Thread: Implicit differentiation applications (exam tomorrow!)

1. 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.

2. 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.

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

$\displaystyle 2C*\frac{dC}{dt}=2B*\frac{dB}{dt}-8\cos{(120)}$

3. Okay, I got what you are doing, but I don't understand how to solve for dC/dt

The question is asked to find dC/dt when 3 mins after, so I will have find the value of B and C 3 mins later in order to find the rate of change of C.

B after 3 mins is 72km, what is C, 5488km?