若MN=atan0+btanr
證明dr/d0=(-asec^2)/(bsec^2r)
f4,,a-maths微分應用
2007-04-23 3:05 am
回答 (2)
2007-04-23 3:15 am
✔ 最佳答案
MN is just a constant.MN = a tan θ + b tan r
Differentiating both sides with respect to θ,
0 = a sec^2 θ + (b sec^2 r)(dr/dθ)
-a sec^2 θ = (b sec^2 r)(dr/dθ)
dr/dθ = (-a sec^2 θ)/(b sec^2 r)
2007-04-23 3:14 am
MN=atan0+btanr
0 = asec^2(0) + bsec^2r．(dr/d0)
dr/d0 = -asec^2(0) / bsec^2r
2007-04-22 19:17:14 補充：
the above deduction occurs:MN , a , b are constant0,r are variable
0 = asec^2(0) + bsec^2r．(dr/d0)
dr/d0 = -asec^2(0) / bsec^2r
2007-04-22 19:17:14 補充：
the above deduction occurs:MN , a , b are constant0,r are variable
參考: eason mensa
收錄日期: 2021-04-23 21:09:11
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