I would call the time after passing over the station as t minutes
then by cosine law
D^2 = 4^2 + (5t)^2 - (2)(4)(5t)cos125
= 25t^2 * (40cos125)t + 16
I usually don't take the square root, but rather do it implicitly, so
2D(dD/dt) = 50t + 40cos125
dD/dt = (25t + 20cos125)/D
when t=10
D^2 = 2500 - 400cos125 + 16
D = 47.818
and dD/dt = (250 + 20cos125)/47.818
= 4.988 km/min
A plane flying with a constant speed of 5 km/min passes over a ground radar station at an altitude of 4 km and climbs at an angle of 35 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 10 minutes later?
I got it down to [250-20cos(125)]/[sqrt(40000-400cos(125))] but it's wrong.
4 answers
Not right for some reason.
re-checked my calculations, found a few typing and calculator errors. The old brain just doesn't work properly after midnight
NEW VERSION:
I would call the time after passing over the station as t minutes
then by cosine law
D^2 = 4^2 + (5t)^2 - (2)(4)(5t)cos125
= 25t^2 - (40cos125)t + 16
I usually don't take the square root, but rather do it implicitly, so
2D(dD/dt) = 50t - 40cos125
dD/dt = (25t - 20cos125)/D
when t=10
D^2 = 2500 - 400cos125 + 16
D = 47.818
and dD/dt = (250 - 20cos125)/52.3969
= 4.99 km/min
check my calculations
NEW VERSION:
I would call the time after passing over the station as t minutes
then by cosine law
D^2 = 4^2 + (5t)^2 - (2)(4)(5t)cos125
= 25t^2 - (40cos125)t + 16
I usually don't take the square root, but rather do it implicitly, so
2D(dD/dt) = 50t - 40cos125
dD/dt = (25t - 20cos125)/D
when t=10
D^2 = 2500 - 400cos125 + 16
D = 47.818
and dD/dt = (250 - 20cos125)/52.3969
= 4.99 km/min
check my calculations
Ahh, there we go. Can't believe such a little number could cause it to be wrong. Thanks.