题目

Problem

A dose of antibiotics is given to a patient.

The amount of the antibiotic, $x$ milligrams, in the patient’s bloodstream $t$ hours after the dose was given, is found to satisfy the equation

\log_{10}x=2.74-0.079t

(a) Show that this equation can be written in the form

x=pq^{-t}

where $p$ and $q$ are constants to be found. Give the value of $p$ to the nearest whole number and the value of $q$ to 2 significant figures.

(4)

(b) With reference to the equation in part (a), interpret the value of the constant $p$ .

(1)

When a different dose of the antibiotic is given to another patient, the values of $x$ and $t$ satisfy the equation

x=400\times 1.4^{-t}

\frac{dx}{dt}

when $t=5$

(3)

题目中文翻译

给一位病人服用了一剂抗生素。

给药后 $t$ 小时，病人血液中的抗生素含量为 $x$ 毫克，并满足方程

\log_{10}x=2.74-0.079t

(a) 证明该方程可以写成

x=pq^{-t}

的形式，其中 $p$ 和 $q$ 是需要求出的常数。把 $p$ 的值取最接近的整数，把 $q$ 的值保留 2 位有效数字。

(b) 结合 (a) 中的方程，解释常数 $p$ 的含义。

当给另一位病人服用不同剂量的抗生素时， $x$ 和 $t$ 满足方程

x=400\times 1.4^{-t}

\frac{dx}{dt}

的值，答案保留 2 位有效数字。

解答

We are given

\log_{10}x=2.74-0.079t

Raise $10$ to the power of both sides:

x=10^{2.74-0.079t}

x=10^{2.74}\cdot 10^{-0.079t}

Now

10^{-0.079t}=(10^{0.079})^{-t}

Therefore

x=10^{2.74}(10^{0.079})^{-t}

This is of the form

x=pq^{-t}

where

p=10^{2.74}, \qquad q=10^{0.079}

Now

p=549.540\ldots

and

q=1.199\ldots

Hence

\boxed{p=550,\qquad q=1.2}

x=pq^{-t}

when $t=0$ ,

x=p

Therefore $p$ represents the amount of antibiotic, in milligrams, in the patient’s bloodstream immediately after the dose was given.

For the second patient,

x=400\times 1.4^{-t}

求导得

\frac{\mathrm{d}x}{\mathrm{d}t} =400\cdot(-\ln1.4)\cdot1.4^{-t}

\frac{\mathrm{d}x}{\mathrm{d}t} =-400\ln1.4\cdot1.4^{-t}

When $t=5$ ,

\frac{\mathrm{d}x}{\mathrm{d}t} =-400\ln1.4\cdot1.4^{-5}

Thus

\frac{\mathrm{d}x}{\mathrm{d}t} =-25.02\ldots

To 2 significant figures,

\boxed{\frac{\mathrm{d}x}{\mathrm{d}t}=-25}

This means the amount of antibiotic is decreasing at about $25$ milligrams per hour when $t=5$ .