Solving differential equation of equation to find the tangent point.

Question

Normally it is easier to differential the equation that has only one variable but I am stuck with this equation that has two variables of different degree. $x^3+3y^2-6xy=0,$

The answer was mentioned as $3x^2 + 6yy′ − 6y − 6xy′ = 0$ which I quite do not understand.

When I use wolfarm it gives me reuslt such as the one below which just adds more confusion: $\frac {d}{dx}(x^3 - 6 y x + 3 y^2) = 3 (x^2 - 2 y)\\ \frac {d}{dy}(x^3 - 6 y x + 3 y^2) = 6 y - 6 x$

Could anyone please explain clear method of finding the point tangent to x-axis by explaining the differentiation of equation above?

Answer 1

Think of $y$ not as a variable, but as a function of $x$.

We might not know how do describe that function (yet) nonetheless it is a function and its derivative is $y'$

$x^3+3y^2−6xy$ becomes a composition of functions.

$(x^3+3y^2−6xy) = 0\\ \frac {d}{dx}(x^3+3y^2−6xy) = \frac {d}{dx} 0 = 0\\ \frac {d}{dx}(x^3+3y^2−6xy) = \frac {d}{dx}x^3+\frac {d}{dx}3y^2−\frac {d}{dx}6xy = 0$

Lets look at each term separately.

We will apply the power rules, product rules, and chain rules step by step to each term as we go.

The first term is easy. $\frac {d}{dx}x^3 = 3x^2$

$\frac {d}{dx}3y^2$ we apply the chain rule $\frac {d}{dx} f(y(x)) = \frac {df}{dy}\frac{dy}{dx}$

$\frac {d}{dx}3y^2 = 6yy'$

Same again for $-\frac {d}{dx}6xy,$ first the product rule then the chain rule

$-\frac {d}{dx}6xy = -6y - 6xy'$

now when you have:

$3x^2+6yy'−6y - 6xy' = 0$

It is common to collect the $y'$ terms on one side of the equation, and then express $y'$ as a function of $x$ and $y.$

$6yy' - 6x y' = -3x^2 + 6y\\ y' = \frac {-x^2 + 2y}{2(y-x)}$

Regarding the Wolfram-Alpha answers, don't worry that the answers don't tie out. Wolfram is applying a different sort of operation. It is applying "partial differentiation" which you probably have not learned yet. Under partial differentiation, $y$ is treated truly as a variable and $y$ is assumed to be independent from $x.$

Answer 2

The derivative of $x^3+3y^2-6xy=0$ with respect to $x$ is computed as follows. $$\frac{d}{dx}(x^3+3y^2-6xy) = \frac{d}{dx}(x^3)+\frac{d}{dx}(3y^2)-\frac{d}({dx}(6xy) = 3x^2+6y\frac{dy}{dx}-6y-6x\frac{dy}{dx}=0.$$

Here, I have used the chain's rule (e.g. in computing $\frac{d}{dx}(3y^2)=\frac{d}{dy}(3y^2)\frac{dy}{dx}$.)

Consequently, $$\frac{dy}{dx}=\frac{x^2-2y}{2y-2x}.$$ What Wolfram alpha computed are the partial derivatives. A partial derivative of a function of two or more variables is the derivative of the function with respect to one variable such that the other variable(s) are treated as constant. For example, $\frac{\partial}{\partial x}(x^2+xy) = 2x +y$.

Answer 3

For derivation of implicit function $x^3+3y^2-6xy=0$, we have two ways:

By formula: For function $F(x,y)=0$ $$y'=-\dfrac{F_x}{F_y}$$ here $F_x$ obtain when we suppose $y$ is as a number and $F_y$ obtain when $x$ to be supposed as a number as well. With $F(x,y)=x^3+3y^2-6xy=0$, $$y'=-\dfrac{F_x}{F_y}=-\dfrac{3x^2+0-6y}{0+6y-6x}$$

By implicit derivation: Derivate function $F(x,y)=0$ when we suppose $y'=y$ and $x'=1$ then use derivation rules. With $F(x,y)=x^3+3y^2-6xy=0$, $$3x^2x'+3\times2yy'-6(x'y+xy')=0$$ or $$3x^2+6yy'-6y-6xy'=0$$

Answer 4

Primes indicate differentition with respect to $x$. When $x,y$ appear together implicitly, the chain rule is used thus: $y$ or $f(y)$ is treated as constant when differentiating wrt $x$... $x$ or $f(x)$ is treated as constant when differentiating wrt $ y$.

Next Wolfram differentiation is partial differentiation.Stricly it should be $\partial y/ \partial x $ rather than $dy/dx.$

Answer 5

To prove that the curve is tangent to $x-$axis you substitute $x=0$ in the equation of the curve getting $3y^2=0$ which is a double root so the curve is actually tangent to the $x-$axis.

There is another tangent at $(0,0)$ because the origin is a singular point for the given curve as both derivatives vanish.

We can write the general equation of a line passing through the origin $y=mx$ and consider the system formed by the equation of the curve $x^3-6 x y+3 y^2=0$ and the line. Substitute $y=mx$ in the curve equation to get

$3 m^2 x^2-6 m x^2+x^3=0$ that is $x^2 \left(3 m^2-6 m+x\right)=0$

which is solved if $3 m^2-6 m=0$ to get multiple root at $x=0$

$3m(m-2)=0$ if $m=0$ which gives us $y=0$ the equation of $x-$axis already found and another solution $m=2$ which gives $y=2x$ that is the other tangent at $O$

Hope this helps