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IAL 2021 Oct Q7

A Level / Edexcel / P3

IAL 2021 Oct Paper · Question 7

题目

Problem

The mass, MM kg, of a species of tree can be modelled by the equation

log⁡10M=1.93log⁡10r+0.684\log_{10}M=1.93\log_{10}r+0.684

where rr cm is the base radius of the tree.

The base radius of a particular tree of this species is 4545 cm.

According to the model,

(a) find the mass of this tree, giving your answer to 2 significant figures.

(2)

(b) Show that the equation of the model can be written in the form

M=prqM=pr^q

giving the values of the constants pp and qq to 3 significant figures.

(3)

(c) With reference to the model, interpret the value of the constant pp.

(1)
题目中文翻译

某种树的质量 MM（单位：kg）可由方程

log⁡10M=1.93log⁡10r+0.684\log_{10}M=1.93\log_{10}r+0.684

建模，其中 rr cm 是树干底部半径。

某一棵这种树的底部半径是 4545 cm。

根据该模型，

(a) 求这棵树的质量，答案保留 2 位有效数字。

(b) 证明模型方程可以写成

M=prqM=pr^q

的形式，并把常数 pp 与 qq 的值保留 3 位有效数字。

(c) 结合模型，解释常数 pp 的含义。

解答

(a)

解法一

思路

展开

直接把 r=45r=45 代入对数模型，先求 log⁡10M\log_{10}M，再取以 1010 为底的反对数。

答题过程

展开

Substituting r=45r=45 into the model gives

log⁡10M=1.93log⁡1045+0.684.\log_{10}M =1.93\log_{10}45+0.684.

Therefore

log⁡10M=3.8747…,\log_{10}M=3.8747\ldots,

so

M=103.8747…=7492.9….M=10^{3.8747\ldots} =7492.9\ldots.

Hence, to 2 significant figures,

M=7500 kg.\boxed{M=7500\text{ kg}}.

解法二

思路

展开

官方评分资料也接受使用 (b) 得到的幂函数模型。为避免过早舍入，计算时保留 p=100.684p=10^{0.684} 的准确形式。

答题过程

展开

Using the equivalent model from part (b),

M=100.684r1.93.M=10^{0.684}r^{1.93}.

When r=45r=45,

M= 100.684×451.93= 7492.9….\begin{align*} M=&\,10^{0.684}\times45^{1.93} \\[4mm] =&\,7492.9\ldots. \end{align*}

Therefore, to 2 significant figures,

M=7500 kg.\boxed{M=7500\text{ kg}}.

(b)

解法一

思路

展开

把等式右边两项分别写成常用对数，再用对数加法公式合并，最后比较 M=prqM=pr^q 的系数和指数。

答题过程

展开

Starting from

log⁡10M=1.93log⁡10r+0.684,\log_{10}M =1.93\log_{10}r+0.684,

use the laws of logarithms:

log⁡10M= log⁡10(r1.93)+log⁡10(100.684)= log⁡10(100.684r1.93).\begin{align*} \log_{10}M =&\,\log_{10}\big(r^{1.93}\big) +\log_{10}\big(10^{0.684}\big) \\[4mm] =&\,\log_{10}\big(10^{0.684}r^{1.93}\big). \end{align*}

Therefore

M=100.684r1.93.M=10^{0.684}r^{1.93}.

Comparing this with M=prqM=pr^q gives

p=100.684=4.831…andq=1.93.p=10^{0.684}=4.831\ldots \qquad\text{and}\qquad q=1.93.

Hence, to 3 significant figures,

p=4.83,q=1.93.\boxed{p=4.83,\qquad q=1.93}.

(c)

解法一

思路

展开

在 M=prqM=pr^q 中令 r=1r=1，即可看出 pp 对应哪一种树的质量。

答题过程

展开

When r=1r=1,

M=p(1)q=p.M=p(1)^q=p.

Therefore pp represents the predicted mass, in kilograms, of a tree of this species whose base radius is 11 cm.