Related rates: The town of Redville and a plane

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wind

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The town of Redville is 1 km east of the regional airport's control tower. A road passes throught Redville, on which cars can travel north and south. A plane is flying north directly abouve the road. When the plain passes over Redville, its altitude is 2 km, its altitude is increasing at a rate of 20 m/s, and its speed is 300 km/h. Determine the rate at which the distance from the plane to the control tower is increasing when the plane is directly over Redville.

I drew a picture and there is a right triangle between the town, plane, and control tower, right?

Let c represent the distance between the control tower and the plane.
Let p represent the distance between the plane and Redville.
Let r represent the distance between Redville and the control tower.

. . .r^2 + p^2 = c ^2
. . .1+ 4 = c^2
. . .2.24 = 4

. . .d/dt 2r + d/dt 2p = d/dt 2c
. . .d/dt 2r + (20)(4) = d/dt 2c

Now what? I don't have dr/dt.

Thanks
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Edited by stapel -- Reason for edit: spelling, punctuation, capitalization, etc
 
It appears this is just another Pythagorean related rates problem.

Distance between town and tower=x

Height of plane=y

Distance between plane and tower=z


\(\displaystyle \L\\x^{2}+y^{2}=z^{2}\)

\(\displaystyle \L\\x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}\)

Enter in your given info and solve for dz/dt.
 
Given:

x = 1 km
y = 2 km
z = 2.24 km

20 (meters / second) = 72 km/h

dx/dt = 70 km/h
dy/dt = 300km/h
dzx/dt = ?

Enter those?
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Edited by stapel -- Reason for edit: spelling
 
Convert the 20 m to km: .02 km or 1/50 km

x is a constant.

\(\displaystyle \L\\(1)(0)+2(\frac{1}{50})=\sqrt{5}\frac{dz}{dt}\)
 
What difference does that make?. Put it into meters if you like.
 
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