So… I am in this “problem solving” group at school, and we spent today trying to come up with a lesson centered around problem solving that we could use for one of our classes.
I’ve been really hankering to make one of these hyperboloids out of skewers:
and I thought it would be a great investigation for my multivariable class to figure out if indeed that was a hyperboloid of one sheet. I figured it would take a number of days — at least one to create one of our own, and a good number to figure out how in the world we would come up with the equation to define that beast. [1]
Of course one of the things we talked about in our problem solving group is how to bring the questions down to simpler questions — and then generalize. So I immediately thought of these drawings I spent hours of my childhood making:
[Yes, clearly my mother was happy that I found these to amuse myself with, instead of whiiiiiining “I’m so BORED… we have NOTHING to do in this house” as I did way too often.]
If you look, they define a really nice gently sloping curve.
So my question is: what is the equation (written in terms of x and y) for the curve above?
The first segment goes from (0,5) to (0,0). Then another segment might go from (0,4) to (0,1). Another segment might go from (0,3.5) to (0,1.5). (So however much down you go on the y-axis, you go that much right on the x-axis.)
I haven’t solved the harder 3-d question yet, but I had a heck of a time solving this 2-d question.
Since I had so much fun, I thought I’d share the problem with you!
I’ll post my solution later, but if you want to throw your solution down in the comments and how you came up with it (or blog about it), awesome. Just like with this “circles, circles everywhere” problem where someone posted the most elegant solution EVAR.
Continuous Everywhere but Differentiable Nowhere 
Dammit, Shah, I’ve got things I gotta do!!! Like sleep!
I just come up with a big mess of a differential equation. As in
(y – x y’)(1 – 1/y’) = 5
Maybe you can untangle it, but I sure can’t.
Sam,
I was just showing my 6-year-old son how to draw those pictures the other day. He thought it was really cool.
Dave
Well, we have a family of lines between (0,k) and (5-k,0), which can be seen to have equations y=k-[k/(5-k)]x. The curve is the maximum of those, so the equation of the curve is given by
y = max(0<k<5) (k-k/(5-k)x).
The value of k producing that maximum can be found by finding the zeroes of d/dk (k-kx/(5-k)) = 1 – 5x/(5-k)^2, which are 5 ± sqrt(5x). Since k<5, we'll use k=5-sqrt(5x), so
y = 5-sqrt(5x)-(5-sqrt(5x))/sqrt(5x)x = 5-2sqrt(5x)+x
That was a bit uglier than I expected, and I might have screwed it up.
The answer matches what I was getting (in a different form), so that’s awesome. But I’m trying to understand what you did, and I’m getting lost… When you say “the curve is the maximum of those” — if you have time and come back here, could you explain that conceptually? I totally get the algebra, but I don’t get why you’re doing that doing…
To me it seems that by finding d/dk of y, it’s like we’re finding the k value (in terms of x) that gives you a maximum y. But that doesn’t make sense to me… why we’d want that?
Best,
Sam
Well, the curve lies on top of that family of lines, so at each individual x value, the y value will be dictated by whichever line in the family is highest at that particular point; that is, whichever line in the family happens to attain the highest value of [k-k/(5-k)x] for a specific x. The selection of a member of the family of lines is essentially equivalent to selecting a value of k, so ‘selecting the line with the highest value of y for a particular x’ is the same as ‘selecting the value of k maximizing [k-k/(5-k)x] for a particular x’.
I’m assuming a more faithful (and more adaptable, as regards 3-dimensional analogues) consideration of the spirit of the problem would involve finding the line in the family such that it is tangent to the curve, but I assume that’s what gets you the messy differential equation Matt presented.
Hi Jake,
That makes so much sense (“selecting the line with the highest value of y for a particular x”). What an elegant solution. Makes mine look so ugly. Awesome! Thanks.
Sam
Okay I’m totally late to the party but OH SNAP! I assigned THIS EXACT PROBLEM (well, except that the lines had x and y intercepts adding to 6 instead of 5, and what I asked for was the area bounded by the curve and the axes) to my Calculus classes at CRLS in 2003-4 and 2004-5.
Consequently, I can give you not only my own solution but the solutions of several of my students.
N’s solution: identical to Jake’s. Fix a value of x and consider the height of each of the family of lines at that particular x value. The point where the maximum height (for each individual value of x) is reached is on the curve.
A’s solution (not rigorous, but very cool): Extend the famiy of lines past k=0 and k=5. For example, consider the line containing (6,0) and(0,-1), the one containing (7,0) and (0,-2), etc. and the line containing (-1,0) and (0,6), etc. Don’t just draw the segments, but the whole lines (to infinity). Once you draw the broader family of lines, it’s pretty visually clear what kind of curve it is. The problem becomes to find the curve of that nature that passes through (0,5), (5,0), and (1.25,1.25).
K’s solution (manages not to use any calculus): Call the family of all lines whose intercepts sum to 5 “the Family.” All points below the curve are at the intersection of 2 lines from the Family. All points above the curve are not on any lines from the Family. All points on the curve itself are on exactly one line from the Family – the one tangent to the curve at that point. This is reminiscent of solving quadratics: if the discriminant is > 0, there are 2 solutions; k. It went something like:
k – [k/(5-k)]x = k* – [k*/(5-k*)]x so
[k*/(5-k*) – k/(5-k)]x = k* – k
[(5k* – kk* – 5k + kk*)/(5-k)(5-k*)]x = k* – k
i.e. [5(k* – k)/(5-k)(5-k*)]x = k* – k
Divide both sides by k*-k
5x / (5-k)(5-k*) = 1
so x = (5-k)(5-k*)/5
Clearly this approaches (5-k)^2/5 as k* -> k, so that’s the x-coordinate of the point on the line with y=k-[k/(5-k)]x as a tangent. Plug into this equation to find y:
y = k – [k/(5-k)][(5-k)^2/5]
= k – k(5-k)/5
= (5k – 5k + k^2 )/5
So y is k^2/5
How can we express the relation between x=(5-k)^2/5 and y=k^2/5?
Since k and 5-k add to 5, taking square roots seems warranted:
sqrt(x) = (5-k)/sqrt(5)
sqrt(y) = k/sqrt(5)
so sqrt(x) + sqrt(y) = 5/sqrt(5) = sqrt(5)
Something wrong happened just now. A whole middle part didn’t come out when I submitted it. Where it says “if the discriminant is > 0, there are 2 solutions; k” THERE WAS A WHOLE LOT IN HERE before “It went something like:…” I explained the rest of K’s solution, gave one other kid’s solution, and then gave my original solution, which is what the part that goes “It went something like:…” is all about. I have to get off the computer now though so I can’t fix.
Okay just b/c I spent like the whole afternoon writing all that down and then the internet ate it, I have to finish. (You may have noticed I hate not finishing once I start trying to communicate some math ;)
K’s solution was to pick an arbitrary point (x,y) in the plane and observe that the point is on the curve if and only if it is on exactly one line of the form y=k-[k/(5-k)]x. Points below the curve are on two such lines and points above it are on no such lines. In other words, treat x and y as fixed; then the point (x,y) is on the curve if the equation y=k-[k/(5-k)]x has exactly 1 solution for k. (Below the curve if there are 2 solutions for k and above the curve if there are no real solutions.) This equation is quadratic in k:
y = k – [k/(5-k)]x
(5-k)y = k(5-k) – kx
k^2 + (x – y – 5)k + 5y = 0
And it has exactly 1 solution if and only the discriminant
(x-y-5)^2 – 20y
is zero.
Thus, an equation for the curve is
(x-y-5)^2 – 20y = 0
i.e.
x^2 + y^2 – 2xy – 10x – 10y + 25 = 0
or
(x-y)^2 – 10(x+y) + 25 = 0
My original solution was to observe that the intersection of any two lines of the form y = k – [k/(5-k)]x occurs below the curve, but as the two values of k get closer, the intersection point approaches the curve. Thus, we can find a point (x,y) on the curve where the line y = k – [k/(5-k)]x is tangent to it, by finding the intersection between this line and a close-by line y = k* – [k*/(5-k*)]x, and then finding the limit of this intersection point as k* -> k. This is what I was doing above, after the words “It went something like:…” It yields
sqrt(x) + sqrt(y) = sqrt(5) as the final equation for the curve.
p.s. Awesome to meet you on Wednesday! Sorry to blow up your comment thread.