[3ACTS] Coffee Traveler

See the task page.

I set up the problem and then had a whale of a fun time figuring out an answer. I suspect I used a railroad spike where a penny nail would have sufficed, though, so I’d like to see how you’d solve it. Leave your method or a link to a scanned scribble sheet in the comments.

BTW: This is another example of the advantages of the digital medium I’m working with. The student sees two images. One looks almost identical to the other. With the first image, I can ask the student to guess where the water levels falls in the rotated traveler. Then we lay down a mathematical structure on the image and the student works on a more abstract task. But I’ll wager that when people use this task they’ll just print out the second image because, wow, that’s a lot of paper to use for something as fleeting as a guess. That’s an advantage of digital media: students can work on more concrete tasks using more concrete representations, then abstract tasks using more abstract representations. At no extra charge.

2012 Jun 16. From Discovering Geometry, Fifth Edition, pg. 548:

A sealed rectangular container 6 cm by 12 cm by 15 cm is sitting on its smallest face. It is filled with water up to 5 cm from the top. How many centimeters from the bottom will the water level reach if the container is placed on its largest face? I'm Dan and this is my blog. I'm a former high school math teacher and current head of teaching at Desmos. He / him. More here.

1. Alex

June 5, 2012 - 6:09 am -

I like the question and I like your emphasis (as always) on not laying down the mathematical framework immediately. One comment about the video: your animation of the water level appears to rise at a constant rate. The rate should slowly decrease due to the increasing cross section of the container. I know this isn’t essential for the question you have in mind but it did catch my attention. Do you think a changing rate would lead students towards a question that they aren’t ready to discuss without calculus? If so, would this be a bad thing?

2. Jay

June 5, 2012 - 9:19 am -

Solved by calculating the area (proportional to volume) covered as (17+11)*21/2=294. When flipped the rectangle bottom’s area is 275. If water level above the rectangular bottom is x, then the area above is given by 25*6/2 – (6-x)*(25/6 * (6-x))/2. This should be equal to 19 which gives approx. x=0.82.

Not really elegant or easy, but the first approach that came to mind.

3. mr bombastic

June 5, 2012 - 10:41 am -

I think the animation detracts from the problem quite a bit, but it is still very, very nice. With the animation, it seems less real, and I don’t get the sense of the coffee sloshing around until it reaches a new equilibrium. Can’t really see a good work around.

Similar triangles is one way to go. Once you flip the carrier, the triangular cross section above the rectangle has an area of 87.5, the triangular section corresponding to the unfilled part has an area of 68.5, so the scale factor between the two triangles is (68.5/87.5)^.5.

I would like to view the problem in terms of marginal changes, seeing what happens as you move little slices of coffee from one location to another, but haven’t come up with anything.

4. Jay

June 5, 2012 - 11:16 am -

I see now that my first solution has some of the numbers wrong. Anyway, i think using the areas of the empty parts of the cross section is probably easier. Before flipping, the empty part is a trapezoid with area (17+18)*4/2=70. After flipping, the empty part is a right-angle triangle (assuming that we fill up the bottom rectangle, which can be easily checked) that is similar to a triangle with catheti of length 7 and 25. Calling the short catheti of this triangle x and the long one y we get x/y=7/25 (similarity) and xy/2=70 (area). Solving this gives us x=14/sqrt(5)=6.26. So the level of coffee will be 18-6.26=11.74.

5. Ryan

June 5, 2012 - 11:45 am -

I think we got punk’d a la http://www.ocf.berkeley.edu/~wwu/images/riddles/triangle_area.gif

1:4 is not 6:21 is not 7:25 (similar triangles on the so-called triangle part made by the blue line)

More likely than mal-intent, we just assumed the blue line had a nice round number. Rather, it’s 16.88.

Giving approx. 11.75 with Jay’s method and the same with mine.

6. Jon Voisey

June 5, 2012 - 12:43 pm -

Not going to sit down and actually push through all the numbers but here’s my method for solving it (NOTE: I’m making the assumption that the depth (not shown) is constant so, except for a factor of the unit depth, the volume and area are correlated and I use them rather interchangeably in my description):

1) I find the volume of the liquid by breaking it into a rectangle (right) and a triangle (left) both of which are easy to work with.

2) Once the container is rotated, the volume of the liquid hasn’t changed. Again, I break the container into a rectangle (bottom) and triangle (top).

3) I calculate the volume of the rectangle first. If this is greater than the volume of the liquid, then I know it doesn’t even hit the triangle portion and I figure out what the height would have to be based on the volume and the base.

4) If it exceeds the volume of the rectangle, then the liquid is filling the triangular portion. I’d subtract out the volume of the rectangle so I could concentrate on the triangle. The volume remaining would fill a percentage of the overall volume so determine what that percentage is (vol remaining after subtracting out the rectangle/vol total of triangle). 100% – that = the amount of air left in there which forms a pocket that is a similar triangle. Use that with the percentage just determined to figure out its size, subtract its height off the larger triangle, and then add that to the height of the rectangle and you’re done.

7. James Key

June 5, 2012 - 12:46 pm -

The water forms a prism with a trapezoidal base. The area of the trapezoid in part 1 is (1/2)(11+17)(21) = 294 square cm.

Provisional assumption: the area is invariant between parts 1 and 2 (more on this later). That being said, the area of the trapezoid in part 2 is (1/2)(11+h)(25) = 294. Solving yields h = 12.52.

Now for the assumption: the original prism has V1 = B1H1, and the second prism has V2 = B2H2. I wondered why Dan did not specify the depth of the container. (Should students be given this information, letting them figure out that it doesn’t matter?) V1 = V2 because the water is the same before/after. H1 = H2 b/c the container is the same. Hence B1 = B2.

Dan’s “railroad spike” comment had me thinking that there might be a simpler way, but when I read the comments, I saw that mine was simple indeed. :-)

I tried computing the area of the “space” in the container, thinking it might be simpler to deal with the space than the “stuff” (i.e. water). It wasn’t. I got tripped up for a long time b/c it looked as though the water was max 17 units wide, with the container max 18 units wide. I computed the total area of the base of the container, the area of the portion covered by water, and the area not covered by water. When I used subtraction to check my work, I saw that something wasn’t working out. It took me *quite a while* to realize what was going on: either the 17 or the 18 must not be quite accurate.

Once I cleared that hurdle, I realized that “space triangle 2” had no known dimensions, although you can figure out the relationship between the dimensions if you were motivated to do so. That’s when I gave up, as I saw that solving (1/2)(11+h)(25) = 294 is about a million times easier. :-)

8. Garth

June 6, 2012 - 6:15 am -

1. Draw the filled portion on a piece of heavy paper and trim off excess.
2. Weigh it.
3. On a second piece of the same paper draw the whole container.
4. Rotate the paper.
5. Start cutting thin horizontal strips and weigh after each cutting.

Not the greatest solution if you are trying to teach area formulas but excellent solution if you are teaching problem solving.

9. Elaine Watson

June 6, 2012 - 1:47 pm -

I copied the picture onto GeoGebra and used the arbitrary measurements that showed up. (Also, the lines were not exactly parallel. I went from dots I created at the vertices of the original container). I used GeoGebra to find the lengths of the sides and the areas of polygons (rectangles and trapezoids).

If the short side of the trapezoid (base when filling) is 1.13 units, my calculations showed that when the container was turned sideways, the water would go past the 1.13 units and start creeping up the slanted side a length of approximately .43 units. On the other side of the container, the water would rise approximately 1.25 to 1.3 units.

I didn’t check my answer and am anxious to see what others came up with.

10. brooke

June 9, 2012 - 9:00 am -

Did this with my 5th grader. I had him first just estimate where he thought the line would be. (looked at it as area since volume couldn’t be ascertained)

Then we did area of the rectangle, then the triangle. Added together. Then we divided that (294) by the base of the new shape to find how high the level would go. He did most of it in his head.

Then his 7th grade brother wandered by and totally ridiculed us both for doing math on a Saturday. ;)

11. Dan Meyer

June 11, 2012 - 8:49 pm -

Bruce James:

Dan- Can you say more about this…”students can work on more concrete tasks using more concrete representations, then abstract tasks using more abstract representations.”

This will be the subject of a longer series this summer so I’ll keep my powder a little bit dry. In short, I find it difficult to ask students concrete questions about an abstract representation. ie. It’s hard to ask students to guess, predict, and debate a context when its already been distilled into a graph, a table, and an equation.

Likewise, it’s downright impossible to ask students to generalize for a certain variable of a context without them having abstracted the context first. ie. If I’m just staring at a concrete photo of a basketball, I can’t answer the question, “How much air will fill basketballs of different sizes?” without abstracting the context first, turning the whole thing into variables and expressions.

Paper is the problem. It forces the designer to choose a certain level of abstraction for a task whether or not that level matches the questions being asked. Multimedia lets me change the representation as we change the level of abstraction.

12. Brendan Murphy

June 14, 2012 - 7:53 am -

I’ve always taught my students that volume is the area of the base times height. Knowing this helps me to realize that volume doesn’t matter in this case because the height will always be the same.

Area: the first area is a trapezoid. I can never remember the formula so I break it into a rectangle and triangle, but you don’t have to do that. When I tilt I see that the bottom part is a rectangle so I take my original area and set it as the answer to the area of rectangle formula. I know the base and solve for the height.

In my case with my estimations (My eyes had trouble counting the tiny squares) that was that. But if the liquid rose above the rectangle I would then have simply used the remaining area with the formula for a triangle.

This is a great questions no only for area but also for estimation. While finding exact numbers would add some complexity for students who finish early.

13. Dan Meyer

June 14, 2012 - 8:56 am -

Brendan:

But if the liquid rose above the rectangle I would then have simply used the remaining area with the formula for a triangle.

Yeah, this is key, ain’t it? I’m realizing right now that a student could probably divide the area by the base of the rectangle, get a height that’s within the right range of the problem, and write it down confidently. That’s going to be a tricky conversation to manage.

14. mr bombastic

June 14, 2012 - 4:30 pm -

@Brendan
I like base times height as an underlying idea, but not as a formula. How do your students react when you ask them to find the volume of a less regular shaped object – one where the volume is not base times height?

Also, why do you feel this is a better problem for pre-calc? The crux of the “area” problem is easily and clearly stated. A 6th grade class could approach it through guess and check – at the very least you provide a reason to practice otherwise boring computations. For an algebra class, this is a practical problem, where, for once, algebra actually is a good way to approach the problem. The vast majority of students are capable of thinking about complicated problems like this one – they have just been trained not to.

15. Brendan Murphy

June 14, 2012 - 5:41 pm -

@mr bombastic

No I think this is a great problem for first year algebra students. However, if the water level is greater than the area of the side rectangle then the exact area would be more difficult to prove making it beyond the patience of a normal 8th or 9th grade student. Most but not all.

When we find the volume of cones and such basically we build a cylinder around it and find the volume of that and see how many cones it takes to fill it up. It is an extension of the way we learned area, by breaking the area into smaller parts we know or drawing boxes around and subtracting the bits we know.

How to set this problem up do students “discover” it on their own is a bit more involved.

16. mr bombastic

June 15, 2012 - 7:31 pm -

If you have the coffee level end up below the rectangle, you have a nice exercise to add to a worksheet, but very little to discuss. There is one approach to take, either you see it or you don’t. There are no tempting mistakes to be made and no misunderstandings to explore.

A class may well need some guidance and hints to solve the original problem. But, isn’t it a win if we sacrifice a little of the “discovery” approach in exchange for students working at a higher cognitive level – higher than seeing how many cones can be poured into a cylinder for example.

I do like the “filling up” exercises you mentioned for building intuition on volume. An extension to breaking area up into familiar pieces might be something like looking at the volume of an office building whose top is slanted like a cheese wedge. Or, estimating the cone volume by slicing it up into small pieces, each of which is nearly a cylinder.

17. Brendan Murphy

June 15, 2012 - 8:06 pm -

Mt Bombasitc

my usual judge of the difficulty of a problem is the time it takes to figure out. I spent about 5 minutes on the problem so that would mean my average student would need about 15 minutes to get a decent estimate.

For me this problem was nothing more than triangles and rectangles. When I imagined the water rising above the bottom rectangle I saw another triangle but didn’t see a simple way to measure the height. (mainly because I wasn’t letting it shift in my minds eye). I was thinking at first the exact solution would requite some messing around with tangents and not having time I dropped the matter.

But when I do let the triangle shift I see it becomes a triangle with a base of the rectangle and a height measured from the highest point straight down to the imaginary rectangle underneath. And we are back to a simple area problem.

In the end one of those not very difficult problems as long as you can get your head out of the box so to speak.