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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.)

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    $\begingroup$ after I answered, I found here that the answer is to fill it by the fraction $1-2\cos(\frac49\pi)$ $\endgroup$ – J. W. Tanner Sep 22 '19 at 21:51
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    $\begingroup$ 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. ;) $\endgroup$ – Xander Henderson Sep 23 '19 at 13:55
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    $\begingroup$ @RandomAspirant Do you need to comment that on every answer as well? $\endgroup$ – Todd Sewell Sep 23 '19 at 14:38
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    $\begingroup$ Just use a second spoon - fill the first one completely, then pour from it into the second until they're even... $\endgroup$ – twalberg Sep 23 '19 at 16:55
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    $\begingroup$ Do we allow housework problems? $\endgroup$ – Acccumulation Sep 23 '19 at 21:53
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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 $a\approx 65\%.$

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    $\begingroup$ I actually got $65.27...\%$ like other answers, but I don't think you could eyeball that precisely $\endgroup$ – J. W. Tanner Sep 22 '19 at 21:42
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    $\begingroup$ What the hell, call it $\frac23$. Everybody likes vanilla. $\endgroup$ – TonyK Sep 22 '19 at 21:46
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    $\begingroup$ And they say calculus is of no use in real life... $\endgroup$ – RandomAspirant Sep 23 '19 at 13:41
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    $\begingroup$ @TonyK eff it. Call it one teaspoon, LOL. Any good vanilla will only further enhance a recipe if you add a bit extra. $\endgroup$ – Doktor J Sep 24 '19 at 17:12
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    $\begingroup$ @DoktorJ but that principle leads to a divergent sequence! $\endgroup$ – JosephSlote Sep 24 '19 at 19:19
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There is actually an analytic solution to the problem, as shown below.

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

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

Set $V$ to half of the semisphere volume $\frac{2\pi}{3}r^3$ to obtain,

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

Compare with the identity $4\cos^3 x -3\cos x -\cos 3x=0$ and let $r/h = 2\cos x$ to obtain $x=40^\circ$.

Thus, the depth $h$ as a fraction of the radius $r$ Is

$$\frac hr = \frac{1}{2\cos40^\circ}$$

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    $\begingroup$ 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). $\endgroup$ – TonyK Sep 23 '19 at 16:16
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    $\begingroup$ @TonyK - I knew of the close-form result, but was also surprised of a different form from his, until convinced myself numerically $\endgroup$ – 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 $65.27\%$.

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    $\begingroup$ "It makes things a bit simpler if we turn your measuring spoon upside down" Then there isn't any liquid in the hemisphere. $\endgroup$ – Acccumulation Sep 23 '19 at 21:47
  • $\begingroup$ @Acccumulation One could easily turn it upside-down to measure it, then turn it right-side-up when filling it $\endgroup$ – user45266 Sep 23 '19 at 23:03
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    $\begingroup$ 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 :) $\endgroup$ – Ruslan Sep 24 '19 at 6:27
  • $\begingroup$ @Ruslan How did you find that? $\endgroup$ – Ovi Sep 25 '19 at 0:39
  • $\begingroup$ @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). $\endgroup$ – Ruslan Sep 25 '19 at 5:32
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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.

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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.

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    $\begingroup$ -1. Holly doesn't have two teaspoons. $\endgroup$ – TonyK Sep 26 '19 at 22:19
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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.

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