微積分-微分量 求詳細過程及解答... 一牽引機以 11.8ft/s 之速率離開一高 90.8ft之建築物.則當牽引機離建築物之底部 151ft 時它與建築物之頂端距離變化率為何?

2015-11-23 3:53 pm

回答 (2)

2015-11-24 1:19 am
✔ 最佳答案
dh=[90.8^2+(151+11.8dt)^2]^(1/2)-(90.8^2+151^2)^(1/2)
=[90.8^2+151^2+2*151*11.8dt+(11.8dt)^2]^(1/2)-31045.64^(1/2)
~(31045.64+3563.6dt)^(1/2)-176.1977
=(31045.64)^(1/2)*[1+(3563.6/31045.64)dt]^(1/2)-176.1977
~176.1977*[1+(3563.6/31045.64)dt/2]-176.1977
~176.1977+176.1977*(3563.6/62091.28)dt-176.1977
=(176.1977*3563.6/62091.28)dt
=10.1125dt
dh/dt=10.1125ft/sec
2015-11-24 11:40 am
10.1125ft/sec


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