Two weeks ago on car talk, Click and Clack heard an interesting gas tank turned geometry problem that was simple to state, but not so easy to solve. One of their callers had a cylindrical gas tank in his truck, and because the gas gauge was broken, this caller would drop a stick into the tank and check the gas level. The question was where he needed to mark his stick so that he could properly measure a quarter tank of gas.
Now if you’re thinking, that’s easy, it’s just a quarter of the height of the stick, then a picture will make you feel less sure:
The key is that the tank is not standing upright like a soda can, but is instead laying on its side, like a toilet paper roll. So let’s draw the cylindrical tank laying on its side, as the caller described. Then draw a stick that’s stuck in there vertically. If the tank is 20 inches in diameter, then a 20 inch stick will go from top to bottom, and 10 inches will surely be the ½ full mark. But where is the ¼ full mark?
Well, Click and Clack rightfully turned the problem into a 2-dimensional one, by rotating the tank around so that all you could see was the circular base. That turns the volume problem into a circle area problem, which will make life easier. Then, as they often do, they simplified the problem by splitting the circle in half, and focused on solving for the height at which the semicircle is split into two.
That’s called progress! But then they got stuck. They started talking about how you needed calculus to solve this problem, and then suggested that the caller find someone with a tank that is a quarter full, and use that to calibrate the stick 
In response to what must have been overwhelming fan-mail, this past week Tom and Ray returned to the problem, and went over some interesting insights that a variety of their listeners and family members had shared. Yes, calculus will help to solve the problem (Wired does a nice analysis). However, trigonometry is really all that is needed (Dr. Math to the rescue!).
Here at Virtual Nerd we love to see problems like this generate so much discussion, and we can’t help but throw in our 2 cents. As you’ll see below, we’ve got an approach that gets pretty darn close, and doesn’t require any trig at all!
Let’s redraw the diagram and label and shade some things, like the radius (r), the height (h) from the center of the circle to the level that splits the semicircle in half, and the two equal regions A and B that we get after we split the semicircle:
We know that the area of our semicircle is ½ of the area of the whole circle. The whole circle has area πr², and so our semicircle has area ½πr². Now the shaded regions A and B both are supposed to have half the area of this semicircle. We can write that in math terms, by writing B=½•(½πr²). That is one equation that we have, but so what?
Well now we need to find the area of that shaded region B, and that is tough, because it’s a weird shape. There is a way to find the exact area using some trigonometry, but what if we did something silly. What if we stepped back and said, you know what, that area kind of looks like a rectangle!
Yeah, it kind of does. It’s a rectangle with a height of h, and a base of 2r, and the area of a rectangle is base times height. That means that our approximate value for the area of region B is B=2rh.
Now something magical happens! We can set our two equations for B equal to each other:
To solve this equation, we want to get h by itself, but we’ve got a lot of stuff going on. It turns out that there are many ways to proceed, but let’s just try to make life easier by considering the numbers first, and dealing with the variables second. We’ve got 2 on the left hand side. It would be nice to get it to the right side, away from the h. Well we can divide both sides by 2, or multiply both sides by ½ it’s really comes down to whatever you prefer. Multiplying will work nicely, so lets do that.
½•2 cancels out on the left side, and now the right side has ½•½ and then another ½ in the parentheses. Since we’re multiplying all the way through, it’s safe to drop the parentheses and simply evaluate ½•½•½, which equals 1/8.
Now we have h almost by itself, except there is a pesky r still with us on the left side. Well, that’s ok, because we can just divide both sides by r!
That leaves h on the left side, and since r²/r = r, we are left with (1/8)•π•r on the right side.
So, if our radius is 10 inches, we take out our calculator, type in 1/8•3.14159•10, and we get h= 3.93 inches. That means that if we mark the stick 3.93 inches from the bottom, we should have a pretty good measure of 1/4 tank of gas. Now that’s pretty cool, because the exact answer to this problem, as others have found and explained, is about 4.04 inches!
This is a great example of how simple estimation can go a long way to solving a complicated problem. Sure region B is not a rectangle, but it’s close, and so a rectangular approximation works pretty well. Plus, we know that region B is a little less than the area of the rectangle that we found, so with a little head scratching we can conclude that our estimate for height is a bit on the low side. Some of the extra area in the rectangle we drew that isn’t in region B will go into bumping the height up.
It turns out we can do even better, without touching trig or calculus, by finding another geometrical approximation to the area in region B, as you can see in the diagram below. We can go into details about solving it that way later (still no trig necessary!), but you should try it yourself, and see if you get any closer to h=4.04 inches!
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