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I've been trying to figure out how can I solve this exercise but I haven't had much luck so far. Do you think you can help me out a bit? Pointing out what might I possibly be doing wrong?

The exercise is as follows:

Find the coordinates where the tangent to the curve is horizontal

$$x^3+3xy+2y^2+4y=1$$

Given that it's difficult to solve for either x or y. I decided to differentiate implicitly. And here's what I got:

$$- {3x^2+3y\over 4y+3x+4}=0 $$

In order to find the horizontal tangents, the first order differential must be zero, and for this case particularly:

$$ 3x^2+3y=0 $$

Now, solving for x:

$$x=\sqrt{-y}$$ $$x=-\sqrt{-y}$$

Which tells me that y must be positive. (Real field)

But now I'm stuck there. Just looking at the answers I can't think of anything else but some numbers that might satisfy the equation; $(1,-1)$, $(-1,-1)$,$(0,0)$ But I wouldn't know how to get there, nor I know if those are the right coordinates. Can you help me out? Thanks in advance.

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    You've made a sketch of this curve of yours, by any chance?2012-11-08
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    I actually have, and I think I'm really far from finding the correct answer.2012-11-08

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Your first steps seem right, but then it is much more natural to solve for $y$: $~~y = -x^2$. Of course, the points have to lie on the original curve, so substitute this back into the original equation to get $$ 2x^4 - 2x^3 - 4x^2 - 1 = 0 ~. $$ So now you need the real roots of this quartic. This is slightly tricky though; it certainly has real roots, but I don't think it has any rational roots (assume it does, derive a contradiction), for example...

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    I used Derive in order to find the roots to the quartic because using synthetic division wasn't helpful, and I got this, which are the answers: $x=-1.12 , x=2.039$ so the answers would be $(2.039, f(2.039))$ and $(-1.12,f(-1.12))$ But now I ask, how can I solve for that quartic? I don't know any other method but synthetic division.2012-11-08
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    Yes, those are the solutions, where $f(x) = -x^2$. You can only use polynomial division to simplify the problem if you already know one of the roots. The general solution (http://en.wikipedia.org/wiki/Quartic_polynomial#Solving_a_quartic_equation) of a quartic, which I guess is what you're asking about, is horribly messy, so I would settle for the numerical solutions if I were you! Where did this problem come from? Usually examples like this are cooked up to have nice analytic answers.2012-11-08
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    Actually it was one of the problems on my Calculus quiz. And I had asked my teacher about it in class, but he didn't actually answer because he was a bit busy. And I was surprised when I saw it on my quiz, anyway, I talked to him and he agreed on just leaving it till the implicit differentiation.2012-11-08