题目
Problem
The variables x and y satisfy the equation y3=Aekx, where A and k are constants. The graph of lny against x is a straight line passing through the points (−1.32,5.09) and (0.25,1.16).
Find the values of A and k.
(5)
题目中文翻译
变量 x 和 y 满足方程 y3=Aekx,其中 A 和 k 为常数。以 x 为横坐标、lny 为纵坐标的图像是一条经过点 (−1.32,5.09) 和 (0.25,1.16) 的直线。
求 A 和 k 的值。
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
解法一
思路
展开
纵坐标是 lny。对原方程取对数后,直线的斜率为 k/3、截距为 lnA/3。
答题过程
展开
Taking logarithms gives
3lny=lnA+kx.
The slope of the given line is
3k=0.25−(−1.32)1.16−5.09=−1.573.93.
Thus k=−7.509554…≈−7.51. Using (0.25,1.16),
lnA=3(1.16)−0.25k=5.357388….
Hence A=e5.357388…=212.170…≈212.
解法二
思路
展开
也可先由图中两点恢复 y,直接代入指数方程。将两个方程相除,消去 A。
答题过程
展开
The two values of y are e5.09 and e1.16. Substituting them into the original equation gives
e15.27=e3.48=Ae−1.32k,Ae0.25k.
Dividing the second by the first gives e−11.79=e1.57k, so k=−11.79/1.57≈−7.51. Then A=e3.48−0.25k≈212.