Expert: Scotto - 4/14/2008

Question
I have come to a bit of a standstill with the following problem, in fact I have spent the past 4 days toiling with this.

Part c of a question states "Find patrametric equations for the line of flight of the aeroplane. Your equations should be in terms of the parameter t, and should be such that the aeroplane is at city A when t=0 and at city B when t=1".

The aeroplane is at (300,-200) at city A and at (-100, 600) at city B. I have worked out the equation to be:

x= -400t + 300 and y= 800t - 200

Howver part c (ii) states "write down an expression, in terms of t, for the square of the distance between the air traffic control centre (300,0) and the aeroplane at the point with parameter t on the line of flight of the aeroplane. Simplify your answer.

(The distance required here, and in part c (iii) is the horizontal distance, that is, the distance between the air traffic control centre and the point on the ground immediately below the aeroplane).

This leads on to c (iii). Use your answer to part c (ii) and the method of competing the square, to determine the distance, to the nearest kilometre, between the air traffic control centre and the aeroplane at the point on the line of flight of the aeroplane where it is closest to the air traffic control centre.

If you could point me in the right direction here, I would be grateful for any help.

Regards

Ray Carter

Answer
This is a right triangle.  The square of the distance can be found as x^2 + y^2, putting in what x and y are in terms of t.

Take the square root of this function at time t to determine how far away the plane is at any time.

Hopefully this is enough to get you going.  If not, feel free to write again.  Best of luck!