Dear Uncle Colin,

I have the equation of a curve, $\frac{2x+3y}{x^2 + y^2} = 9$. If I differentiate implicitly using the quotient rule, I get $\diff{y}{x} = \frac{2(x^2 + 3xy - y^2)}{3x^2 - 4xy - 3y^2}$. If I rearrange first to make it $2x + 3y = 9\left(x^2 + y^2\right)$, I find $\diff{y}{x} = \frac{2-18x}{18y-3}$. Those are the same on the curve - but they’re not algebraically identical! (For example, at $(1,1)$, the first gives $-\frac{3}{2}$ and the second $-\frac{16}{15}$). Why the difference?

- Curves Obviously Not Turning Out Uniformly Right

Hi, CONTOUR, and thanks for your message - what an interesting question!

On Twitter, @_WTProject gave one excellent explanation, but my answer is a little different.

Functions of two variables

As long as you restrict your analysis to the curve, everything comes out in the wash: under the condition that $\frac{2x+3y}{x^2+y^2} = 9$, the two expressions for the derivative work out to be the same. (It’s worth noting that the original curve is undefined at (0,0), but the transformed one gives (0,0) as a point on the curve unless we specify otherwise.)

The trouble is, once you come off of the curve, you’re no longer looking at a simple link between two variables; you’re looking at a function of two variables:

  • In the first case, $f_1(x,y) = \frac{2x+3y}{x^2 + y^2} - 9$
  • In the second, $f_2(x,y) = -2x- 3y + 9x^2 + 9y^2$

The curve we’re looking at can be written as $f_1(x,y)=0$ or as $f_2(x,y) = 0$ - perhaps obviously, since the two functions are multiples of each other. You can see the curve as a contour ((hey! that’s your name!)), showing where the value of the function is 0.

But there’s nothing special about 0 ((Well… in this case, at least.)) We could just as easily consider $f_1(x,y) = 1$. We’ve just added a constant to an equation of the curve, so the derivative should work out to be the same as for $f_1(x,y)=0$

But that’s also true for (say) $f_2(x,y) = 1$ – when you work out $\diff{y}{x}$ implicitly, you’ll get the same expression as you did for the curve $f_2(x,y) = 0$.

The trouble is…

… $f_1(x,y)=1$ and $f_2(x,y) = 1$ are completely different curves. The first is inside the original curve (although they get mighty close when they approach (0,0)), and the second is outside.

So, the reason the two different equations of the same curve have different derivatives off of the curve is that the curves are contours of different functions of two variables (which coincide at the 0 level).

Hope that helps!

- Uncle Colin