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Yacht Distances and Vectors


Date: 09/05/98 at 20:43:47
From: Zoe Stokes
Subject: Distances between lines, vectors

I have been given the following assignment:

  Yacht A has initial position (-10,4) and has velocity vector [2,-1].
  Yacht B has initial position (3,-13) and has velocity vector [-1,3].

In this investigation you will plot the path of each yacht and 
determine the time when they are nearest and the shortest distance 
they are apart.

1. Explain why the position of each yacht at time "t" is given by 
   r(A)=[-10,4] + t[2,-1] and
   r(B)=[3,-13] + t[-1,3].
2. On squared paper plot the path of the yachts when t=O,1,2,3,4,5,...
3. Find the position vector of B relative to A.
4. Use 3. to show that if "d" is the distance between the yachts at 
   any time "t" then d^2 = 25t^2 - 214t + 458.
5. Show that d^2 is a minimum when t = 4.28.
6. Hence, find the time when the "d" is a minimum and then find the 
   shortest distance.

I have already done questions 1 and 2, but I am having problems 
understanding the rest. Could you please help me?

Thank you,
Zoe


Date: 09/06/98 at 11:32:01
From: Doctor Anthony
Subject: Re: Distances between lines, vectors

Question 3:

  The position of B relative to A is r(B) - r(A) = [13,-17] + t[-3,4], 
  which in (x,y) coordinates will be [(13-3t), (-17+4t)].

Question 4:

  The formula for finding the distance apart is:

     d = sqrt[(13-3t)^2 + (-17+4t)^2]

   Thus: d^2 = 169 - 78t + 9t^2 + 289 - 136t + 16t^2
             =  25t^2 - 214t + 458  

Question 5:

  To find the minimum of d^2, set the derivative equal to 0.
  So 50t - 214 = 0. Thus t = 214/50 = 4.28.

Question 6:

  The time when d is to be a minimum is the same time as when d^2 is a 
  minimum, so closest approach occurs at t = 4.28. So put t = 4.28 
  into the expression for d:

     d = sqrt[(13-3t)^2 + (-17+4t)^2]
       = sqrt[(0.16)^2 + (0.12)^2]
       = sqrt(0.04)
       =  0.2 miles 

- Doctor Anthony, The Math Forum
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Associated Topics:
High School Calculus
High School Linear Algebra

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