### Today’s Lemniscate Problem

… is essentially an example from the text, first of all. So hopefully, that’ll inspire some reading (nobody had the right slope on this one). Here are some other remarks.

In Calc III (here’s an old blog by me), one considers polar co-ordinates (so skip the rest of this paragraph if you prefer not to think about ’em yet). Our curve $(x^2+y^2)^2 = 50xy$ “translates” into $r^4 = 50(r\cos(\theta))(r\sin(\theta))$, so we can put $r =5\sqrt{2\cos\theta\sin\theta}$ into a grapher. The “Draw” feature can then be used to produce the desired tangent line ( $(x,y) = (2,4)$ gives us $\theta =\arctan({4\over2})$; input this at the prompt); this feature also produces the display ${{dy}\over{dx}}=.181818$ (or words to that effect); being translated (via the “Frac” feature, say), one has y’=2/11.

The Calc I version: differentiating both sides of the given equation gives $D_x[(x^2+y^2)^2] = D_x[50xy]$ $2(x^2+y^2)\cdot D_x[x^2+y^2] = 50x\cdot D_x[y] + D_x[50x]\cdot y$ $2(x^2+y^2)(2x + 2yD_x[y])=50xD_x[y] + 50y\,.$
One now easily isolates the $D_x[y]$ (or ${{dy}\over{dx}}$) terms, “factors out” ${{dy}\over{dx}}$ (and “cancels” a 2): dividing by the other factor on both sides of the equation gives ${{dy}\over{dx}} = {{25y - (x^2+y^2)\cdot 2x}\over{(x^2+y^2)\cdot 2y - 25x}}$. Finally, substituting the given values gives ${2\over 11}$—by Calc I methods (and without a calculator). “Plug in” on the point-slope form; done.