Related Rates: find location on track closest to position in

[Idealistic]

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?

.

 

[Deleted member 4993]

Re: Related Rates

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.

If a person is standing at (a,b) - and the equation of the ellipse is x^2/a^2 + y^2/b^2 = 1, the person cannot be inside the racetrack (ellipse). Please clarify

 

[Idealistic]

Re: Related Rates

point (a, b) can be anywheres within the race track, the center of the ellipse is like the origin, where both x and y equal zero. Point (a, b) is not on the track because when you plug a and b in for x and y respectively it does not satisfy the equation of the ellipse:

a^2/a^2 + b^2/b^2 does not equal one. Am I suposed to isolate y (from ellipse function) and make L a function of x? Then take the derivative and set dL/dx = 0?

 

[Deleted member 4993]

Re: Related Rates

point (a, b) can be anywheres within the race track, the center of the ellipse is like the origin, where both x and y equal zero.

Point (a, b) is not on the track because when you plug a and b in for x and y respectively it does not satisfy the equation of the ellipse:

But it [point (a,b)]is not inside the track - which your problem statement requires to be inside. Make sure your problem statement/interpretation is correct.

a^2/a^2 + b^2/b^2 does not equal one. Am I suposed to isolate y (from ellipse function) and make L a function of x? Then take the derivative and set dL/dx = 0?

 

[Idealistic]

Re: Related Rates

Is there a reason why it cannot be inside the track? Does it have to do with the equation of the ellipse? Also is the question legit?

 

[Idealistic]

Re: Related Rates

Ohh I see now, point (a, b) is outside the ellipse. But I still need to find out which point on the ellipse is going to minimize L.

 

[galactus]

Just as an aside, this is not a related rates nor a DE.


Solve the ellipse equation for y:

\(\displaystyle y=\pm\frac{b}{a}\sqrt{a^{2}-x^{2}}\)

Use the distance formula, \(\displaystyle L=\sqrt{(x-a)^{2}+(y-b)^{2}}\), and plug in y.

There is a trick we can use to simplify the computations when we want to minimize or maximize a distance.

It is based on the observation that the distance and the square of the distance have their min or max at the same point.

Therefore, we can do it sans radical.

\(\displaystyle S=L^{2}=(x-a)^{2}+(y-b)^{2}\)

\(\displaystyle S=(x-a)^{2}+(\frac{b}{a}\sqrt{a^{2}-x^{2}}-b)^{2}\)

\(\displaystyle \frac{dS}{dx}=\frac{2b^{2}x}{a\sqrt{a^{2}-x^{2}}}-\frac{2b^{2}x}{a^{2}}+2x-2a=\frac{2(((a^{2}-b^{2})x-a^{3})\sqrt{a^{2}-x^{2}}+ab^{2}x)}{a^{2}\sqrt{a^{2}-x^{2}}}\)

Now, finish up by setting to 0 and solving for x?. This may prove to be slightly daunting to solve for x.

This problem is rather confusing with the a,b for the point and the a,b in the ellipse equation.

Maybe I am off base here entirely.