Darwin’s theory of evolution tells us that nature always selects the species best adapted to the environment from among numerous organisms, granting them a higher chance of survival. Over time, through hundreds of millions of years of "survival of the fittest," these species have evolved into the diverse and peculiar organisms we see today. Undoubtedly, through long-term selection, superior traits are accumulated. In other words, these species have reached an optimal or near-optimal state in terms of certain environmental adaptation capabilities (which is, again, an extremum problem). Now, let us consider the mushroom.
A mushroom is a fungal organism that generally grows in dark and humid environments. Since it prefers moisture, it naturally does not want to lose excessive water. Therefore, it strives to adjust its shape to minimize "water loss." Assuming the rate of water loss per unit area of the mushroom is consistent, the problem transforms into making the surface area of a solid of a given volume as small as possible. Furthermore, considering the issue of horizontal isotropic growth, the ideal mushroom shape should be a solid of revolution of a planar figure. So, what is this solid of revolution? Have you, being clever, thought of it as (a part of) a sphere?
Unfortunately, the answer is not a sphere. Let us analyze this problem: which curve, when rotated, produces the minimum surface area, given that the curve passes through (x_1, y_1) and (x_2, y_2).
If the curve y=f(x) is known to satisfy the conditions, and since the two bases are circles of known area, we only need to consider the lateral surface area. The formula for calculating the lateral surface area of a solid of revolution is: \begin{aligned}S&=\int_{x_1}^{x_2} 2\pi x\sqrt{dx^2+dy^2} \\ &=2\pi\int_{x_1}^{x_2} x\sqrt{1+\dot{y}^2}dx\end{aligned}
Here \dot{y}=\frac{dy}{dx}. The meaning of the above expression is to infinitely divide the solid, treat each part as a frustum of a cone, approximate the lateral surface area using the frustum’s formula, and accumulate them.
At this point, the problem becomes finding a function y=f(x) that makes the integral S=2\pi\int_{x_1}^{x_2} x\sqrt{1+\dot{y}^2}dx an extremum. According to the Euler-Lagrange equation, we should have: \frac{d}{dx}\left(\frac{\partial \left(x\sqrt{1+\dot{y}^2}\right)}{\partial \dot{y}}\right)=\frac{\partial \left(x\sqrt{1+\dot{y}^2}\right)}{\partial y}=0
Therefore, we have \frac{\partial (x\sqrt{1+\dot{y}^2})}{\partial \dot{y}}=\frac{x\dot{y}}{\sqrt{1+\dot{y}^2}}=C_1, where C_1 is an integration constant. From this, we can derive: \dot{y}=C_1\sqrt{\frac{1}{x^2-C_1^2}}
By integrating both sides, we obtain: y=C_1 \ln\left|x+\sqrt{x^2-C_1^2}\right|+C_2. Alternatively, using hyperbolic functions, this can be rewritten as: x=C_1 \cosh\left(\frac{y-C_2}{C_1}\right)
As we can see, this is not a circle. What kind of shape is this? If we set C_1=2, C_2=0, we can plot it using The Geometer’s Sketchpad:
Clearly, this shape is quite close to that of a mushroom. On the other hand, the hyperbolic function contains the magical constant e, the base of the natural logarithm. This means that e originates from nature and fully embodies nature! What a harmonious world!
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