Plane

[legacyofpiracy]

Sorry to keep bugging all of you guys, I am just having trouble with this problem.

A plane flying with a constant speed of 24 km/min passes over a ground radar station at an altitude of 15 km and climbs at an angle of 45 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 3 minutes later?



there was a hint thing at the bottom of the problem that said: The law of cosines for a triangle is

c^2=a^2+b^2-2abcos(theta)

where is the angle between the sides of length a and b.
- but I havent gotten to that point yet.

Firstly I was just wondering if this picture would work for the problem

 

[stapel]

It looks like some stuff got dropped during your copy-n-pasting. Working backwards from the picture, I will be taking the exercise as being the following:

A plane flying with a constant speed of 24 km/min passes over a ground radar station at an altitude of 15 km and climbs at an angle of [can't figure this out] degrees. At what rate, in km/min, is the distance from the plane to the radar station increasing minutes later?

Please reply with confirmation and/or correction.

Thank you!

Eliz.

 

[legacyofpiracy]

whoops sorry about that, I edited the numbers in :roll:

 

[stapel]

Thank you!

Okay, if the angle is 45°, then we know that the sides of the triangle will be increasing 1/sqrt(2)-times as fast as the hypotenuse is growing. That is, if we label the height of the triangle as "y" and the base of the triangle as "x", with the hypotenuse being "h", then dh/dt = 24 while dy/dt = dx/dt = 24/sqrt(2).

If you draw in another "hypotenuse" line D from the station (lower left) to the plane (upper right), you'll have a base equal to x and a height equal to 15 + y, with dy/dt = dx/dt = 24/sqrt(2).

Use this info to find the position (in terms of the horizontal and vertical distances x and 15 + y) of the plane with respect to the station three minutes after passing overhead.

Then note that, whatever the angle \(\displaystyle \theta\) is at the station:

. . . . .\(\displaystyle \Large{\tan{(\theta)}\mbox{ }=\mbox{ }\frac{15\mbox{ }+\mbox{ }y}{x}}\)

Find the value of \(\displaystyle \theta\) at time t = 3. Differentiate the above with respect to time t, plug in all the known values, and solve for the angle in the angle.

Hope that helps.

Eliz.

 

[Gene]

From the hint I think they want you to use the law of cosines where
theta = 90+45=135° (not the 45° you have shown)
a=15
b=24t
c=the distance from station at time t
Then you have to find dc/dt