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So I am given a function of $x$ and $y$:

Question

The partial derivative with respect to $x$, and the partial derivative with respect to $y$ are found. But there are a few things in the solution given that don't make sense to me.

Solution

Firstly, what is the thinking behind eliminating y (combining the results of both partial derivatives). Is this done due to the partial wrt $x$ representing all of the $x$ points where there is a gradient of zero, and the partial wrt to $y$ representing all of the $y$ points where the gradient is zero, so combining the results will give you the $(x, y)$ points of where the gradient is zero?

Secondly, once $y$ is eliminated to find the values of $x$ that satisfy both partial derivatives, how is the factorizing done. It's an easy cubic but what is the actual method for figuring out where to go from $x^4 = x$ ?

Also, once the factorizing is done, it says there are two $x$ points that satisfy the equation.

$x = 0$ due to the $x$ term being multiplied to everything ]

$x = 1$ due to the root of $(x - 1)$

but there is another factor, $(x^2 + x + 1)$. Is the reason there are no solutions from this factor due to the fact that if the quadratic formula is used to factorize this factor, we will end up with complex roots?

Thanks

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  1. Eliminating one variable to solve the system of two equations with two variables is a typical way. What you said is close. It basically means you want to find $(x,y)$ that satisfies both of the two equations.

  2. Once you get a polynomial equation like $x^4=x$, to solve it, you can usually first try if you can factorize it. The equation $x^4-x$ has a common factor $x$ among the two terms. So $x(x^3-1)$. Then a factorization formula gives you $x^3-1=(x-1)(x^2+x+1)$.

  3. Correct. You can use quadratic formula to see that there is no real root.

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  • $\begingroup$ So the factorisation formula you used was the "difference of cubes"? $\endgroup$ – lgdl.y Dec 19 '16 at 17:31
  • $\begingroup$ Yes, if you meant $x^3-y^3=(x-y)(x^2+xy+y^2)$. $\endgroup$ – KittyL Dec 19 '16 at 17:34
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Once the partial derivatives are found here, we have a system of two equations to solve: $$\left\{\begin{aligned} y&=-x^2,\\ y^2&=x. \end{aligned}\right.$$ The reason for setting it up is the definition of stationary points. But once we did set it up, from this point on, finding $x$ and $y$ is a question from elementary algebra. Roughly speaking, for the time being we can forget all about functions of several variables and gradients and stationary points and such. We just need to solve this system of two simultaneous equations.

Your first question: no. (And your description is a bit confused, I'm afraid.) See what I said above. We're just looking for ways to solve this system. It's not a system of linear equations, so there are no standard algorithms of formulas. But there are common techniques, and elimination is one of them. Conveniently, the first equation is solved for $y$ and the second equation is solved for $x$, so we an choose to eliminate either one. For no special reason, this solution chose to proceed by eliminating $y$.

Your second question: standard techniques for solving polynomial equations. We should collect all terms on the left hand side, so that we have a polynomial equal to $0$, and then try to factor it. So $x^4=x$ becomes $x^4-x=0$, which is a degree $4$ polynomial, not cubic. Fortunately, this one is easy to factor, and that's what they did.

Your third question: yes.

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