
If ye chart a course round that bit, ye’ll follow the points what have coordinates on that tharr equation.
That curve be circlin’ round, always the same distance from the centarr. Any lubber what’s seen a circle afore can see why. This line be curving through the shape of a piece of eight, an’ it be stayin’ one day’s voyage from the middle.
Now that we’re familiarr with the shape of things, check the other curve to see if they be all similarr. If they be the same, then arr second curve be parrt o’ a circle as well.
When ye look at a map, the distance between two points be √((x-x1)^2+(y-y1)^2)
The point in the center be a point full o’ scurvy blistering scuttlehounds with a value o’ zero. So the distance to any point on tha curve be √((1/√(1+s2)-0)2+(s/√(1+s2)-0)2) =√((1+s2)/(1+s2))=√1 =1.
So everythin’ on that second shape be one unit distance from the centarr. It be a part o’ tha same circle.
If I were a yellow-bellied flea-ridden coward, I’d stop here. But I gotta show it be the same part o’ tha same circle. The second ship sails from the same start as tha first, cause cos(0)=1=1/√(1+02) and sin(0)=0=0/√(1+02), an’ it makes for the same bay on account o’ how cos(π/4)=1/√2=1/√(1+12) and sin(π/4)=1/√2=1/√(1+12).
But the cap’n asked us to show that these two ships be sailin’ the same course. There’s more than one way to skin a cat-o’-nine-tails (as ye’ll find out if ye think ye be done). We gots to prove they take the same path exactly. An’ we’ll do it with calculus.
Some o’ ye cod-faced scallywags turned tail and ran when I threatened ye with calculus. For those of you with more backbone than a jellyfish, dx/ds= -s/(1+s2)3/2 and dy/ds= -s2/(1+s2)3/2+1/(1+s2)1/2. 1/(1+s2)1/2 be bigger’n the mutiny-minded -s2/(1+s2)3/2, so dy/ds be positive when s is in our range. That be range o’ tha parametrization, not range o’ the cannons.
dx/ds is always less than nothing; how’s that for worthless! So this ship be sailing from the same start point as the other, then veering hard a’port and for’ard till they make landfall. And since all the functions be continuous, it’s sailin’ straight there like a ship should and not disappearin’ like the Flying Dutchman. And, if ye look back at the charts, these be the same directions the first ship followed. Yarr!
2. Use the arclength formula on each parametrization and verify that they are equal.
As any landlubber could tell you, L=∫√((dx/dt)2+(dy/dt)2)dt. The lubber could also tell you that dx/dt=sin(t) and dy/dt=-cos(t), but only if they’ve got the brains to tell a keel from a kelp or a squall from a squid. And unless they be smarter than the usual rotten-rigged deck swabbers usually on me crew, I wouldn’t count on that. Since sin2(t)+cos2(t)=1, L=∫√(1)dt from 0 to π/4=π/4. But that was the easy one, so if ye be ready for a fight then run up the Jolly Roger and start calculatin’. No quarter given to the enemy and I’ll shoot any one of ye that asks for any.
The formula be longer than a shot from a high-angle cannon.
L=∫√((-s/(1+s2)3/2)2+((-s2/(1+s2))3/2+1/(1+s2)1/2)2)ds. Avast, formless thing! I’ve known buccaneers who’d rather face a kraken in single combat than go up against something with that many parentheses, but it’s going to get worse before it gets better. L=∫√(s2/(1+s2)3+s4/(1+s2)3-2s2/(1+s2)2+1/(1+s2))ds. Arr, now find a common denominator or walk the plank. We’ll place some jury-riggged factors of (1+s2) athwart those terms until it’s in one piece. L=∫√((s2+s4-2s2(1+s2)+(1+s2)2)/(1+s2)3)ds. This may look timber-shiveringly scary, but from here on it’s clear sailing by multiplyin’ and addin’.
L=∫√((s2+s4-2s2-2s4+s4+2s2+1)/(1+s2)3)ds=∫√((s2+1)/(1+s2)3)ds. A swipe o’ me cutlass shortens that to ∫√(1/(1+s2)2)ds=∫1/(1+s2)ds.
Me spyglass shows an inverse trig identity on the horizon. Solution ahoy!
∫1/(1+s2)ds (0≤s≤1)=arctan(s)=arctan(1)-arctan(0)=π/4. Aye, that be exactly what the other one was, and if you didn’t expect that then you’re due for a keelhauling.