Related Rates: A plane flying with a constant speed passes

[chucknorrisfish]

A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 7 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 2 minutes later?
I know you use law of cosines.

The answer i got was 14.22. But it isnt right.

dx/dt = 14 km/min
dy/dt = ?

x^2=y^2 + 7^2 - (14 y cos135) is what i started with.

 

[skeeter]

if the plane's speed is 14 km/min and it climbs at a 45 degree angle, then

\(\displaystyle \L \frac{dx}{dt} = \frac{dy}{dt} = \frac{14}{\sqrt{2}}\) km/min

but, I wouldn't go about it using those values. as you stated, use the law of cosines ...

let a = distance between the plane and where it started the climb
and \(\displaystyle \L \frac{da}{dt} = 14\) km/min

let b = distance between the plane and the radar station

\(\displaystyle \L b^2 = a^2 + 7^2 - 2(a)(7)\cos{(135)}\)

\(\displaystyle \L b^2 = a^2 + 7^2 + 7\sqrt{2}a\)

\(\displaystyle \L 2b\frac{db}{dt} = 2a\frac{da}{dt} + 7\sqrt{2}\frac{da}{dt}\)

\(\displaystyle \L \frac{db}{dt} = \frac{\frac{da}{dt}(2a + 7\sqrt{2})}{2b}\)

at t = 2 min, a = 28 km and b = approx 33.3 km

I get db/dt = approx 13.8 km/min

just fyi, the rate of climb (dy/dt) of this plane in more familiar units would be greater than 32000 ft/min ... we're talking advanced fighter aircraft performance here, especially at the altitude where it started.

 

[chucknorrisfish]

Thanks!

Thanks a lot. These types problems get tricky for me.