Okay we're going to compute the arc length of an ellipse in this problem
and here's the equation for the ellipse and i just wanted to point out when you're looking at
the equation for an ellipse the game i like to play is to just say what if y was equal to zero
then i should be able to find these two values of x so i plug in y equals zero
and i get x over three squared equals one take the square root x over 3 is plus or minus 1.
multiply by 3 x is plus or minus 3. so i see that this thing intersects
the x-axis negative 3 and 3. you can do something similar with the y-coordinate
you go what if x is zero then i would be at these points and if i square root this and
multiply by two i can see that y is plus or minus two so it's it's nice just to remind
yourself how easily the formula for the ellipse connects to what you're seeing in the graph
so that's a bit beside the point
but just a valuable reminder i thought so we're going to go back to our equation i'm going to
leave these little intersections on here because i'm going to need them as limits of integration
and i'm going to go ahead and solve this for y and i'm going to keep only
the positive y's so we're going to look at the upper half of the ellipse just this
and then we're going to double the result so i have y over 2
quantity squared is 1 minus x over 3 squared which is x squared over 9.
and i take the square root of both sides and i get y over 2 equals the square root
of 1 minus x squared over 9. and i multiply by 2 on both sides and i get y equals 2 square roots
1 minus x squared over 9. now we're going to plug into the arc length formula
where we have our total arc length is going to be the integral on whatever interval we're looking
at so let's say x1 x2 that's negative 3 to 3. square root 1 plus f prime of x
squared dx so we've got to differentiate this thing
and then square the derivative and this is going to start to get really ugly so i have y prime
is 2 and then the derivative of this
it's the quantity 1 minus x squared over 9 to the one-half so the derivative is one-half
1 minus x squared over 9 to the negative one half and the chain rule says you've got attack
on the derivative of the this interior function which is negative two over nine
times x clean things up a little bit i get negative two x over nine
maybe i'll put the square root in the denominator here
1 minus x squared over 9. so now in this integral i have to square that thing
clean up my integral sign a little bit
and i'm inside this square root which in practice makes these integrals
almost always impossible to do symbolically and i end up with a 1 plus
and my fraction here is 4x squared
over 81 times 1 minus x squared over 9.
so i squared that square root to undo it square the 9 to get 81 and so on
okay good luck figuring that thing out symbolically these are usually actually
impossible there is no symbolic solution so we go to a utility to get it done and
i happen to have already put this thing in i wrote it in a slightly different form
but i don't i don't want you to have to watch me type this in for several minutes
so i already put it in i used the command integrate in maxima and then i typed in that
whole function i didn't do the squaring prior to putting it in maxima when i did it earlier
and i hit shift enter and maxima just produces another copy of the integral
and that's maximus way of saying there is no symbolic solution to this integral
so what we do is we use the numerical approximation algorithm so it's quad qag
and then i'll copy
the integral that i just put in and paste it down here now this is not quite it
when you use quad qag you have to select which integration algorithm to use and we don't care
which one it is they're all accurate enough for us so i'm just going to use number one
shift enter so this is an interesting error on the issue here is that when the three gets plugged
into the denominator you end up with a zero denominator in here and this thing blows up so i
guess one strategy to deal with this is we could just go really really close to negative three
and the positive three has the same problem so we'll go really really close to the positive three
i'm going to round my answer to a few decimal places anyway so this would be good enough
so i get an arc length of 7.932 and that's just the upper half of the ellipse why don't
i just save myself some work and multiply by 2 in maxima and i get for the whole ellipse 15.865
so oh i want to get this number real quick
just so i can write it in the answer
so 7.932
was my approximate arc length but remember that represents only the upper half
so then the entire ellipse is just twice that and it was 15.865 or 15.87 is good
enough
all right finally we should check for the reasonableness of our solution
so if i look at the graph the distance from here to here
is 6. so i can see that this chunk here is substantially bigger than six it's like well
i know that's seven point nine ish but a straight line would be six and this is
definitely a little bigger than that another six down here but a little bigger than that so is it
plausible that the arc length is a little bigger than 15 yeah it looks about right
you