这是一份exeter埃克塞特大学MTH1001/MTH1001-JD作业代写的成功案例
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
$$
f(t)=n f\left(\frac{1}{n} t\right)
$$
which read
$$
f\left(\frac{1}{n} t\right)=\frac{1}{n} f(t)
$$
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-JD COURSE 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
$$
f(h)=\frac{1}{2} b h .
$$