求曲線r=aθ ，在θ=π/2 的斜率?
請問如何下手？
微積分『斜率』
2011-05-28 7:40 pm
回答 (4)
2011-05-28 8:43 pm
✔ 最佳答案
dy/dx = [(dr/dθ)sinθ + rcosθ]/[(dr/dθ)cosθ - rsinθ]
= [asinθ + aθcosθ]/[acosθ - aθsinθ]
= [sinθ + θcosθ]/[cosθ - θsinθ]
= (1 + 0)/(0 - π/2)
= -2/π
2011-05-28 23:17:25 補充：
Thank you for linch
2011-06-12 11:15 pm
提示y=rsinθ, x=rcosθ
2011-05-29 5:20 am
r = f(θ)
dy/dx
= (dy/dθ) / (dx/dθ)
= [d/dθ (r sinθ)] / [d/dθ (r cosθ)]
= [ (dr/dθ) * sin θ + r cosθ] / [ (dr/dθ) * cosθ - r sinθ ]
dy/dx
= (dy/dθ) / (dx/dθ)
= [d/dθ (r sinθ)] / [d/dθ (r cosθ)]
= [ (dr/dθ) * sin θ + r cosθ] / [ (dr/dθ) * cosθ - r sinθ ]
2011-05-28 9:33 pm
請問myisland大：
[(dr/dθ)sinθ + rcosθ]/[(dr/dθ)cosθ - rsinθ]
這行我實在不懂
麻煩講解一下
謝謝
[(dr/dθ)sinθ + rcosθ]/[(dr/dθ)cosθ - rsinθ]
這行我實在不懂
麻煩講解一下
謝謝
收錄日期: 2021-04-26 14:07:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110528000015KK02980