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darksider

原帖由 不死的ProG 于 2008/9/6 08:28 PM 发表


只有这种方法吗?

应该是吧。但我很肯定这个做法会比较容易。

详圣

这个也可以～不过他的比较容易～
cos3X- cos(90-X) = 0
-2sin((3X + 90 - X)/2)sin((3X - 90 + X)/2) = 0
sin [(2X +90)/2] = 0 or sin [(4X -90)/2] = 0
(2X + 90)/2 = 180n
(4X -90) /2 = 180n

[ 本帖最后由 详圣 于 6-9-2008 09:21 PM 编辑 ]

mathlim

90 ≠ 90°

90 = 90×(180°/π) ≈ 5156.62°

darksider

A hemisphere bowl of radius 12cm is initially full of water. Water runs out of a small hole at the bottom of the bowl at a rate of 48pi cm^3 s^-1. When the depth of the water is x cm , show that the depth is decreasing at a rate of 48/[x(24-x)] cm s^-1
Also, find the rate at which the depth is decreasing when

a) The bowl is full.
b)The depth is 6cm.

Another question is in this picture



Thanks in advance! Really urgent