Answer:
D^2 = (x^2 + y^2) + z^2
and taking derivative of each term with respect to t or time, therefore:
2*D*dD/dt = 2*x*dx/dt + 2*y*dy/dt + 0 (since z is constant)
divide by 2 on both sides,
D*dD/dt = x*dx/dt + y*dy/dt
Need to solve for D at t =0, x (at t = 0) = 10 km, y (at t = 0) = 15 km
at t =0,
D^2 = c^2 + z^2 = (x^2 + y^2) + z^2 = 10^2 + 15^2 + 2^2 = 100 + 225 + 4 = 329
D = sqrt(329)
Therefore solving for dD/dt, which is the distance rate between the car and plane at t = 0
dD/dt = (x*dx/dt + y*dy/dt)/D = (10*190 + 15*60)/sqrt(329) = (1900 + 900)/sqrt(329)
= 2800/sqrt(329) = 154.4 km/hr
154.4 km/hr
Step-by-step explanation:
