Idealistic
Junior Member
- Joined
- Sep 7, 2007
- Messages
- 97
A racetrack is in the shape of an ellipse (Super Speedway). Find the location on the track that is closest to a person standing at a postion (a, b) inside the racetrack.
the equatuion of an ellipse is x[sup:1523qgwf]2[/sup:1523qgwf]/a[sup:1523qgwf]2[/sup:1523qgwf] + y[sup:1523qgwf]2[/sup:1523qgwf]/b[sup:1523qgwf]2[/sup:1523qgwf] = 1
L[sup:1523qgwf]2[/sup:1523qgwf] = x[sup:1523qgwf]2[/sup:1523qgwf] + y[sup:1523qgwf]2[/sup:1523qgwf], Is the the length "L" constant in this expression? or is it changing? When I differentiate, will my new equation be:
2L(dL/dt) = 2x(dx/dt) + 2y(dy/dt)?
Im not sure exatly what I subsitute in order to minimize the distance, "L".
Also, L is not equal to zero so the location isn't (a, b).
Do I need to get valus for "x" and "y" so I can plug them into my location equation, x[sup:1523qgwf]2[/sup:1523qgwf]/a[sup:1523qgwf]2[/sup:1523qgwf] + y[sup:1523qgwf]2[/sup:1523qgwf]/b[sup:1523qgwf]2[/sup:1523qgwf] = 1?
.
the equatuion of an ellipse is x[sup:1523qgwf]2[/sup:1523qgwf]/a[sup:1523qgwf]2[/sup:1523qgwf] + y[sup:1523qgwf]2[/sup:1523qgwf]/b[sup:1523qgwf]2[/sup:1523qgwf] = 1
L[sup:1523qgwf]2[/sup:1523qgwf] = x[sup:1523qgwf]2[/sup:1523qgwf] + y[sup:1523qgwf]2[/sup:1523qgwf], Is the the length "L" constant in this expression? or is it changing? When I differentiate, will my new equation be:
2L(dL/dt) = 2x(dx/dt) + 2y(dy/dt)?
Im not sure exatly what I subsitute in order to minimize the distance, "L".
Also, L is not equal to zero so the location isn't (a, b).
Do I need to get valus for "x" and "y" so I can plug them into my location equation, x[sup:1523qgwf]2[/sup:1523qgwf]/a[sup:1523qgwf]2[/sup:1523qgwf] + y[sup:1523qgwf]2[/sup:1523qgwf]/b[sup:1523qgwf]2[/sup:1523qgwf] = 1?
.