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本人对高中课本数学第二册P。182例8的结论有一个疑问。例3的原题是:“化圆的直角坐标方程χ~2+y~2-2aχ=0为极坐标方程”。答案是:p=2acoSθ,这是没有问题的。问题在最后的小结:“这个方程和上节例2圆的极坐标方程是相同的。”我们认为这个方程和上节例2圆的极坐标方程是不相同的。“上节例2”是指该书P。179的例2,其原题是:“求圆心是C(a,o),半径是a的圆的极坐标方程。”这和例3不同之点在于指出了a为半径——a是非负实数!而例3中的a∈R(R为实数集),由于这个区别,所提及例2和倒3的圆的半径分别为a和|a|,其图形也不相同如下图,左是例2的图,右是例3的图。当然,这个问题有的读者会认为不值一谈,
I am in high school textbook mathematics second volume. There is a doubt about the conclusion of 182 cases. The original question of Example 3 is: “The rectangular equation of the circle χ~2+y~2-2aχ=0 is a polar coordinate equation.” The answer is: p=2acoSθ, which is no problem. The final summary of the question: “This equation is the same as the polar coordinate equation of the circle in the previous section.” We think this equation is not the same as the polar coordinate equation of the circle in the previous section. “Case 2” refers to the book P. In Example 2 of 179, the original title is: “The center of the circle is C (a, o), and the radius is the polar coordinate equation of a.” This difference from Example 3 is that a is a radius - a is non-negative Real number! In example 3, a∈R (R is a set of real numbers). Due to this difference, the radiuses of the circles mentioned in example 2 and inverted 3 are a and |a|, respectively. It is a diagram of Example 2, and right is a diagram of Example 3. Of course, some readers of this issue will find it worth noting