# A girl facing North is standing next to a river which flows East. She tosses a stick into the water exactly 4 meters North of where she stands. The river carries the stick East at the constant rate of 3 m/s. How fast is the stick moving away from the girl after 2 seconds? The stick does not move directly "away" from the girl, the direction from the girl to the stick changes. Because of this, the question "how fast" should be clarified. For example, "what is the speed of change of the distance between the girl and the stick?" or "what is the projection of the stick's velocity onto the line that goes from the girl to the stick?".

To answer the first version of the question, determine the distance between the girl and the stick, which is `d ( t ) = sqrt( 4^2 + ( 3 t )^2 ) ` by the Pythagorean theorem. Here t is in seconds and d is in meters.

Then the speed of change of this distance is its first derivative with respect to `t :`

`d ' ( t ) = ( 18 t ) / ( 2 sqrt ( 16 + 9 t^2 ) ) = ( 9 t ) / sqrt ( 16 + 9 t^2 ) .`

At `t = 2 s ` it will be `18 / sqrt ( 16 + 36 ) = 9 / sqrt ( 13 ) approx 2.5 ( m / s ) .`

To answer the second question, note that the velocity of the stick relative to the ground is the constant vector `lt 3, 0 gt ` if the girl stands at the origin and the x-axis is parallel to the river. The direction to project onto is `lt 3t, 4 gt ` and one can use the formula for a projection for `t = 2 . ` A link about projections is attached.