Question

Two airplanes are flying in the air at the same height. Airplane A is flying east at 300mi/h and airplane B is flying north at 200mi/h. If they are both heading to the same airport, located 90 miles east of airplane A and 120 miles north of airplane B, at what rate is the distance between the airplanes changing

Answer 1

Answer:

$\frac{dAB}{dt}=340 mil/h$

Step-by-step explanation:

The change of distance over time of the plain A is 300 mi/hour and 200 mi/hour for plane B. O is the point of the airport.

So, the distance from A to O AO = 90 miles and BO = 120 miles.

Now, we have a right triangle here.  We can use the Pythagorean theorem, so the distance between the planes will be:

$AB^{2} =AO^{2}+BO^{2}$ (1)

$AB =\sqrt{AO^{2}+BO^{2}}=$

$AB =\sqrt{90^{2}+120^{2}}=150 miles$

If we take the derivative of the equation (1) we could find the change of the distance between planes.

$2AB\frac{dAB}{dt}=2AO\frac{dAO}{dt}+2BO\frac{dBO}{dt}$

$2*150\frac{dAB}{dt}=2*90*300+2*120*200=102000 mil/h$

$\frac{dAB}{dt}=\frac{102000}{150*2}$

Finally,

$\frac{dAB}{dt}=340 mil/h$

I hope it helps you!