Question 88·Hard·Advanced Math

The water forms a smaller cone with the same tip as the whole cone. A vertical slice through both makes similar triangles, triangles with the same shape. Their matching radius-to-height ratios are equal. Call the water’s surface radius $r$. For $x>0$, write: $\frac{r}{x}=\frac{9}{12}$.

The water’s radius is not always the cone’s $9$-centimeter rim radius. Multiply the proportion by $x$: $r=\frac{9x}{12}$. Reduce the fraction: $r=\frac{3x}{4}$. The water’s radius grows in the same proportion as its depth.

A cone’s volume is $V=\frac13\pi r^2h$, where $h$ is its height. The water’s height is $x$, not $12-x$; $12-x$ is the empty height above it. Substitute $r=\frac{3x}{4}$ and $h=x$: $V(x)=\frac{\pi}{3}\left(\frac{3x}{4}\right)^2x$.

Separate the numbers from the powers of $x$: $V(x)=\pi\left(\frac13\left(\frac34\right)^2\right)x^2x$. Type $\frac13\left(\frac34\right)^2$ in Desmos. It shows $0.1875$; the fraction button gives $\frac{3}{16}$. That is the coefficient multiplying $\pi x^2x$.

The squared radius contributes $x^2$, and the water’s height contributes another $x$. Multiply the powers: $x^2\cdot x=x^3$. So the volume of water is $V(x)=\frac{3\pi}{16}x^3$ cubic centimeters, for $0\le x\le12$. Choice D.

Instead of finding the water’s radius, compare the two similar cones. The linear scale factor, the ratio of matching lengths from the whole cone to the water, is $\frac{x}{12}$. Volume uses its cube because it measures three dimensions. If $V_{\text{full}}$ is the whole cone’s volume, then: $V(x)=V_{\text{full}}\left(\frac{x}{12}\right)^3$.

The whole cone has radius $9$ and height $12$. Type $\frac13(9)^2(12)$ in Desmos for the numerical part of its cone volume. Desmos shows $324$, so $V_{\text{full}}=324\pi$.

Cubing $\frac{x}{12}$ gives $\frac{x^3}{12^3}$. Type $\frac{324}{12^3}$ on a new Desmos line; it shows $0.1875$, or $\frac{3}{16}$ with the fraction button. Keep the $x^3$ from the scale factor, so the water’s volume is $V(x)=\frac{3\pi}{16}x^3$ cubic centimeters. Choice D.