Problem : Equation of a circle using implicit differentiation.

Question

This problem is found in the Shaums's outline Calculus (Ch.11. ex.9). As I am rather new to implicit differentiation I'm not really sure where to start with this kind of problem. The only theoretical background and practice I have is about solving implicit problems containing only x and y.

For the circle $x^2+y^2=r^2$, show that $∣\frac{y''}{[1+(y')^2]^{3/2}}∣$ $=\frac{1}{r}$

Is $x$ a constant as it disappear from the final equation? Should I only try to differentiate $y$ and $r$?

Thanks for any hints.

Answer 1

HINT

We have that

$$x^2+y^2=r^2 \implies 2xdx+2ydy=0 \implies y'(x)=-\frac xy\implies y''(x)=-\frac1y-\frac{x^2}{y^3}$$

Answer 2

Hint:

Compute the first derivative first, differentiating the circle equation: $$(x^2+y^2)'=0\iff 2x+2yy'=0\iff y'=-\frac xy.$$ We deduce, by the usual rules, $$y''=-\biggl(\frac xy\biggr)'=-\frac{1\cdot y- xy'}{y^2}=-\frac{y+ \cfrac{x^2}{\smash y}}{y^2}=-\frac{r^2}{y^3}.$$ Can you proceed?

Answer 3

Just diffeerentiate: $$x^2+y^2=r^2$$ $$2x+2yy'=0 \;\; ( \implies y'=-x/y)$$ $$2+2(y'^2+yy'')=0$$ $$y'^2+yy''=-1$$ $$\dfrac {x^2}{y^2}+yy''=-1$$ $$\implies y''=-\dfrac 1y -\dfrac {x^2}{y^3}$$

Answer 4

The first differentiation is straightforward,

$$x+yy'=0.$$

Then we will use a trick to speed up the computation:

$$\left(\frac1{\sqrt{1+y'^2}}\right)'=-\frac{y'y''}{(1+y'^2)^{3/2}}.$$

But$$\frac1{\sqrt{1+y'^2}}=\frac y{\sqrt{x^2+y^2}}=\frac{y}r$$

immediately giving us

$$-\frac{y'y''}{(1+y'^2)^{3/2}}=\frac{y'}r$$

from which the claim.

Answer 5

The concluding steps might also be taken this way. The first level of implicit differentiation gives us $ \ x^2 + y^2 \ = \ r^2 \ \rightarrow \ y' \ = \ -\frac{x}{y} \ \ , $ as shown in most of the posted answers. Since we have a specified expression we wish to prove, we can then produce $$ ( \ y' \ )^2 \ \ = \ \ \left( -\frac{x}{y} \right)^2 \ \ = \ \ \frac{x^2}{y^2} \ \ \Rightarrow \ \ 1 \ + \ ( \ y' \ )^2 \ \ = \ \ \frac{y^2}{y^2} \ + \ \frac{x^2}{y^2} \ \ = \ \ \frac{x^2 \ + \ y^2}{y^2} \ \ = \ \ \frac{r^2}{y^2} \ \ , $$ upon applying the circle equation. We could then either differentiate implicitly this last equation, $$ \frac{d}{dx} \ [ \ 1 \ + \ ( \ y' \ )^2 \ ] \ \ = \ \ \frac{d}{dx} \ [ \ r^2 \ · \ y^{-2} \ ] \ \ \Rightarrow \ \ 0 \ + \ 2 · y' · y'' \ \ = \ \ r^2 \ · \ (-2y^{-3} \ · \ y') $$ $$ \Rightarrow \ \ y'' \ \ = \ \ r^2 \ · \ (-y^{-3}) \ \ = \ \ -\frac{r^2}{y^3} \ \ , $$

or we can apply the "quotient rule", as Bernard does.

The expression sought is thus $$ \left|\frac{y''}{[ \ 1 \ + \ (y')^2 \ ]^{3/2}} \right| \ \ = \ \ \left|\large{\frac{-\frac{r^2}{y^3}}{\left(\frac{r^2}{y^2} \right)^{3/2}}} \right| \ \ \large{= \ \ \frac{ \frac{r^2}{y^3}}{\frac{r^3}{y^3}}} \ \ \normalsize{= \ \ \frac{1}{r} } \ \ . $$

Incidentally, the expression you are being asked to calculate gives the curvature of a curve at a particuar point. That we found a constant value, the reciprocal of the radius, for the circle shows that circles have constant curvature.