Question

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?

Answer 1

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)`.

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`