derivative word problem

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PaulKraemer

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Hi,

I am stuck on the following problem:
Find the point on the graph of y = x^2 + 1 that is closest to the point (3,1).

I figured the following:

a = x^2 (where a is the vertical distance from 3,1 to any point on the function)
b = 3-x (where b is the horizontal distance from 3,1 to any point on the function)

I then figured that the total distance c between 3,1 and any point on the function would be defined by:

c^2 = a^2 + b^2 = x^4 + (3-x)^2 = x^4 + x^2 - 6x + 9
..so...
c = sqrt (x^4 + x^2 - 6x + 9)
...now I figure that I would have to take the first derivative c ' and find the zeros...
This looks like a much tougher formula to find the zeros in than any other problem I've seen so far, so I was wondering if anyone could tell me if I am heading in the right direction or if I made a mistake somewhere?

Thanks in advance,
Paul
 
You're thinking is OK. But, there is a trick to these.

Use the distance formula. Where \(\displaystyle y=x^{2}+1\)

\(\displaystyle D=\sqrt{(x-3)^{2}+(y-1)^{2}}\)

But, we are given \(\displaystyle y=x^{2}+1\), sub it in for y into D:

\(\displaystyle D=\sqrt{(x-3)^{2}+x^{4}}\)..........you got here OK.

A trick when dealing with minimum distance problems is to note that the distance and the square of the distance occur at the same points.

This allows us to eliminate the radical.

And write:

\(\displaystyle S=D^{2}=(x-3)^{2}+x^{4}\)

Now, finish?. Do the differentiating thing, set to 0, yada yada yada.
 
Hi galactus,
Taking the derivative I come up with 4x^3 + 2x - 6. With the power of 3, I am stuck on how to find the zeros(s). If u can give me one more hint I'd really appreciate it.
Thanks again,
Paul
 
Factor it.

\(\displaystyle 2(x-1)(2x^{2}+2x+3)\)

There is only one real zero.
 
PaulKraemer said:
Hi galactus,
Taking the derivative I come up with 4x^3 + 2x - 6. With the power of 3, I am stuck on how to find the zeros(s). If u can give me one more hint I'd really appreciate it.
Thanks again,
Paul
Click to expand...
Obviously you can try graphing the sucker.
Moreover, there are at least four analytic ways to go.
The first is to see if you see an obvious factoring. I didn't.
Another is to use the formula for the general cubic, which I have to look up and is ugly to boot.
Another is to use the formula for the depressed cubic, which I also don't remember.
Fourth is to use Newton's method of successive approximation. Newton's method converges quickly if your initial approximation is reasonably good. Of course it is not an exact answer, but it may give you a great hint for factoring.

I decided to try the approximation method initially because, first, I do not have a graphing calculator, and second, I memorize as few formulas as possible. So, being cheap and lazy, I thought about what my initial guess might be. 4x[sup:1tlrcsal]3[/sup:1tlrcsal] + 2x is - 6 is negative if x <= 0 so the initial guess must be a positive number. But 4x[sup:1tlrcsal]3[/sup:1tlrcsal] + 2x - 6 > 0 even if positive x is fairly small. Aha, now the factoring became immediately apparent to me, but I suspect it would have in any case, had I taken an approximation or two. Why don't you try? As usual Galactus beat me to it.
[spoiler:1tlrcsal](x - 1) * (4x[sup:1tlrcsal]2[/sup:1tlrcsal] + 4x + 6) = 4x[sup:1tlrcsal]3[/sup:1tlrcsal] + 4x[sup:1tlrcsal]2[/sup:1tlrcsal] + 6x - 4x[sup:1tlrcsal]2[/sup:1tlrcsal] - 4x - 6 = 4x[sup:1tlrcsal]3[/sup:1tlrcsal] + 2x - 6
You should be on solid ground now.
And thanks Subhotosh. You taught me a method that I never learned (or else forgot.)[/spoiler:1tlrcsal]
 
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