How do write parametric equations for Margot's position t seconds after s if Margot is walking in a straight line from a point 30 feet due east of a statue from a point 24 feet due north of the statue and walks at a constant speed of four feet per second?

1 Answer
Jan 29, 2017

The equations are:

# x(t) = -30 - (20sqrt(41))/41 t #
# y(t) = (16sqrt(41))/41 t #

Explanation:

I prefer to use vectors for this type of question;

Assume the statue is at the origin #(0,0)#, Margot starts at a position #30#ft due east, or coordinate #(30,0)# and walks to a position #24#ft due north, or coordinate #(24,0)#.

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The direction of travel is given by the vector:

# ( (-30),(24) )#, or # vec(d)= ( (-5),(4) )#

In order to find the velocity we need to scale this direction vector by an appropriate factor, #lamda#, such that # abs(lamda vec(d)) # = 4 (the given speed).

ie we require:

# sqrt((-5lamda)^2 + (4lamda)^2)) = 4 #
# :. sqrt(25 lamda^2 + 16 lamda^2) = 4 #
# :. sqrt(41 lamda^2) = 4 #
# :. sqrt(41) lamda = 4 #
# :. lamda = (4sqrt(41))/41 #

And so our velocity vector is given by:

# vec(v) = (4sqrt(41))/41( (-5),(4) )#

And so we get the vector equation for the displacement using:

# "displacement" = "initial position" + "velocity" xx "time" #

to give:

# vec(s) = ( (-30),(0) ) + (4sqrt(41))/41( (-5),(4) )t #

We can now easily for parametric equation:

# x(t) = -30 + (4sqrt(41))/41(-5) t #
# \ \ \ \ \ \ \= -30 - (20sqrt(41))/41 t #

And;

# y(t) = 0 + (4sqrt(41))/41(4) t #
# \ \ \ \ \ \ \= (16sqrt(41))/41 t #