Related Rates Problem

2008-10-18 10:10 am

I am having a lot of difficulty with this problem: A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is pi/3, this angle is decreasing at a rate of pi/6 rad/min. How fast is the plane traveling at that time?

thanks!

回答 (1)

用戶10012

2008-10-18 2:54 pm

✔ 最佳答案

Let the distance when the plane is right above the tracking telescope be zero and the distance always from the telescope be s.
Let the angle of elevation = x. Therefore,
5/s = tan x
(-5/s^2)(ds/dt) = (1/cos^2 x )(dx/dt).
Now x = pi/3, dx/dt = -pi/6, s = 5/tan x = 5/(tanpi/3) = 5/sqrt3. cos^2(pi/3) = 1/4.
Therefore,
(-5)(3/25)(ds/dt) = 4(-pi/6)
ds/dt = 20pi/18.
That is the speed of plane is 10pi/9 km/min.

2008-10-18 06:58:13 補充：
Correction: 2nd line should be 'distance away from the telescope be s.' That is let the telescope be the origin, coordinates of the plane be (s, 5) at all time.

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