# Curves intersecting level surfaces at right angles

I've been trying to solve a question in my textbook for a little while now, and I can't seem to get the reasoning behind the answer. I'll quote the question, followed by my attempted solution, then finally followed by the answer in the book.

Exercise/Question:

Find an equation of the curve in the xy-plane that passes through the point $$(2,-1)$$ and that intersects every curve with equation of the form $$x^2y^3 = K$$ at right angles.

My attempted solution:

Start off by recognizing the question is asking about level curves, i.e. equations of those forms are indeed level curves. As such we denote the equation of our curve as a level curve as well, $$g(x,y) = L$$. Since the gradient of a function is normal to it at all points, we simply require $$\nabla f \perp\nabla g$$. So, $$(\nabla f|\nabla g) = 0 \\ \iff \Big((2xy^3, 3x^2y^2)\Big|(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y})\Big) = 0 \\ \iff 2xy^3\frac{\partial g}{\partial x} + 3x^2y^2\frac{\partial g}{\partial y} = 0$$

And at this point I get stuck. It's a PDE (Which I'm not very familiar with at all), but its also in several variables meaning I'm completely out of my comfort zone.

Solution in the textbook:

Let the curve be $$y=g(x)$$. At $$(x,y)$$ this curve has normal $$\nabla(g(x)-y) = (g'(x), -1)$$. The curve $$x^2y^3 = K$$ has normal $$(2xy^3,3x^2y^2)$$. These curves will intersect at right angles if their normal are perpendicular, that is, if $$2xy^3\frac{dy}{dx} - 3x^2y^2 = 0 \\ \iff \frac{dy}{dx} = \frac{3x}{2y} \\ \iff 2y dy = 4x dx \\ \iff y^2 = \frac{3}{2}x^2 + C$$ Since the curve must also pass through $$(2,-1)$$ we have that $$C=-5$$, so the required curve is $$3x^2-y^2 = 10$$.

My problem with the solution in the book:

So what I don't understand about the solution is how we are allowed to assume our curve has the form $$y = g(x)$$. That would suggest that it can be described as a function of one variable, yet the answer we get out is definitely not a function of one variable (Considering the plus/minus square root). Yet this assumptions leads to this curve. It seems much more natural to define our curve as a level surface of a function of two variables, but doing that, as you've seen, doesn't work, and I don't quite know why.

• In your original question statement you want the curve to be perpendicular to $x^4+y^2=K$, but your own solution and the one in the textbook seem to work with $x^2y^3=K$ in stead. Which one is it? – Servaes Feb 26 '19 at 19:26
• Sorry, It's $x^2y^3=K$, I edited my original post as well. – Peatherfed Feb 26 '19 at 19:48
• Well I think the problem is inherently a PDE, but considering the PDE you found in the case $xy\neq0$ yields the much simpler $$2y\frac{\partial g}{\partial x} =-3x\frac{\partial g}{\partial y},$$ which might be solved by educated guessing. – Servaes Feb 26 '19 at 19:59
• On the other hand, the solution in the textbook either presupposes a lot of calculus, or uses a lot of handwaving. What does the expression $$2ydy=4xdx,$$ mean? If the reader is supposed to be familiar with differential forms, then I would expect such a PDE to be no problem. – Servaes Feb 26 '19 at 20:01
• Also, the first equivalence in the textbook solution does not hold; it fails when $xy=0$. – Servaes Feb 26 '19 at 20:04

So I did some more looking around at similar problems on the internet, as well as looked at some other books, and I've come up with an answer I feel is much less shady. I'll post it here if anyone has a comment on it, but also so that if anyone else has a similar question to mine perhaps this answer will be helpful.

So I will basically take the advice of one of the comments above and just solve the PDE.

We have our original function $$f(x,y) = x^2y^3$$ whose level curves we wish to be perpendicular to. As such we set up the function $$g$$ and take its level curve at $$g(x,y)=0$$, as so: $$g(x,y) = L$$. We need them to intersect orthogonally, i.e. their gradients are orthogonal. So

$$\nabla f \perp \nabla g \\ \iff \Big(\nabla f \Big| \nabla g\Big) = 0 \\ \iff 2xy^3\frac{\partial g}{\partial x} + 3x^2y^2\frac{\partial g}{\partial y} = 0$$ Note that if $$x = 0$$ or $$y = 0$$ the level curve of $$f(x,y) = K$$ is just a point, so we can safely assume $$x,y\neq 0$$. Hence the above becomes

$$2y\frac{\partial g}{\partial x} + 3x\frac{\partial g}{\partial y} = 0$$

This holds when $$\frac{\partial g}{\partial x} = \frac{1}{2y}$$ and $$\frac{\partial g}{\partial y} = -\frac{1}{3x}$$. We integrate both of these: $$g(x,y) = \frac{x}{2y} + \phi(y) \\ g(x,y) = -\frac{y}{3x} + \psi(x)$$

Where $$\phi$$ and $$\psi$$ are some arbitrary functions of one variable. Combining these two equations for $$g$$ we find that $$g(x,y) = \frac{x}{2y} - \frac{y}{3x} + C$$. Since $$g$$ passes through $$(2,-1)$$ we have that $$C= -5$$. Finally we take the level curve when $$g(x,y)=0$$, which implies the curve is $$3x^2-y^2=10$$. //

I prefer this answer because it seems much less shady than the one in the book, but also it's quite nice since it actually reveals that there are an infinite amount of these level curves, which can be trivially found by setting $$g(x,y) = L$$ for different values of $$L$$. It should be noted that I'm very inexperienced with PDE's, so my solution might have some shady steps.