Problem with minimum (it's short)

[Born2Perform]

Homework Statement

"find a point P on the parabola $$x^2-4y = 0$$ for which is minimum the distance PO/PF, where O is the origin and F the focus of the parabola"

the solution is P(0;0), coincides with the origin, this is also clear graphically.

The Attempt at a Solution

Please follow these lines:

1) i set up the function PO/PF, and i call it "y". [O(0;0), F(0;1), P(x;x²/4)]

$$y = \frac{\sqrt{x^2+x^4/16}}{\sqrt{x^2+(\frac{x^2}{4}-1)^2}}$$

simplified:
$$y = \sqrt{\frac{16x^2+x^4}{x^4+8x^2+16}}$$

2) now i set up the derivative:

$$y' = \frac{\frac{(32x+4x^3)(8x^2+x^4+16) - (16x+4x^3)(16x^2+x^4)}{(x^4+8x^2+16)^2}}{2\sqrt{\frac{16x^2+x^4}{x^4+8x^2+16}}}$$

3) i put the derivative equal to zero:(some stuff simplifies)

$$y' = 0 = (32x+4x^3)(8x^2+x^4+16) - (16x+4x^3)(16x^2+x^4)$$

solving: $$x^4-4x^2-32 = 0$$

4) this thing should give me as result x=0, in order to have P(0;0), but it doesnt...
i checked and checked again, could you find the error inside?
thanks.

 

[neutrino]

Looking at it (without checking) I'd say that you forgot about the square root when taking the derivative.

 

[Born2Perform]

Looking at it (without checking) I'd say that you forgot about the square root when taking the derivative.

Hi: i had some problem with latex, but now i edited the post and added the complete steps so i hope someone could follow on line my procedure.
thanks.

 

[AKG]

The minimum doesn't always occur where the derivative is zero. Consider the function $f : \{x \in \mathbb{R} : x \geq 0\} \to \mathbb{R}$ defined by $f(x) = \sqrt{x}$. This function never has zero derivative, but you know it has a minimum and you know where it is.

 

[Born2Perform]

The minimum doesn't always occur where the derivative is zero. Consider the function $f : \{x \in \mathbb{R} : x \geq 0\} \to \mathbb{R}$ defined by $f(x) = \sqrt{x}$. This function never has zero derivative, but you know it has a minimum and you know where it is.

this was unespected, however can't it be solved by putting the derivative equal to infinite? for example if you derive $f(x) = \sqrt{x}$ and put the derivative equal to infinite, x=0 is the only value for which the relation can be true.

but in my derivative, step 2, when i put it equal to infinite, x=0 gives an indeterminate form. there is an error!

are there some problem with max and min that can't be solved algebrically?

 

[AKG]

Okay, what about |x|? And what about x³? Not all local extrema are points where the derivative is zero, and not all points where the derivative is zero are local extrema.

Your function y is the square root of something (a rational expression in x). Therefore, the smallest possible value it can attain is 0 (square roots are always non-negative). So when is the square root of something 0? When that something itself is 0. And it's clear that that rational expression you have is zero iff the numerator is zero and the denominator is non-zero iff x is zero (restricting our possibilities to real values).

But why are you doing all this work anyways? You know the origin lies on the parabola, and you know that the distance from the origin to itself is 0, so when P = (0,0), PO/PF = 0 which is the smallest it could possibly be (the ratio of two distances is a non-negative real, as long as the denominator is non-zero). Why even bother computing the derivative?

In order to find local extrema, you normally need to check two places: i) where the derivative is 0, and ii) where the derivative does not exist. In your function, the derivative does not exist at x=0, it's like the function |x|. And that's where the minimum is. Also, the derivative does not exist at the boundary points of functions defined on closed intervals. If you have a function defined on [a,b] then the derivative is said to not exist at a or b. Even if you have a function defined on [a, infinity), you get no derivative at a. So you have to check these kinds of places too.

 

[D H]

Your mistake is in this step.

3) i put the derivative equal to zero:(some stuff simplifies)

$$y' = 0 = (32x+4x^3)(8x^2+x^4+16) - (16x+4x^3)(16x^2+x^4)$$

solving: $$x^4-4x^2-32 = 0$$

What does substituting $x=0$ in your unsimplified polynomial yield?

Suggestion: When dealing with minimum distance problems, it is often much easier to minimize the square of the distance rather than the distance itself. In this case, minimizing $PO^2/PF^2$ avoids those nasty radicals.