Find a Curve that is perpendicular to another curve

Question

We want to find a curve in $xy$ plane that passes through $(1,1)$ is perpendicular to all of contour curves of form $f(x,y) = x^4 + y^4$.

For the solution, I considered the curve as $r = (x(t) , y(t))$ so $r' = (x'(t) , y'(t))$ and the $\nabla f(x,y) = (4x^3 , 4 y^3)$. so I think we must say that $r' . \nabla f =0$ and from this with some calculations we get the line $y=x$ as the answer? Is this right? What is the general method to solve this type of question?

Answer 1

$$x^4+y^4=c$$

$$4x^3+4y^3 \frac {dy}{dx}=0$$

$$\frac {dy}{dx}= \frac {-x^3}{y^3}$$

For the orthogonal curves we get

$$\frac {dy}{dx}=\frac {y^3}{x^3}$$

$$\frac {dy}{y^3}= \frac{dx}{x^3 }$$

$$\frac {-1}{y^2}=\frac {-1}{x^2}+c $$

$$y^2=\frac {x^2}{1+cx^2}$$

Answer 2

The gradient of a function $f$ is already perpendicular to all contour curves $f(x, y) = c$. So what you really want is for $r'$ and $\nabla f$ to be parallel, not perpendicular.

And yes, a straight line $r(t) = (t, t)$ will do that, as $r'(t) = (1, 1)$ is parallel to $\nabla f(x, y) = (4x^3, 4y^3)$ at any point along $r$ (except perhaps at the origin, that's a matter of definition). And clearly this line will also go through $(1, 1)$, so it's a solution to your problem.

This also turns out to be the only solution (at least for $t>0$, as the origin is a bit problematic, and of course different parametrisations of the same line will give different solutions), but that's an entirely thing to show.

Answer 3

@Mohammad Riazi-Kermani's answer is a special solution.

Given a curve $f(x,y)=0$, finding a perpendicular curve $g(x,y)=0$ can be translated to a problem of solving a partial differential equation. In other words, curve $f(x,y)=0$ must be dual to curve $g(x,y)=0$, i.e. $d f\wedge * d g=0$, which gives $$ f_x g_x+f_y g_y=0 $$ In your case, we have $$ x^3 g_x+y^3 g_y=0 $$ According to the method of characteristics, we can transform it into a system of equations $$ \begin{cases} \frac{d g}{d s}=0 \\ \frac{d x}{d s}=x^3 \\ \frac{d y}{d s}=y^3 \end{cases} $$ the solution of which is $$ \begin{cases} g=g_0 \\ x=-\frac{1}{ \sqrt{x_0-2s}} \\ y=-\frac{1}{ \sqrt{-2s}} \end{cases} $$ Combining the 2nd and 3rd gives $$ x_0=\frac{1}{x^2}-\frac{1}{y^2} $$ therefore we have a final solution $$ g=g_0(x_0)=g_0\left(\frac{1}{x^2}-\frac{1}{y^2}\right) $$ here $g_0$ is an arbitrary function, namely any function with an argument $\frac{1}{x^2}-\frac{1}{y^2}$ perpendicular to $f=0$.