# RP^2=RQ^2 + QP^2 -2(RQ)*(QP)cosQ (1)

QP = 24 (km/min) t (2)

Diffentiate both sides w.r.t. the time t:

2 RP* d Rp/dt = 2 t [24 (km/min)]^2 -2(RQ)* 24 (km/min) cosQ (3)

Using (1) and (2) you find out what RP is at t = 4 minutes. You plug that into (3) and solve for d Rp/dt.

A plane flying with a constant speed of 24 km/min passes over a ground radar station at an altitude of 14 km and climbs at an angle of 25 degrees.

The situation is depicted in the picture below where the radar station is the point R and the point Q is the airplane as it passes the station.

The point P indicates the position of the airplane at some later time.

At what rate, in km/min, is the distance from the plane to the radar station increasing 4 minutes later?

draw the triangle
RQP

YOu know the angle Q, it is constant. (90+25).

Law of cosines:

RP^2=RQ^2 + QP^2 -2(RQ)*(QP)cosQ
Take the derivative of all this with respect to time, and you should be able to solve for dRP/dt.

thanks....what confuses me...is how do I solve for it...with there being no change on the one side that has 24 km/min...

what do I use for the rate there?

No. QP is the distance of QP at four min (24km/min)4min. dQP/dt is the rate 24km/min.

Here, I will diff it for you.

RP^2=RQ^2 + QP^2 -2(RQ)*(QP)cosQ (1)
2RP*dRP/dt = 0 + 2QP dQP/dt -2RQ*dQP/dt*cosQ
check that.
everything is known except dRP/dt, and that is what you are looking for. You will have to calculate the distance RP from the triangle (at the 4 min mark).

I got it guys, thanks. :-D