数学结构 Mathematical Structures MTH1001/MTH1001-JD

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Let
$$s=f(t)$$
be the distance as a function $f$ of the time $t$. We just noticed:
$$f(n t)=n f(t) \quad \text { for } n \in \mathbf{N} .$$
Replace $t$ with $\frac{1}{n} t$.Then
$$f(t)=n f\left(\frac{1}{n} t\right)$$

$$f\left(\frac{1}{n} t\right)=\frac{1}{n} f(t)$$
yields
in $\frac{1}{n}$ of the time $\frac{1}{n}$ of the distance is covered.
If in the last formula $t$ is replaced with $m t(m \in N)$ one gets
$$f\left(\frac{m}{n} t\right)=\frac{1}{n} f(m t)=\frac{m}{n} f(t)$$

MTH1001/MTH1001-JDCOURSE NOTES ：

Take similar triangles with their areas as a function $f$ of their – variable height $h$,
$$f(h)=\alpha h^{2} \quad \text { with fixed } \alpha ;$$
the difference
$$f(h+\delta)-f(h)$$
is the area of a strip of height $\delta$ on the base, the difference quotient
$$\frac{f(h+\delta)-f(h)}{\delta}$$
approximately equals the base, and the differential quotient
$$\frac{\mathrm{d} f}{\mathrm{~d} h}=2 \alpha h$$
equals the base,
$$b=2 \alpha h .$$
Thus indeed
$$f(h)=\frac{1}{2} b h .$$