Related Rates Quesiton

Question

A road running north to south crosses a road going east to west at the point P. Car A is driving north along the first road, and an airplane is flying east above the second road. At a particular time the car is 15 kilometers to the north of P and traveling at 60 km/hr, while the airplane is flying at speed 190 km/hr 10 kilometers east of P at an altitude of 2 km. How fast is the distance between the car and the airplane changing?

Answer 1

To solve this problem, you need to first address the geometry. Point P is a reference distance point with value equal to 0 km. The north-south road the car is driving north on is defined as the ordinate or y-axis. At time 0, the north-bound car is at y = 15 km (directly north of P) with a velocity or rate, dy/dt = 60 km/hr.  The east-west road that the plane is flying east-bound 2 km directly above is defined as the abscissa or x-axis. At time t = 0, the east bound plane is at x = 10 km east of P and 2 km above at a velocity or rate, dx/dt = 190 km/hr.  To define the distance between the points on the two roads at t = 0, where the car is on the y-axis and the plane directly above the x-axis, with a right angle at P, using Pythagorean's theorem or identity, c^2 = x^2 + y^2, were x is x-axis, y is y-axis, and c is the hypotenuse.  However, we need to solve the rate of distance between the plane 2 km directly above the x-axis and the car on the y-axis.  So we need another correlation to solve problem. There is another right angle between the point directly below the plane and the c hypotenuse. The axis to the plane is defined as the z-axis. The plane is flying parallel to the x-axis at z = 2 km (constant).  Therefore, again using Pythagorean's theorem, with D defined as the hypotenuse or distance between the plane and car, z is the z-axis, and c is the lkower plane hypotenuse at a right angle to the z-axis.

Therefore, D^2 = c^2 + z^2, where we need to solve for dD/dt.

Substituting in for c^2 = x^2 + y^2 from first identity into second identity:

 

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