The thing to do first is to find the time when they are closest.
Since the dog is moving at 2 m/s, her distance is (15-2t).
The squirrels distance is 4t
Therefore the time they're closest together is by ol' Pythagoras.
After 3/2 seconds they are closest together.
If the dog is moving at 2 m/s for 3/2 seconds then she moves 3 feet.
Therefore, she is 15-2(3/2)=12 feet from the tree.
The squirrel is 4(3/2)=6 feet up the tree,
The distance between them along the hypoteneuse is