# Implicit differentiation applications (exam tomorrow!)

• April 26th 2006, 03:23 PM
Implicit differentiation applications (exam tomorrow!)
Q: A plane flying with a constant speed of 24 km/min passes over a ground radar station at an altitude of 4 km and climbs at an angle of 30 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 3 minutes later?

A: I let the trangle sides be A, B, and C; A is the initial distance between
radar and the plane (4km); B is the distance between initial position of
the plane to 3 mins later (24km x 3min = 72); C is the final distance between radar to plane.

so I have C^2 = 4^2 + 72^2 - (2)(4)(72)cos(120 degrees)

Solved for C = 5488km, but the answer is wrong...

What was wrong? Thank you.
• April 26th 2006, 06:24 PM
Jameson
I think you're close. Remember you're solving for a rate, not a distance. A is the only value in the triangle that's constant, so B and C are variables as the plane flies. You have used the law of cosines correctly, but keep B as a variable and differentiate to find the rate of change.

$C^2=4^2+B^2-2(4)(B)\cos{(120)}$

$2C*\frac{dC}{dt}=2B*\frac{dB}{dt}-8\cos{(120)}$
• April 26th 2006, 08:45 PM