题目
Problem
The resting heart rate, h beats per minute, of a particular group of mammals, can be modelled by the equation
log10h=2.4−0.25log10m
where m kg is the mass of the mammal.
The mass of one of these mammals is 3 kg.
According to the model,
(a) find the resting heart rate of this mammal. Give your answer to 3 significant figures.
(2)
(b) Show that the equation of the model can be written in the form
h=mqp
giving each of the values of the constants p and q to 3 significant figures where appropriate.
(3)
(c) With reference to the model, interpret the value of the constant p.
(1)
题目中文翻译
某一类哺乳动物的静息心率 h(单位:次/分钟)可由下式建模:
log10h=2.4−0.25log10m
其中 m kg 表示该哺乳动物的质量。
现有一只该类哺乳动物,其质量为 3 kg。
根据模型,
(a) 求这只哺乳动物的静息心率,答案保留 3 位有效数字;
(b) 证明模型方程可写成
h=mqp
的形式,并求常数 p,q 的值,在适当处保留 3 位有效数字;
(c) 结合模型实际意义,解释常数 p 的含义。
解答
(a)
解法一
思路
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把 m=3 代入模型:
log10h=2.4−0.25log103
然后用 10(⋅) 消去 log10。
答题过程
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Substitute m=3 into the model.
log10h=2.4−0.25log103
Therefore
h==102.4−0.25log103190.862…
So the resting heart rate is
191
beats per minute, to 3 significant figures.
(b)
解法一
思路
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用对数法则:
klog10m=log10(mk)
并且
2.4=log10(102.4)
把右边合并成一个对数后,就可以去掉 log10。
答题过程
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Starting from
log10h=2.4−0.25log10m
write
2.4=log10(102.4)
and
−0.25log10m=log10(m−0.25)
So
log10h==log10(102.4)+log10(m−0.25)log10(102.4m−0.25)
Hence
h=102.4m−0.25
Since
102.4=251.188…
we get
h=m0.25251
to 3 significant figures where appropriate.
Therefore
p=251,q=0.25
(c)
解法一
思路
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在
h=mqp
中,若 m=1,则 mq=1,所以 h=p。因此 p 表示质量为 1 kg 时模型预测的静息心率。
答题过程
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When m=1,
h=1qp=p
Therefore p represents the resting heart rate, in beats per minute, of a mammal of mass 1 kg according to the model.