已知直角坐标求极坐标怎么判断极径的正负 直角坐标中的直线化极坐标时极径积分限函数如何表达

\u6781\u5750\u6807\u4e2d\u7684\u6781\u5f84\u600e\u4e48\u6c42\uff1f

\u6781\u5f84\u5c31\u662f\u6781\u5750\u6807\u4e0a\uff0c\u5bf9\u4e8e\u5e73\u9762\u4efb\u610f\u4e00\u70b9M,\u8fde\u63a5OM\u7684\u90a3\u6761\u7ebf\uff0c\u4e3a\u6781\u5f84\uff0c\u82e5\u77e5\u9053OM\u4e0eOX\u7684\u5939\u89d2A\uff0cX\u503c=\u6781\u5f84*cosA,Y\u503c=\u6781\u5f84*sinA

\u6cd5\u4e00\uff1a
\u5148\u5199\u51fa\u5706\u7684\u76f4\u89d2\u5750\u6807\u65b9\u7a0b\uff1a(x-a)²+y²=a²\uff0c\u5373x²+y²-2ax =0
\u518d\u5229\u7528\u5173\u7cfb\u5f0f\uff1a\u03c1²=x²+y²\uff0ctan\u03b8=y/x\uff08x\u22600\uff09\uff0c\u901a\u5e38\u53d6\u03c1\u22650,0\u2264\u03b8<2\u03c0.
\u4ee3\u5165\u53d8\u5f62\u5f97\u5230\u5706\u7684\u6781\u5750\u6807\u65b9\u7a0b\uff1a\u03c1=2acos\u03b8.

\u6cd5\u4e8c\uff1a
\u5728\u76f4\u89d2\u5750\u6807\u7cfb\u4e2d\u4f5c\u51fa\u4ee5M\uff08a\uff0c0\uff09\u4e3a\u5706\u5fc3\uff0c\u4ee5a\u4e3a\u534a\u5f84\u7684\u5706\uff0c
\u8bbe\u5750\u6807\u539f\u70b9\u4e3aO\uff0c\u5706\u4e0ex\u8f74\u6b63\u534a\u8f74\u4ea4\u4e8e\u70b9A\uff0c\u5706\u4e0a\u4e00\u70b9P\uff0c\u8fde\u7ed3OP\uff0cAP\uff0c
\u5219OA=2a\uff0cOP=OAcos\u2220POA=2acos\u2220POA\uff08*\uff09\uff0c
\u4ee5O\u4e3a\u6781\u70b9\uff0c\u4ee5x\u8f74\u6b63\u534a\u8f74\u4e3a\u6781\u8f74\uff0c\u5efa\u7acb\u6781\u5750\u6807\u7cfb\uff0c
\u8bbe\u70b9P\uff08\u03c1\uff0c\u03b8\uff09
\u5219\u03c1=OP\uff0c\u03b8=\u00b1cos\u2220POA\uff0c
\u4ee3\u5165\uff08*\uff09\uff0c\u5f97\u03c1=2acos\u03b8.

\u6cd5\u4e00\u5b9e\u9645\u662f\u4e00\u79cd\u6362\u5143\u601d\u60f3\uff0c\u4fbf\u4e8e\u8fd0\u7528\uff1b\u6cd5\u4e8c\u662f\u7ed3\u5408\u51e0\u4f55\u610f\u4e49\uff0c\u9700\u8981\u5f88\u597d\u7684\u7406\u89e3\u6781\u5f84\u548c\u6781\u89d2\u7684\u6982\u5ff5.\u5e0c\u671b\u5bf9\u4f60\u6709\u6240\u5e2e\u52a9.


如图所示。



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