# If $$( x-a)^2 + (y-b)^2 = c^2$$, for some $$c > 0$$, prove that $\Large\frac{\begin{bmatrix} 1 + \left(\frac{dy}{dx}\right)^2 \\[0.3em] \end{bmatrix}^{\frac{\Large 3}{\Large 2}}}{\Large\frac{d^2y}{dx^2}}$is a constant independent of $$a$$ and $$b$$.

## 1 Answer

Toolbox:
• $\large\frac{d}{dx}$$(x^{\large n})=nx^{\large n-1} Step 1: (x-a)^2+(y-b)^2=c^2-------(1) Differentiating (1) we get 2(x-a)+2(y-b)\large\frac{dy}{dx}$$=0$
$\Rightarrow (x-a)+(y-b)\large\frac{dy}{dx}$$=0--------(2) Step 2: Differentiating (2) we get 1+(y-b)\large\frac{d^2y}{dx^2}+\big(\large\frac{dy}{dx}\big)^2$$=0$
$(y-b)=-\begin{bmatrix}\large\frac{1+\big(\Large\frac{dy}{dx}\big)^2}{\Large\frac{d^2y}{dx^2}}\end{bmatrix}$-------(3)
Step 3:
Substituting the value of $(y-b)$ in (2) we get
$(x-a)=\begin{bmatrix}\large\frac{1+\big(\Large\frac{dy}{dx}\big)^2}{\Large\frac{d^2y}{dx^2}}\end{bmatrix}\big(\large\frac{dy}{dx}\big)$-------(4)
Putting these values of $(y-b)$ and $(x-a)$ from (3) and (4) in (1) we get
$\large\frac{[1+\big(\Large\frac{dy}{dx}\big)^2]^2}{(\Large\frac{d^2y}{dx^2})^2}\big(\large\frac{dy}{dx}\big)^2+\large\frac{[1+\big(\Large\frac{dy}{dx}\big)^2]^2}{(\Large\frac{d^2y}{dx^2})^2}$$=c^2 \Rightarrow \begin{bmatrix}1+\big(\large\frac{dy}{dx}\big)^2\end{bmatrix}^3=c^2\begin{bmatrix}\large\frac{d^2y}{dx^2}\end{bmatrix}^2 \Rightarrow\large\frac{ \begin{bmatrix}1+\big(\Large\frac{dy}{dx}\big)^2\end{bmatrix}^{\Large\frac{3}{2}}}{\Large\frac{d^2y}{dx^2}}$$=c$ which is a constant independent of a & b.
answered May 14, 2013

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