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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?

1 Answer
Feb 16, 2018

#154.37# #Km#/#hr#

Explanation:

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The picture below depicts the the locations of the Car and the Airplane.

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Please note the purple arrow showing the direction of our view as we look at the right triangle #Delta ABD#. That view is shown below:

enter image source here

From the top picture, we can see that #Delta APD# is a right triangle. We can use the Pythagoras' formula to calculate the distance #AD#:

#(AD)^2=(AP)^2+(PD)^2# #color(red)(Equation 1)#

#(AD)^2=15^2+10^2=225+100=325#

#AD=sqrt325# #Km#

Now, we can use the lower picture and calculate the distance #AB# which is the distance between the Car and the Airplane at the moment described in the problem:

#(AB)^2=(AD)^2+(BD)^2# #color(red)(Equation 2)#

#(AB)^2=325+4=329#

#AB=sqrt329# #Km#

Let's plug the right hand side of equation 1 into equation 2 to substitute for #(AD)^2#:

#(AB)^2=(AP)^2+(PD)^2+(BD)^2#

For simplicity, let's let:

#AB=l#

#AP=y#

#PD=x#

#BD=z#

Then our equation becomes:

#l^2=y^2+x^2+z^2#

We need to find the rate of change of distance #l# at the moment specified in the problem. As such, we will take the derivative of the entire equation (both sides) with respect to time #t#:

#2l(dl)/dt=2y(dy)/dt+2x(dx)/dt+2z(dz)/dt#

We know that, at the specified moment, the car is traveling at #60# #Km#/#hr# and is #15# #Km# north of point #P#. Therefore:

#y=15# and #dy/dt=60#

The airplane is traveling at #190# #Km#/#hr# and is #10# #Km# east of point #P#. Therefore:

#x=10# and #dx/dt=190#

The airplane is #2# #Km# above the ground and that altitude is constant. Therefore:

#z=2# and #(dz)/dt=0#

Now, let's plug in known values at the specified moment:

#2(sqrt329)(dl)/dt=2(15)(60)+2(10)(190)+2(2)(0)#

#2sqrt329(dl)/dt=1800+3800+0=5600#

#(dl)/dt=5600/(2sqrt329)=154.37# #Km#/#hr#

At the specified moment, the distance between the car and the airplane is changing at the rate of #154.37# #Km#/#hr#.