# Really quick question about minimum of function

1. Feb 5, 2008

### Saladsamurai

!!!Really quick qestion about minimum of function

I ADDED ANOTHER QUESTION IN POST #3 ! Thanks for looking :)

I am told the a jet's follows the curve $y=20(10^{-6})x^2+5000$ where y AND x are in feet. Basically all I need help with is for the first part of the problem, I need to find out where the jet is at its LOWEST point on the curve.

I am pretty sure this is at x=0. But how can I prove it? Usually I would just plug in zero for x. But this is not a function of time, it is of position, so I am not quite sure how to show that if I plug in a neg number I would not get a number less than 5000. I mean I know since it's an even function thath even numbers<0 for x will yield numbers>5000.

I was hoping to show this with Calculus rather than analysis, though maybe the latter is more reasonable in an engineering course?

Any thoughts are welcome, I can move on withthe problem regardless (since I could use the analysis) but I just want some input.

Last edited: Feb 5, 2008
2. Feb 5, 2008

### Dick

In general find where dy/dx=0 and then check if that's a minimum. It doesn't matter that the equation is not a function of time. You are just finding the minimum point on a curve. In this case though you hardly even need to do that. You know x^2>0 unless x=0. So the lowest point is clearly at x=0, isn't it?

3. Feb 5, 2008

### Saladsamurai

Like I said, I know that it is at x=0, but I'm an idiotm so I would to prove that it is (to myself). So I see your point. 1st Derivative test does work.

ANOTHER QUESTION:

How about this now: I need to find the normal acceleration $a_n$ of a point traveling along said curve AT X=0.

Now I know that $$a_n=\frac{v^2}{\rho}$$

My teacher gave the hint $ds=rd\theta$. How can I find velocity from this? I would nee to write theta as a function of x?

Any ideas?

4. Feb 5, 2008

### Saladsamurai

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5. Feb 5, 2008

6. Feb 6, 2008

### Dick

??? Gravity has absolutely no effect on the acceleration of the plane. It's not in free fall. It's following a fixed trajectory. You can't find the normal acceleration from the information you are given. To use v^2/R you would need to find the radius of curvature, R, of the curve at x=0 and then find v. But you have no way of finding v without knowing something about the position of the plane as a function of t. For all you know, it could just be sitting at x=0 without accelerating at all.

Last edited: Feb 6, 2008
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