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I have a baking recipe that calls for $1/2$ tsp of vanilla extract, but I only have a $1$ tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere.

My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely $1/2$ tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?

(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)

bob
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Holly
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  • A bit more than halfway. I seriously doubt you'll be able to eyeball it perfectly, but you can set up a calculus problem to get the exact answer if you would like. – xxxxxxxxx Sep 22 '19 at 20:58
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    after I answered, I found here that the answer is to fill it by the fraction $1-2\cos(\frac49\pi)$ – J. W. Tanner Sep 22 '19 at 21:51
  • Related video: https://www.youtube.com/watch?v=Ut_0LvLQmI4 How deep is the liquid in a half-full martini glass? – Eric Duminil Sep 23 '19 at 13:19
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    And they say calculus is of no use in real life... – Mr.HiggsBoson Sep 23 '19 at 13:41
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    Not really relevant to the mathematics, but: just eyeball it. Half a teaspoon of vanilla extract, one way or another, ain't gonna make that much difference. In fact, just put in a full teaspoon. Then, do yourself a favor and add some mace and clove, too. ;) – Xander Henderson Sep 23 '19 at 13:55
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    @RandomAspirant Do you need to comment that on every answer as well? – KarelPeeters Sep 23 '19 at 14:38
  • @ToddSewell Sorry , I will delete , but the thought of it was so funny to me , I literally fell of my chair...:P – Mr.HiggsBoson Sep 23 '19 at 15:29
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    Just use a second spoon - fill the first one completely, then pour from it into the second until they're even... – twalberg Sep 23 '19 at 16:55
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    The shape is called a spherical cap, you can find all the information you need on wikipedia. https://en.wikipedia.org/wiki/Spherical_cap – Brady Gilg Sep 23 '19 at 17:06
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    Cooking is not as exact a science as people make it out to be. Being "off" by a little isn't likely going to ruin anything. Many of the greatest chefs (and my mom) don't even use recipes all the time. How do you think recipes get created in the first place? Most of it is "to taste" anyway, especially vanilla and other flavoring. – computercarguy Sep 23 '19 at 20:10
  • @twalberg: The OP says that her small measuring spoon is in the dishwasher, so I doubt that she has a second big spoon available. – TonyK Sep 23 '19 at 20:26
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    Do we allow housework problems? – Acccumulation Sep 23 '19 at 21:53
  • @J.W.Tanner Close except that in this case the cap is inverted. – Beanluc Sep 23 '19 at 22:18
  • Or use a kitchen scale and get it to 2.1 grams. – Viktor Mellgren Sep 24 '19 at 15:14
  • As Xander said, use a full teaspoon. And make sure it's real vanilla extract or vanilla paste, not the imitation crap. An extra half teaspoon of real vanilla extract will never detract from any recipe, and in many cases will only serve to enhance it, so long as you're using a good quality extract! I pay through the nose for the stuff I use, but it does last a while, and is totally worth it. The paste even has vanilla bean flecks in it which are small enough they don't impact the actual texture (i.e. not gritty) but add a wonderful aesthetic flourish to vanilla cupcakes or frosting! – Doktor J Sep 24 '19 at 17:11
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    @Acccumulation see the FAQ post about this: "As a general rule, we do not discourage any specific category of questions, as long as it is mathematical" – Ruslan Sep 25 '19 at 06:42
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    @Ruslan It's a joke. Oftentimes, homework problems aren't really fit for the site because they're not posted clearly or may be dupes. "housework" sounds similar to "homework", but... yeah, the meaning is different. – EKons Sep 25 '19 at 20:24
  • @RandomAspirant How long did it take to hit the floor? –  Sep 26 '19 at 05:16
  • @Holly, Give it to your brother and tell him to fill it all the way. That guy quits halfway through everything. –  Sep 26 '19 at 05:18

6 Answers6

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Assuming the spoon is a hemisphere with radius $R$,

let $x$ be the height from the bottom of the spoon, and let $h$ range from $0$ to $x$.

The radius $r$ of the circle at height $h$ satisfies $r^2=R^2-(R-h)^2=2hR-h^2$.

The volume of liquid in the spoon when it is filled to height $x$ is $$\int_0^x\pi r^2 dh=\int_0^x\pi(2hR-h^2)dh=\pi Rh^2-\frac13\pi h^3\mid_0^x=\pi Rx^2-\frac13\pi x^3.$$

(As a check, when the spoon is full, $x=R$ and the volume is $\frac23\pi R^3,$ that of a hemisphere.)

The spoon is half full when $\pi Rx^2-\frac13\pi x^3=\frac13\pi R^3;$ i.e., $3Rx^2-x^3=R^3;$

i.e., $a^3-3a^2+1=0$, where $a=x/R$.

The only physically meaningful solution of this cubic equation is $a\approx 65\%.$

J. W. Tanner
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It may be surprising that the problem actually admits an analytic solution.

A spherical cap is the difference between two overlapping cones, one with a spherical bottom and the other with a flat bottom, i.e.

$$ V = \frac{2\pi}{3}r^2h - \frac{\pi}{3}(2rh-h^2)(r-h) =\frac{\pi}{3}(3rh^2-h^3)$$

which, with half of the semisphere volume $V=\frac{2\pi}{3}r^3$, becomes

$$\left(\frac rh \right)^3 - 3\frac rh+1=0$$

Let $\frac rh = 2\cos x$ and compare with $4\cos^3 x -3\cos x -\cos 3x=0$ to obtain $x=40^\circ$. Thus, the depth $h$ as a fraction of the radius $r$ is

$$\frac hr = \frac{1}{2}\sec 40^\circ$$

Quanto
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    That was a big surprise for me! Also a bit surprising was that your $\dfrac{1}{2\cos 40^\circ}$ is equal to J.W.Tanner's $1-2\cos(\frac49\pi)$ (in a comment to the OP). – TonyK Sep 23 '19 at 16:16
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    @TonyK - I knew of the close-form result, but was also surprised of a different form from his, until convinced myself numerically – Quanto Sep 23 '19 at 16:34
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It makes things a bit simpler if we turn your measuring spoon upside down, and model it as the set of points $\{(x,y,z):x^2+y^2+z^2=1, z\ge 0\}$. The area of a cross-section at height $z$ is then $\pi(1-z^2)$, so the volume of the spoon between the planes $z=0$ and $z=h$ is

$$\pi\int_0^h(1-z^2)dz = \pi\left(h-\frac13h^3\right)$$

The volume of the hemisphere is $\frac23\pi$, and we want the integral to be equal to half this, i.e. $$\pi\left(h-\frac13h^3\right)=\frac{\pi}{3}$$ or $$h^3-3h+1=0$$ This cubic equation doesn't factorize nicely, so we ask Wolfram Alpha what it thinks. The relevant root is $h\approx 0.34730$. Remember that we turned the spoon upside down, so you should fill it to a height of $1-h=0.65270$, or $65.27\%$.

TonyK
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    "It makes things a bit simpler if we turn your measuring spoon upside down" Then there isn't any liquid in the hemisphere. – Acccumulation Sep 23 '19 at 21:47
  • @Acccumulation One could easily turn it upside-down to measure it, then turn it right-side-up when filling it – user45266 Sep 23 '19 at 23:03
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    What do you mean by "doesn't factorize nicely"? In my view, $$h^3-3h+1=\left(h-2\cos\frac{2\pi}9\right)\left(h-2\sin\frac{\pi}{18}\right)\left(h+2\cos\frac\pi9\right)$$ is quite a nice closed-form factorization. Not in radicals, but why would anyone want them :) – Ruslan Sep 24 '19 at 06:27
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    @Ruslan How did you find that? – Ovi Sep 25 '19 at 00:39
  • @Ovi well, I found the first root with Wolfram Mathematica's Solve + FullSimplify, and the second and third by combination of FullSimplify on the additive terms of the solution returned by Solve and then ExpToTrig to get the expressions like $(-1)^{8/9}$ to trigonometric form. But that was a lazy approach. The more general way is to use the algorithm given in this page (page is in Russian, but I guess if you just follow the formulas, you'll get it: the Vieta solution is sufficient here). – Ruslan Sep 25 '19 at 05:32
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Without loss of generality we assume the radius of the sphere to be $1$

The volume of the liquid is found by an integral $$V= \int _{-1}^{-1+h} \pi (1-y^2 )dy$$

and you want the volume of the liquid to be half of the hemisphere which is $\pi/3$

After evaluating the integral and solving the equation I have found $$h=0.65270365$$ That is a little bit more than half as expected.

Mohammad Riazi-Kermani
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Note about eyeballing: Your eye's reference is the surface of the spoon, so when you eyeball you may actually be measuring along the arc from the bottom of the spoon to its top edge.

That is, your eye may be watching the red curve, not the blue line:

half full spoon

Using the 65.27% from other answers, the depth measured along the red curve is $$ \frac{\arccos(1 - 0.6527)} {90\deg }\approx 77.42\%$$

So to the eye, the "depth" of a half-full spoon may look like more like three quarters than two thirds.

Rafi
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Alternative: use two teaspoons.

Use water as you develop your skill. Fill tsp A, and pour into tsp B until the contents appear equal. Each now contains half a tsp. And now you know what half a tsp looks like in practice.

And you don't have to calculate cosines against thumb-sized hardware.